b.sc. physics complete notes of thermodynamics

B.Sc. Physics

Complete Notes of

THERMODYNAMICS

Chapter # 1: KINETIC THEORY OF GASES

Chapter # 2: HEAT AND FIRST LAW OF THERMODYNAMICS

Chapter # 3: STATISTICAL MECHANICS

Chapter # 4: ENTROPY AND SECOND LAW OF THERMODYNAMICS

B.Sc. Physics THERMODYNAMICS in Past Papers of University of

Sargodha

B.Sc. Physics, Paper B, Annual 2017






OTHER UPDATED NOTES FOR STUDENTS OF B.Sc. PHYSICS

ONLY AVAILABLE AT www.HouseOfPhy.Blogspot.com

i. Mechanics

ii. Waves and Oscillations

iii. Electricity and Magnetism

iv. Electronics

v. Modern Physics

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Physics (H.R.K): Thermodynamics Chapter # 23. Kinetic Theory and The Ideal Gas

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CHAPTER # 23: KINETIC THEORY AND THE IDEAL GAS

The basic laws of thermodynamics deal with the relationships between macroscopic properties, such as pressure,

temperature, volume and internal energy of an ideal gas. The laws say nothing about the fact that matter is made up of particles

(atoms or molecules). Owing to the large number of particles involved, it is not practical to apply the laws of mechanics to find the

motion of every particle in a gas. Instead, we use averaging technique to express the thermodynamic properties as the averages of

molecular properties. If the numbers of particles are large, such averages give sharply defined quantities.

In this chapter, we consider an approach to averaging called kinetic theory, in which we follow the motion of representative

particles in a gas and then average this behavior over all particles. Kinetic theory was developed in the 17th to 19th centuries by Boyle,

Bernoulli, Joule, Kronig, Clausius and Maxwell among others. Another approach to averaging is statistical mechanics, in which laws

of probability are applied to statistical distributions of molecular properties.

23.1 Thermodynamics

The branch of Physics which deals with the conversion of heat energy into mechanical energy and

vice versa is called thermodynamics.

23.2 Temperature

It is the degree of hotness or coldness of a body.

It may also be defined as:

Temperature is a physical property of the body which determines the direction of flow of heat from one body

to the other, when they are brought in thermal contact.

Q # 1.One day when temperature is What will be the temperature on Celsius and Kelvin scale?

Given Data: Temperature on Fahrenheit Scale

To Determine: (i) Temperature on Celsius Scale , (ii) Temperature on Kelvin Scale

Calculation: (i) In Celsius scale:

( )

( )

(ii) In Kelvin scale:

23.3 State Variables (Thermodynamic Coordinates)

The physical quantities which describe the state of a system are called state variables or

thermodynamic coordinates e.g., pressure, volume, temperature, internal energy, entropy etc.

23.4 Ideal Gas

The gas which obeys gas laws under all temperatures and pressures is called an ideal gas. The

molecules of an ideal gas have no potential energy but only K.E.

23.5 Equation of State (Ideal Gas Law)

Consider an ideal gas consist of N molecules enclosed in a container of volume V at pressure P and

temperature T.

According to Boyle’s law, volume of gas is inversely proportional to its pressure if its

temperature is kept constant.

------------ (1)

From Charles law, the volume of gas is directly proportional to its absolute temperature if its pressure is

kept constant.

------------ (2)



From Gay-Lussac’s law, volume of gas is directly proportional to number of molecules if pressure and

temperature are kept constant.

------------ (3)

Combining (1), (2) and (3), we have:

Here constant is equal , called Boltzmann constant. It is a universal constant and its value is

If n is the number of moles then . So the above equation becomes:

( )

( )

But , here R is universal gas constant and is value is

This equation is called ideal gas law.

Q # 2. If a helium-filled balloon initially at room temperature is placed in a freezer, will its volume

increase, decrease, or remain the same?

Ans. The volume of the balloon will decrease. The pressure inside the balloon is nearly equal to the constant

exterior atmospheric pressure. Then from PV = nRT , volume must decrease in proportion to the absolute

temperature. Call the process isobaric contraction.

Q # 3. An ideal gas occupies a volume of Find the number of moles of

gas in container.

Given Data:

To Determine: Number of Moles

Calculation: Ideal Gas Law:

Q # 4. A spray Can containing propellant gas at twice atmospheric pressure and volume of

It is then tossed in to an open fire. When temperature of gas reaches What is

pressure inside the Can. Assume the change in volume is negligible.

Given Data:

To Determine: Final Pressure

Calculation: Ideal Gas Law for Initial Case: ( )

Ideal Gas Law for Final Case: ( )

( ) ( )



Q # 5. Verify that the one mole of oxygen occupies the volume of 22.4 at1 atm and 0 .

Given Data: Number of moles ,

To Determine: Volume


Q # 6. The least vacuum that can be attained in the laboratory corresponds to pressure of atm

Or Pa. How molecules are there per cubic container in such vacuum at 22ºC.

Given Data:

To Determine: Number of Moles


23.6 Kinetic Theory of Gases

The kinetic theory of gas is based on following assumptions:

The number of molecules in the gas is large, and the average separation between them is large

compared with their dimensions.

The molecules execute random motion and obey the Newton’s laws of motion.

The molecules make elastic collisions with each other and with the walls of container.

The gas under consideration is a pure substance; that is, all molecules are identical.

23.6.1 Pressure of the Gas

Consider an ideal gas in a cubical container having edge length L. Let m is the

mass of one molecule of gas and N be the total number of molecules in the container.

Let a molecule moves with velocity v and , and are its rectangular

components.

Suppose the molecule moves from one face to face with velocity and

after collision with face its velocity becomes . Consider the molecule reaches face

without any collision with other molecule on the way. After collision with , the

molecule goes back to face .

( )

As ( ) – ( )

( )– ( )

And



The time taken by the molecule in moving from to and back to is

. Putting values in equation

(1), we get:

( )

This is the force exerted by one molecule on . Total impulsive force exerted on by all molecules is

given by:

(

)

Now the pressure of the gas will be:

(

)

(

) -------- (2)

Now as N is the total number of molecules each of mass m, so total mass of gas will be Nm. So the density of

the gas will be:

Equation (2) becomes:

(

) (

)

The quantity within brackets is mean square velocity of gas molecules along x-axis and is denoted by .

Therefore

----------- (3)

For any molecule,

As the molecules have random motion. So,


------------- (4)

Q # 7. In our model of the kinetic theory of gases, molecules were viewed as hard spheres colliding

elastically with the walls of the container. Is this model realistic?

Ans. Collisions between molecules are mediated by electrical interactions among their electrons. On an

atomic level, collisions of billiard balls work the same way. Collisions between gas molecules are perfectly

elastic. So the hard-sphere model is very good. On the other hand, an atom is not ‘solid,’ but has small-mass

electrons moving through empty space as they orbit the nucleus.



23.6.2 Root Mean Square Speed

From equation (4), we have

Taking square root on both sides

√ √

√

Q # 8. Calculate the root mean square speed of hydrogen molecules at and 1 atm pressure,

assuming hydrogen to be an ideal gas. Under these conditions, hydrogen has a density of

.

Given Data: Pressure ,

Density

To Determine: Root Mean Square Speed

Calculation: As √

√

Q # 9. Using that molar mass of helium is kg/mole. Find root mean speed at 20 .

Given Data: Molar Mass ,

To Determine: Root Mean Square Speed

Calculation: As √

√

( ⁄ ) √

√

√

√

23.6.3 Dependence of Speed of Molecules on Temperature

As

( )

If M is the molar mass of the gas and n are the number of moles, then

Equation (5) will become:

( )

From ideal gas law:

( )

Comparing equation (6) and (7), we get:



√

√

√

If and are the speed of molecules of gas at temperature and , then

√

and √

√

Q # 10. If the speed of sound at 273 K is 331 m/s in air, then what will be its speed at 300 K.

Given Data: Initial Temperature , Speed of sound at initial Temperature

Final Temperature

To Determine: Speed of sound at initial Temperature

Calculation: As

√

√

( )√

Q # 11. One container is filled with helium gas and another with argon gas. If both containers are at the

same temperature, which molecules have the higher rms speed? Explain.

Ans. The helium must have the higher rms speed. According to Equation √

, the gas with the

smaller mass per atom must have the higher average speed-squared and thus the higher rms speed.

Q # 12. A gas consists of a mixture of He and molecules. Do the lighter He molecules travel faster

than the molecules? Explain.

Ans. Yes. As soon as the gases are mixed, they come to thermal equilibrium. Equation √

predicts

that the lighter helium atoms will on average have a greater speed than the heavier nitrogen molecules.

Collisions between the different kinds of molecules gives each kind the same average kinetic energy of

translation.

23.6.4 Kinetic Interpretation of Temperature

The pressure of ideal gas on the basis of kinetic theory of gas is given by the expression:

Multiplying both sides by V, we get:

( )

--------------- (8)

Here

If M is the molar mass of gas and n is the number of moles of gas, then

, put in equation (8)



(

)

According to this equation, the average translational K.E per mole of an ideal gas is directly

proportional to the absolute temperature of ideal gas.

23.6.5 Graham’s Law of Diffusion

We know that temperature of gas is directly proportional to K.E per molecule i.e.,

(

)

For two different gases at same temperature T and Pressure P, we can write as:

(

)

(

)

√

√

√

According to this equation, the ratio of root mean square speeds of molecules of two different gases is

equal to the inverse ratio of square root of their masses. This is known as Graham’s law of diffusion.

Q # 13. An ideal gas is contained in a vessel at 300 K. If the temperature is increased to 900 K, by what

factor does each one of the following change? (a) The average kinetic energy of the molecules. (b) The

rms molecular speed. (c) The average momentum change of one molecule in a collision with a wall. (d)

The pressure of the gas.

Ans. (a) Average molecular kinetic energy increases by a factor of 3.

(b) The rms speed increases by a factor of 3 .

(c) Average momentum change increases by 3 .

(d) Pressure increases by a factor of 3.

23.7 Work Done on an Ideal Gas

Consider a gas contained in a cylinder having movable piston. We raise the temperature of gas, the

gas expands and moves the piston up. The upward force exerted by the gas due to its pressure is , where

is the area of the piston.

By Newton’s 3rd

law of motion, the piston exerts an equal and opposite force on the gas. So the

work done on the gas is:



∫

∫ ------------ (1)

Where is the displacement of the piston. In covering

distance , the volume of the gas changes by .

So the equation (1) becomes:

∫

If the volume changes from to , then

∫

| |

( )

This equation shows that:

Work done on the gas during expansion is negative

Work done on the gas during compression is positive.

Q # 14 the gas in a cylinder is at pressure of 800 pa and piston have area of . As heat is slowly

added to the gas the piston is pushed up a distance of 4cm. Calculate work done.

Given Data:

To Determine: Work Done

Calculation: As Change in Volume

Work Done

23.7.1 Work Done at Constant Volume

The work is zero in the process in which volume remains constant i.e., and hence as

shown along path AB and CD.

It should be noted that it is not sufficient that the process should start

and end at the same volume. For the process ABDCA, the volume starts and

ends at but the work done is certainly not zero. The work is zero for

vertical paths AB and DC, because the volume remains constant along these

paths. The process in which volume remains constant is called isochoric

process.

23.7.2 Work Done at Constant Pressure

The work done at constant pressure can be find out using expression:

∫

∫

| | ( )

Along the paths BD and CA, the work is being done at constant pressure.



Work done on the gas is negative for path BD, because the volume increases along BD.

Work done on the gas is positive for path CA, because the volume decreases along CA.

23.7.3 Work Done at Constant Temperature

The process in which temperature remains constant is called isothermal process. This process is represented in

the PV diagram by hyperbolic curve as shown in the figure. This hyperbolic curve is called isotherm.

The work done in isothermal process is calculated by the equation:

∫

By the ideal gas equation:

Therefore,

∫

∫

| |

( )

23.7.4 Work Done at Thermal Isolation (Adiabatic Condition)

The process in which cylinder is completely insulated from its surrounding such that heat can’t enter or leave

the gas is called adiabatic process. The adiabatic process can be represented by the equation:

where K is a constant and

.

On a PV diagram, this adiabatic process is represented by a hyperbolic curve (steeper than an isotherm), called

adiabat.

The work done in adianatic process is calculated by the equation:

∫

As

Therefore,

∫

∫

∫

|

|

[ ]

[

]

Putting

[

]

[

]



Using Ideal Gas Law: , we have

[

]

[

]

This is the expression of work done during an adiabatic process.

23.8 Pressure Force is Non-Conservative

Consider an ideal gas undergo a change from initial condition

( ) to the final condition ( ). We take two different path

between A and D i.e., path 1 (ABD) and path 2 (ACD).

Work Done along Path-ABD

Along the path ABD, first the pressure increases from to at

constant volume along AB and then volume increases from to at

constant pressure along BD. The total work done is equal to the area of

rectangle BDEF (The area below line BD).

----------- (1)

, because i.e., volume is constant.

∫

∫

| | ( )


( ) ( ) ---------- (2)

Work Done along Path-ACD

Along the path ACD, first the volume increases from to at constant volume along AC and then

pressure increases from to at constant volume along CD. The total work done is equal to the area of

rectangle ACEF (The area below line AC)

----------- (3)

, because i.e., volume is constant.

∫

∫

| | ( )


( ) ( ) ----------- (4)

Conclusion

From equation (2) and (4), it is clear:

So work done on the ideal gas along two different paths is different. Hence pressure force is non-conservative.



23.9 The Internal Energy of an Ideal Gas

The sum of all forms of molecular energies of a substance is termed as its internal energy.

In the study of thermodynamics, usually the ideal gas is considered as working substance. The

molecules of an ideal gas are mere point masses which have no force of attraction between them. So the

internal energy of an ideal gas system is generally the translational K.E. The ideal gas has no internal energy

due to vibrational and rotational motion.

Since the temperature of a system is defined as the average K.E of its molecules, thus for the ideal gas

system, the internal energy is directly proportional to its temperature. It does not depends on pressure or

volume of the gas. For the n moles of an ideal gas at temperature T, the internal energy of the ideal gas is

written as:

(

)

We can change the internal energy of the ideal gas by doing work on

it. Suppose the ideal gas in cylinder is insulated from the surroundings and a

work W is done on the gas. By law of conservation of energy:

So the temperature changes are also positive.

Internal Energy of a Diatomic Gas

Let us consider a diatomic gas whose molecules have the shape of a

dumbbell as shown in the figure. In this model, the center of mass of the

molecule can translate in the x, y, and z directions (Fig. a). In addition, the

molecule can rotate about three mutually perpendicular axes (Fig. b). We can

neglect the rotation about the y axis because the molecule’s moment of inertia

and its rotational energy

about this axis are negligible compared with

those associated with the x and z axes. Thus, there are five degrees of freedom

for translation and rotation: three associated with the translational motion and

two associated with the rotational motion:

( )

According to theorem of equipartition of energy, the average energy of

gas molecule per degree of freedom and same and equal to

where is

Boltzmann’s constant and T is absolute temperature.

Because each degree of freedom contributes

of energy per

molecule, the internal energy for a system of N molecules, ignoring vibration for

now, is

(

) (

)



The vibrational motion adds two more degrees of freedom, which correspond to the kinetic energy and the

potential energy associated with vibrations along the length of the molecule. Hence, classical physics and the

equipartition theorem in a model that includes all three types of motion predict a total internal energy of

(

) (

) (

)

The vibrational term in the internal energy is effective at high temperature. At low temperature, the

vibrational and rotational motions can be frozen. So at low temperature, only three translational degrees of

freedom are present. The experiments with gas molecule collision show that the internal energy of an atom is

quantized.

Q # 15. Why does a diatomic gas have greater energy content per mole than a monatomic gas at the

same temperature?

Ans. A diatomic gas has more degrees of freedom—those of vibration and rotation—than a monatomic gas.

The energy content per mole is proportional to the number of degrees of freedom.

Q # 16. Is it possible to convert internal energy to mechanical energy? Explain with examples.

The steam locomotive engine is a perfect example of turning internal energy into mechanical energy. Liquid

water is heated past the point of vaporization. Through a controlled mechanical process, the expanding water

vapor is allowed to push a piston. The translational kinetic energy of the piston is usually turned into

rotational kinetic energy of the drive wheel.

Q # 17. What is the average kinetic energy of oxygen at 300k?

Given Data: Temperature

To Determine:

Calculations:

23.10 The Vander Walls Equations of State

Kinetic theory of gases gives the microscopic description of behavior of ideal gas. But some of the

assumptions of ideal gas are not true for real gases e.g., due to intermolecular forces the molecules have a

small but non zero diameter which is against the assumption of the ideal gas model.

Moreover the range of intermolecular forces may be greater than the size of the molecule which is

against our assumption that intermolecular forces are short range forces.

So the ideal gas equation needs modifications to account for the effect of finite size molecule and

presence of intermolecular forces.

(i) Effect of Finite Size of the Molecule

To consider the effect of finite size of molecule, we

consider the molecule in the form of a hard sphere of

diameter d. The distance between the centers of the two

molecules can’t be less than d. Therefore, the free volume

occupied by molecule is decreased by considering the finite

size of molecule. If b is the decrease in volume due to

molecules in one mole then decrease in volume due to



molecules in n moles is equal to .

If V is the volume of container then the volume available to the gas is not but . So the ideal

gas equation takes the form:

( )

( ) ------------ (1)

This shows that pressure of real gas is increased. Due to smaller volume, the molecules have more collisions

and thus pressure is increased.

(ii) Effect of Intermolecular Forces

To consider the effect of intermolecular forces, we take the range of intermolecular forces equal to

diameter d of molecule. Let a single molecule is about to collide with the wall of container. This molecule also

experiences attractive forces from other molecules lying within a distance d.

The sum of all intermolecular forces gives a resultant force

on the molecule under consideration, which is directed away from

the wall. So the molecule collides with the wall with a smaller

force.

So due to intermolecular forces, the pressure is reduced.

The decrease in pressure is proportional to the square of number of

moles per unit volume (

) . The expression (1) becomes:

( ) (

)

(

)

( )

[ (

)

] ( )

This is called the Vander Walls equation of state. The constant ‘ ’ and ‘ ’ can be determined experimentally

and are called Vander Wall constants.

Physics (H.R.K): Thermodynamics Chapter # 24. Statistical Mechanics


Statistical Mechanics

In previous chapter, we determined the average translational kinetic energy of the molecules of gas.

Knowing the average, however, tells us nothing about how the speeds of individual molecules are distributed

about the average. In some cases, the average might provide sufficient information about the properties of the

gas, such as its temperature. In other cases, it might be necessary to have more information about distribution

of speeds.

In this chapter, we are concerned with distributions of molecular speeds and energies, and their use

in computing the macroscopic properties of collections of molecules. This approach to thermodynamics is

called statistical mechanics.

24.1 Statistical Distributions and Mean Values

Consider a large number of objects, say N, moving with different speeds. In order to calculate the

mean speed, we tabulate the speeds of objects and sort the speeds into groups. The result of such a sorting

might be similar to that shown in the form known as histogram. Each rectangular area has a width equal to the

size of the sorting interval and a height equal to the number of observation or relative frequency of the values

in that interval.

We consider a typical representative value

of the speed in each interval. Each interval has a with

and a height , which is the number of

observations corresponding to the represented speed

. The total number of observations N is then given

expression.

∑

The sum is carried out on all the bins.

To find out average speed , we find the sum

of all obserbations (which is equivalent to the sum of

the product of the representative speed in each

interval with number of observation in that interval)

and divide it by the total number of observations:

∑

∑

∑

We can also define the relative frequency or probability of any value as

Thus equation (1) becomes:

∑

(

)



∑

Problem: The speed of group of 10 molecules are 2,3,4,5,6,7,8,9,10,11 . (a) Find the average speed

of group. (b) Calculate the root mean square speed of the group.

Given Data: Total Molecules , Speed of Molecules

To Determine: (a) Average Speed (b) Root Mean Square Speed

Calculations: (a) As

(b) As

Sample Problem 1: Figure 1 is based on the observation of a total of 1205 cars, whose speed distribution

is as follows:

i Speed Interval (mph)

1 0-5 23

2 5-10 41

3 10-15 54

4 15-20 95

5 20-25 123

6 25-30 142

7 30-35 177

8 35-40 186

9 40-45 170

10 45-50 122

11 50-55 50

12 55-60 15

13 60-65 7

Find the mean value of speed for this distribution, taking the speed at the middle of each interval as

representing the entire interval.

Given Data: Total Cars

To Determine: Average Speed

Calculations: As ∑



24.2 Mean Free Path

The mean free path represent the average distance that a molecule moves between collisions.

To understand the term mean free path, consider a motion of

single molecule of a gas that follow a zigzag path due to collision with

the neighboring molecules. Suppose that we measure the distance

travelled between the collisions and repeat the experiment the several

times. After making N measurements of the distance travelled, we sort

out these distances into bins of equal width The value of the

distance in each bin is Now plot number of molecules per

bins against the distance travelled before colliding as shown in

the histogram.

The average distance is covered is given by:

∑

∑

∑

The distance is called mean free path and is denoted by

To find the continuous function that gives the

probability for the molecules to travel a distance before

collision, consider a beam of molecules having intensity is

incident on a thin layer of gas of thickness

The molecules emerge after passing through of the pass with intensity Those molecules in the

beam that have collisions are scattered into different directions, and thus emerging beam intensity is

determined by the number of molecules that pass through the distance without collision. We assume that

number of molecules in the beam is scattered more then once

by a target molecule.

When we increase the thickness of the gas layer by

amount the come across following observations:

The increase in intensity is proportion to

The increase in intensity is proportional to

additional thickness

We can write these observations as

where c is constant of proportionality. The negative sign means decreased with the addition of thickness

Integrating on both sides with boundary conditions

∫

∫



(

)

That is, the beam intensity decreases exponentially with the thickness of the gas layer.

This calculation suggests an exponential form for and we choose

Where A is another constant to be determined. We can therefore rewrite equation (1), replacing by we

have

∑

∑

∑

∑

Let us now make the width very small, so that we can write the above equation in term of integrals

∫

∫

∫

∫

∫

∫

| (

)|

∫ (

)

|(

)|

(

) [| | ∫

]

( )

| |

[ |(

)|

]

| | (

)| |

| |

Put

in therefore, probability distribution function can be written as

⁄

It is the relation for the mean free path of molecules in terms of probability.

Microscopic Calculation of the Mean Free Path

The mean free path depends upon the following two factors:

The size of molecules

If the molecules are points, they will not collide at all. Hence mean free path is infinite. If

molecules are so big that they completely fill the available space and there will not be any room for

translation and hence the mean free path is zero.

Number of molecules per unit volume

When the number of molecules is large in unit volume, then the probability of molecular

collisions increases.



Consider the molecules of a gas as sphere of diameter

d. The collision of molecules will occur when the center of two

molecules come within a distance d of each other. An

equivalent but more convenient is to think of moving molecule

having diameter 2d and all other molecules are points as shown

in figure.

Now imagine a molecule of equivalent diameter 2d is travelling with speed v through a gas.

The molecule sweeps out a cylinder of base are 2d and length in time t.

The volume of cylinder

The density of particles in cylinder

This is equal to the number of collisions experienced by

the molecule in time t. The cylinder shown in the figure, in fact, a

broken one, changing direction with every collision.

Now as the mean free path is the average distance between

successive collisions. Hence is the total distance covered in

time divided by the number of collisions that take place in that time i.e.,

This equation is based on the picture of a molecule hitting stationary targets. Actually the molecule

hits moving targets. When the target molecules are moving, the two in equation (3) are not the

same. The one in the numerator is the mean molecular speed ( ) with respect to container. The one in

the denominator is the mean relative speed ( ) with respect to other molecules. Moreover, it is the

relative speed that determines the collision rate. On quantitative basis, taking into account the actual

speed distribution of molecules, gives:

√

Thus equation (3) takes the form:

√



Problem: Calculate the mean free path for 35 spherical jelly beans in a jar that is vigorously shaken.

The volume of the jar is 1.0 L and the diameter of jelly beam is 1 cm.

Given Data: Number of Jelly Beans , Volume , Diameter

To Determine: Mean Free Path

Calculation: As

√

√

Problem: At standard temperature and pressure, the mean free path in the helium gas is 285 nm.

Determine (a) the number of molecules per cubic centimeter and (b) effective diameter of helium atom.

Given Data: Temperature , Pressure ,

Mean Free Path

To Determine: (a) Number of molecules per cubic centimeter , (b) Diameter

Calculation: (a) As

(b) As

√

√ √

√

24.3 The Distribution of Molecular Speeds

Maxwell first solved the problem of distribution of speeds in a

gas containing large number of molecules. The Maxwellian speed

distribution, for a sample of gas at temperature T containing N

molecules each of mass m, is

(

)

⁄

(

)

In this equation , which has the dimensions of , is the

number of molecules per unit speed interval having the speeds between

and . is dimensionless quantity, which gives the

number of molecules in gas sample having speed between and

.

It is clear from expression (1) that for a given

gas, the speed distribution depends only on temperature.

Figure shows the Maxwell’s distribution of speeds for

molecules of oxygen at room temperature. This

distribution can be characterized by the most

probable speed , the mean speed and root mean



square speed .

Average Speed

To find the average speed, we must sum all measured speeds. This can be done most simply summing the

products of speed in each interval and the number in that interval . This sum is then divided by total

number of measurements N. in the limit of infinitesimal interval, the sum becomes an integral. Thus average

speed can be described as:

∫

∫

[ (

)

⁄

(

)]

(

)

⁄

∫ (

)

For simplicity, put

, we have:

(

)

⁄

∫ ( )

√

⁄ ∫ ( )

Put

√

⁄ ∫

√

⁄ ∫

√

⁄ [|

|

∫

]

√

⁄ [

∫

]

√

⁄ [

|

|

]

√

⁄

[ ]

√

⁄[ ]

√ (

)

√

√

As the Boltzmann’s constant

, therefore

√ (

)

√

√



Most Probable Speed

The most probable speed is the speed at which has its maximum value. We find it by requiring

that

and solving for .

[ (

)

⁄

(

)]

(

)

⁄

[ (

) (

)

(

)]

[ (

) (

)

(

)]

(

)[

]

(

)

Or √

Since ( √

) is outside the domain, we exclude it. The minimum value of is zero, since , we

will exclude . We can safely assume √

corresponding to maximum value of . Therefore,

the most probable speed is:

√

√

√

Root Mean Square Speed

To find the root mean square speed of gas molecules, we proceed as above except that we find

the average value of by multiplying by all by the corresponding numerical factor and then

divided by the total number of measurements N:

To find root mean square , we need to find :

∫

∫

[ (

)

⁄

(

)]



∫ (

)

⁄

(

)

√ ∫ (

)

⁄

(

)

√ (

)∫ (

)

⁄

(

)

⁄

(

)

(

)

⁄

√ (

)∫ [(

)

⁄

]

(

)

(

)

⁄

Put (

)

⁄ (

)

⁄

√ (

)∫

√ (

)(

√

) ∫

√

√ √

√

√

Problem. What are the average speed , the root mean square speed and the most probable speed

of oxygen molecules at . The molar mass M of oxygen is 0.032

Given Data: Temperature , Molar Mass of Oxygen

To Determine: (a) Average Speed , (b) Root Mean Square Speed , (c) Most Probable Speed

Calculations: (a) Average Speed √

√

(b) Root Mean Square Speed √

√

(c) Most Probable Speed √

√



24.4 Distribution of Energies

An alternate description of motion of molecules can be obtained if we look for distribution in energy

rather than in speeds. For this we find out the distribution . In this distribution gives the number

of molecules with energies between and .

It makes no difference whether we count the molecules by their speeds or by their kinetic energies.

For this, we set the lower and the upper limits of interval to have corresponding speeds and kinetic energies.

The number of molecules in both cases will be equal. That is, the number with K.E between and is

same as the number with speed between and . Mathematically, it is described as:

Sine the energy is only kinetic, we have:

√

√

(

⁄ )

(

)

⁄

(

)

⁄

Also as,

(

)

⁄

(

)

⁄

⁄

⁄

⁄

⁄

( )

⁄

⁄

⁄

⁄

⁄

Putting values in equation (1), we have:

(

⁄

⁄

⁄

⁄

⁄

)(

⁄

⁄

⁄)

⁄

⁄

⁄

⁄

⁄

⁄(

)

⁄

⁄

⁄

⁄

⁄

⁄

This equation is known as Maxwell’s distribution of energies.



Internal Energy of an Ideal Gas

Consider molecules with energy between and and the contribution to the internal energy

of this gas is . The total of all such contributions gives the internal energy of the gas:

∫

∫ [

⁄

⁄

⁄

]

√

⁄∫

⁄

Put

√

⁄∫

⁄

√ ∫

⁄

∫

⁄

√

√ (

√

)

This expression of internal energy of ideal gas derived on the basis of statistical mechanics is same to our

calculations made on the basis of kinetic theory of gases. Thus the Maxwell-Boltzmann distribution is entirely

in consistent with the results derived from kinetic theory of gases.

Sample Problem 7. Find (a) mean energy and (b) the most probable energy of a gas in thermal

equilibrium at temperature T.

Solution:

(a) The mean energy is described by formula:

∫

∫ [

⁄

⁄

⁄

]

√

⁄∫

⁄

Put



√

⁄∫

⁄

√ ∫

⁄

∫

⁄

√

√ (

√

)

(b) For most probable energy , put

. So,

(

⁄

⁄

⁄

)

⁄

⁄[

⁄

(

)

⁄ ]

( )

⁄

(

)

( )

⁄

( )

⁄

( )

⁄

( )

⁄

( )

⁄

Physics (H.R.K): Thermodynamics Chapter # 25. Heat and First Law of Thermodynamics


CHAPTER # 25: HEAT AND FIRST LAW OF THERMODYNAMICS

If the temperatures of the objects that are in thermal contact are initially different, the system will exchange energy until thermal

equilibrium is reached. In this chapter we deal with this energy flow from one body to another, which we call heat. We also describe

the effects of transferring heat to the body, which may include increasing its temperature or changing its state, such as from solid to

liquid or liquid to vapor. Finally, we tie together the concepts of heat, internal energy and work through the first law of

thermodynamics, a statement of law of the conservation of energy.

25.1 Heat

Heat is energy that flows between the system and its environment by the virtue of a temperature

difference between them. Heat is measured in unit of joule or calorie.

25.2 Calorie

It is the amount of heat required to raise the temperature of 1 g

of water from 14.50C to 15.5

0C.

25.3 Mechanical Equivalent of Heat

It is defined as:

“The mechanical work required to produce unit quantity of heat”

It is found that 4.2 J of work produces 1 calorie of heat and

4200 J of work produces 1 kilo calorie of heat. It is denoted by J and its

value is 4.2 J/Cal. Mathematically, it is described as:

25.4 Heat Capacity

It is the amount of heat required to raise the temper of substance through 10C. Mathematically, it is

described as:

Its SI unit is J/0C or J/K

25.5 Specific Capacity (Specific Heat)

It is the amount of heat required to raise the temperature of 1 kg of a substance through 10C.

Mathematically, it is described as:

Its SI unit is J/kg-0C of J/kg-K.

25.6 Molar Heat Capacity

It is the amount of heat required to raise the temperature of one mole of a substance through 10C.




Q # 1. Clearly distinguish among temperature, heat, and internal energy.

Ans. Temperature is a measure of molecular motion. Heat is energy in the process of being transferred

between objects by random molecular collisions. Internal energy is an object’s energy of random molecular

motion and molecular interaction.

25.7 Heat of Transformation

The amount of heat per unit mass transferred during a phase change is called heat of transformation or

Latent heat. It is denoted by L. Thus total heat transferred in a phase change is describe by the formula:

Where m is the mass of the sample.

Latent Heat of Fusion

The latent heat during melting or freezing is called latent heat of fusion. It is denoted by .

Latent Heat of Vaporization

The latent heat during boiling or condensing is called latent heat of vaporization.

Q # 14. When alcohol is rubbed on your body, it lowers your skin temperature. Explain this effect.

Ans. The alcohol evaporates, absorbing energy from the skin to lower the skin temperature.

25.8 Internal Energy

It is the sum of all the energies of all the atoms, molecules or ions within a system.

Internal energy is a definite quantity which depends on the state of system only. So it is also called

state function. The internal energy E of a system cannot be determined. However its change can be

calculated. Change in internal energy depends only on the initial and final state and not on the path through

which it takes place. Thus:

25.9 Relation between and

Consider a cylinder containing n moles of an ideal gas fitted with a piston. Suppose the piston is fixed

i.e., the volume of the gas is constant. When gas is heated at constant volume, all the heat supplied is used

only to increase the internal energy of the gas and no work is done, because the volume is constant. So,

But . Therefore,

----------- (1)

Now consider the same cylinder that is heated at the same pressure i.e., piston is free to move. Now the heat

supplied to the gas is given by:

----------- (2)

Now this heat supplied partly increases the internal energy and remaining heat is used to do work in

moving the piston up:

Since the work done on the gas during expansion is negative, so

----------- (3)

As



So ----------- (4)

Putting vales of equation (1), (2), (4) in equation (3), we get:

( )

Thus the difference of and is equal to molar gas constant.

Special Cases

1. For mono-atomic gas:

And

2. For diatomic gas:

And

3. For polyatomic gas:

And

25.10 First Law of Thermodynamics

In any thermodynamic process between two equilibrium states and , the quantity has the same value

for any path between and . This quantity is equal to the change in the value of a state function, called

internal energy.

Mathematically, it is written as:

This equation is known as the first law of thermodynamics.

Explanation: To better understand these ideas on a quantitative basis, suppose that a system undergoes a

change from an initial state to a final state. During this change, energy transfer by heat Q to the system occurs,

and work W is done on the system. As an example, suppose that the system is a gas in which the pressure and

volume change from and to and . If the quantity is measured for various paths connecting

the initial and final equilibrium states, we find that it is the same for all paths connecting the two states. We

conclude that the quantity is determined completely by the initial and final states of the system, and

we call this quantity the change in the internal energy of the system. Although Q and W both depend on the

path, the quantity is independent of the path.

The mathematical form of 1st law of thermodynamics describes:

Heat absorbed by the system result in increase of internal energy.



Work done on the gas increases the internal energy of the system.

So, the total increase of internal energy is due to heat absorbed plus work

done on the gas. Similarly, when heat is lost by the system or work is done by

the system on its surroundings, both Q and W decrease the internal energy of

the system.

Q # 2: A gas is compressed at constant pressure at 0.3atm from volume of

8L to 3L in this process 400J heat energy flows out of gas. Find the work

done by the gas and the change in internal energy

Q # 3. A gas is enclosed in a container fitted with a piston of cross-sectional area . The pressure

of a gas is maintained at 6000 Pa as the piston moves inward 20 cm. (a) Calculate the work done by the

gas. (b) If the internal energy of the gas decreases by 8 J, find the amount of heat removed from system

during compression.

Given Data: Pressure , Displacement ,

Cross-sectional Area , Change in Internal Energy

To Determine: (a) Work done by the gas , (b) Heat removed from system

Calculations:

(a) Work Done by Gas ( ) ( )

(b) By First Law of Thermodynamics:

Applications of First Law of Thermodynamics

(i) Adiabatic Process

It is the process in which no heat can enter or leave the system. This can be achieved by insulating the

system completely so that .

So the first law of thermodynamics becomes:

So in adiabatic process, the work done on the gas results in increase of internal energy of the gas.

(ii) Isothermal Process

It is the process in which temperature of the system remains constant.

So if temperature remains constant, then for ideal gas, the internal energy also remains constant i.e.,

. So the first law of thermodynamics becomes:

Given Data: ,

,

Change in Volume ,

Heat added to the System

To Determine: Work Done by the Gas , (b) Change in Internal Energy

Calculations: (a) As work done by the gas

(b) By First Law of Thermodynamics



Or

This shows that, during isothermal process, if the work W is done on gas, the gas rejects heat to the

surroundings. So, the work done on the gas does not remain within the gas but flows out as heat to the

surroundings.

(iii) Constant Volume Process

If the volume of the gas is kept constant, then no work can be done on the gas. So the first law of

thermodynamics becomes:

Or

All heat supplied goes to increase the internal energy of the gas.

(iv) Cyclic Process

It is a series of processes after which system goes to its initial state.

Figure shows a three step process. It is a cyclic process, because it starts and ends at the same point A.

After one cycle, the chance in internal energy is zero i.e., .

By the first law of thermodynamics:

or

(Cyclic Process)

The process 1 occurs at the constant volume along

A-B

Process 2 occurs at constant pressure along B-C

Process 3 occurs at constant temperature along C-

A.

The total work done is positive because area under the

curve 3 (AC) is greater than curve 2 (BC).



(v) Free Expansion

A process in which a gas goes from one side of container to the other half initially evacuated side is

called free expansion process.

In this case the gas is initially on one side of the container and when stop cork is open, the gas enters

into the evacuated half side. No weight are raised so no

work is done. The container is insulated, so the process is

adiabatic i.e., .

Thus, the first law of thermodynamics becomes:

(Free Expansion)

As the change in internal energy is zero, therefore

So, during the free expansion, the internal energy of the

ideal gas remains constant. As the internal energy of an ideal gas only depends upon temperature, so

temperature must remain constant during free expansion.

Q # 4. Using the first law of thermodynamics, explain why the total energy of an isolated system is

always constant.

Ans. If the system is isolated, no energy enters or leaves the system by heat, work, or other transfer processes.

Within the system energy can change from one form to another, but since energy is conserved these

transformations cannot affect the total amount of energy. The total energy is constant.

25.11 Proof of for an Adiabatic Process

It is the process in which no heat can enter or leave the system. This can be achieved by insulating the

system completely.

For adiabatic process, . So the first law of thermodynamics becomes:

In differential form:

( )

As the work done on the system of gas is described as:

( ) ( )

The specific heat at constant volume is described by formula:

(

)

(

)

Thus equation (2) becomes:

( )

From ideal gas law:



Differentiating the above equation, we get:

( ) ( )

( )

But

( )

Dividing equation (3) and (4), we get:

Integrating both sides:

∫

∫

| | | |

[ ]

( ) (

)

( ) (

)

( ) (

)

Or

25.12 Show that adiabat is stepper than Isotherm.

Adiabatic process is represented by the equation:

Taking derivative on both sides:

( )

( )



( )

Here

gives the slope of adiabat.

Now isothermal process is described by the formula:

Taking differentials on both sides, we get:

( )

( )

Here

gives the slope of isotherm

Thus from (1) and (2), it can be seen that adiabat is steeper than isotherm.

25.13 The Transfer of Heat

Heat energy can be transferred from one place to the other by three ways:

Conduction

Convection

Radiation

CONDUCTION

In process of conduction, the heat energy is transferred from one point to the other due to vibrations

and collisions of molecules

The method of transfer of heat is prominent in solids (metallic solids) and small in liquids and

negligible in gases.

If one end of rod is heated, the K.E of the molecules at the hot end is increased so they vibrate with

larger amplitude. So they collide with neighboring molecules and give same energy to them. This process

goes on and the energy is transferred from molecule to molecule down the rod. Thus it can be said that

conduction of heat is due to vibration and collisions of molecules.

Factors on which conduction depends

Consider a rod of length L. the two ends of

the rod are at temperature and respectively and

the cross-section of the rod is A. as the two ends of

the rod are at two different temperatures, the heat will

flow from hot end to cold end through the rod. The

heat which is conducted through the rod depends



upon:

i. Area of cross-section A of the rod. Larger te area, greater will he the flow of heat i.e.,

ii. The temperature difference:

The greater the temperature difference between the two ends, greater will be the flow of heat i.e.,

( )

iii. The time of conduction:

More heat will flow in longer time i.e.,

iv. The length of the rod:

Longer the rod, smaller will be the amount of heat across the rod i.e.,

Combining all factors, we get:

( )

( )

Where

is the rate of flow of heat denoted by H.

( )

( )

Where is called thermal conductivity. It depends upon the nature of

the material of rod.

If and are denoted as and then

( )

( )

( ⁄ )

( )

Where

is called thermal resistance or R-value. So for good thermal conductors, the R-value is small and

vice versa.

If we have a small rod of length and temperature then rate of flow of heat is given as:

The negative sign shows that heat flows in the direction of decreasing

, where

is called

temperature gradient.



CONVECTION

It is the process in which heat is transferred from one place to

another by actual movement of the molecule.

As the molecules can move only in liquids and gases (fluids),

so fluids are heated by this process. In solids, heat can’t be transferred

by convection because molecules can’t leave their position.

Factors on which convection depends

Transfer of heat by convection depends on:

i. Temperature difference between the source and the place to

which heat is transferred

ii. Nature of the fluid

iii. Density, viscosity, specific heat and the thermal conductivity

of the fluid

iv. Velocity of the fluid

RADIATION

It is the process in which heat is transferred from one place

to another place in the form of waves without a material medium.

Since there is no medium between the sun and the earth.

Heat reaches the earth from sun by radiation. Heat waves are called

thermal radiation. Thermal radiation belong to the electromagnetic

spectrum of radiation. Both thermal radiation and light have the

similar properties and obey identical laws.

Q # 5. A tile floor in a bathroom may feel uncomfortably cold to

your bare feet, but a carpeted floor in an adjoining room at the

same temperature will feel warm. Why?

Ans. The tile is a better thermal conductor than carpet. Thus, energy is conducted away from your feet more

rapidly by the tile than by the carpeted floor.

Q # 6. Why is it more comfortable to hold a cup of hot tea by the handle rather than by wrapping your

hands around the cup itself?

Ans. The porcelain of the teacup is a thermal insulator. That is, it is a thermal conductor of relatively low

conductivity. When you wrap your hands around a cup of hot tea, you make A large and L small in the

equation

for the rate of energy transfer by heat from tea into you. When you hold the cup by the

handle, you make the rate of energy transfer much smaller by reducing A and increasing L.

Q # 7. Two identical cups both at room temperature are filled with the same amount of hot coffee. One

cup contains a metal spoon, while the other does not. If you wait for several minutes, which of the two

will have the warmer coffee? Which energy transfer process explains your answer?

Ans. The cup without the spoon will be warmer. Heat is conducted from the coffee up through the metal. The

energy then radiates and convects into the atmosphere.

Physics (H.R.K): Thermodynamics Chapter # 26. Entropy and Second Law of Thermodynamics


CHAPTER # 26: ENTROPY AND 2ND

LAW OF THERMODYNAMICS

We can imagine many processes that conserve energy (and thus satisfy the first law of thermodynamics) but they are never observed.

For instance, a glass of cool water is transformed into an ice cube in a glass of warmer water. In such cases, the reverse process is

normally observed. The second law of thermodynamics deals with whether such processes will occur in nature. It is often said that the

second law gives a preferred direction to the “arrow of time”, telling us that systems naturally evolve with time in one direction but

not the other.

In this chapter, we use the second law to analyze engines that convert heat into useful work, and we show that there is an upper limit

to the efficiency at which an engine can operate. The second law leads to a new concept, entropy, just as the zeroth law led to

temperature and the first law to internal energy.

26.1 Heat Engine and 2nd

Law of Thermodynamics

It is a device that converts heat energy into mechanical

energy. Every heat engine takes heat from a hot body (source),

converts a part of it into work and rejects the remaining part to a cold

body (sink).

A heat engine carries some working substance through a cyclic

process during which:

1) The working substance absorbs energy by heat from a high-

temperature energy reservoir

2) Work is done by the engine,

3) Energy is expelled by heat to a lower-temperature reservoir.

Example

Consider the operation of a steam engine, which uses water as

the working substance. The water in a boiler absorbs energy from burning fuel and evaporates to steam, which

then does work by expanding against a piston. After the steam cools and condenses, the liquid water produced

returns to the boiler and the cycle repeats.

Efficiency of Heat Engine

Suppose a heat engine takes heat from hot body at

temperature , converts a part of it into work and rejects the remaining

heat to cold body at temperature .

After completing one complete cycle, the system returns to the

initial state. So its internal energy remains constant in a cycle. So work

done by the engine in one cycle is and the change in

internal energy is zero. So, the net work done W by a heat engine is

equal to the net energy transferred to it.

If the working substance is a gas, the net work done in a cyclic

process is the area enclosed by the curve representing the process on a

PV diagram.

The efficiency of the heat engine is defined as:

“Ratio of work done by the engine in one cycle to the heat absorbed in one cycle”




From this equation, it is clear that efficiency of heat engine

is not or 1. The efficiency can be if i.e., no

heat is transferred to the cold body and heat taken from hot body

is completely converted into work. The engine which will do so is

called perfectly efficient heat engine. But the experiments show that

no such heat engine can be made which can convert whole heat

drawn from a single body to work.

Kelvin Statement of Second Law of Thermodynamics

It is impossible to make a heat engine which operating in a cycle can

go on doing work by taking heat from a single body, without other

body at lower temperature.

This law tells that two bodies at different temperature are

necessary for the working of a heat engine. The hot body is called source while the cold body is called sink.

Sample Problem # 1. An automobile engine, whose thermal efficiency e is 22 % operates at 95 cycles per

second and does work at the rate of 120 hp. (a) How much work per cycle is done on the system by the

environment? How much heat enters and leaves the engine in each cycle.

Given Data: Efficiency , Number of cycles per second = 95,

Rate of Work

To Determine: (a) Work done per cycle , (b) Heat Enters the system , Heat Leaves system

Calculations: (a) Work done per cycle

(b) As

Also as

26.2 Refrigerator and 2nd

law of Thermodynamics

A heat engine run in reverse is known as refrigerator. A refrigerator needs work to transfer heat from

cold body to hot body.

If is the amount of heat removed from the cold body at temperature and is the amount of

heat given to the hot body at temperature , then the difference is the external work required to

drive the refrigerator.



Co-Efficient of Performance

In a refrigerator, instead of efficiency, we use the term co-efficient of

performance K, which is defined as:

i.e., co-efficient of performance is the ration of heat removed from the cold

body to the mechanical work required to operate the refrigerator.

In a perfect refrigerator and so

. So the coefficient of performance of

perfect refrigerator is infinity.

In an ordinary house hold refrigerator, the working substance is the liquid

(Freon) that circulated within the system. The cold body is the interior of the

refrigerator and hot body is the room in which refrigerator is kept. Typical

refrigerator have the coefficient of performance around 5.

Clausius Statement of 2nd

Law of Thermodynamics

It is impossible to cause heat to flow from cold body to the hot body

without expenditure of energy.

So from this, we find that heat can’t flow by itself from cold to hot body.

So we have to do work to make this flow. So this statement can also be stated as:

Perfect refrigerator is impossible

Sample Problem # 2. A household refrigerator, whose coefficient of

performance K is 4.7, extracts heat from the cooling chamber at the rate of

250 J per cycle. (a) How much work per cycle is required to operate the refrigerator? (b) How much

heat per cycle is discharged to the room, which form the high temperature reservoir of the refrigerator?

Given Data: Coefficient of Performance , Heat Extracted

To Determine: (a) Work Done per Cycle (b) Heat Exhausted =?

Calculations: (a) As

(b)

Q # 1. A refrigerator does 153 J of work to transfer 568 J heat from cold temperature reservoir.

Calculate the refrigerator coefficient of performance and how much heat is exhausted to the kitchen.

Given Data: Desired output = 568 J, Required input = 153 J

To Determine: (a) Coefficient of Performance (b) Heat Exhausted =?

Calculations: (a)

(b)

(

) (

)



26.3 Equivalence of Clausius and Kelvin Statement

The two statement of the end law of thermodynamics can be shown to be identical by supposing that

if one of the statements is false, the other is also false.

We suppose that Kelvin statement is false i.e., we can construct a heat engine, which can convert the

heat of a single body into work without a cold body.

Suppose this engine is used to run a

refrigerator as shown in the figure:

In refrigerator, the heat flows from cold body

to hot body. So work done by the engine is used

operate the refrigerator which transfer heat from cold

body to hot body.

Thus, the heat engine and the refrigerator

both from a system which transfer heat from cold

body to hot body without any external work. But it is

against the Clausius Statement. So the Clausius

statement is also false if the Kelvin statement is false. So a violation of Kelvin statement gives the violation of

Clausius statement.

So we see that violation of one statement gives a violation of the other. Hence from the above

discussion, we find that these two statements are logically identical.

26.4 Carnot Engine

It is an ideal heat engine free from all heat loses ad all the process are reversible.

Construction of Carnot Engine

It consist of following parts:

1. A cylinder containing perfect gas having non-conducting walls and piston and only base conducting

2. A source of heat at temperature

3. A sink at temperature

4. A non conducting stand

Carnot Cycle

The operating cycle of Carnot engine is called Carnot cycle. The

Carnot cycle consist of following four processes:



i. Isothermal Expansion The cylinder is placed on the source of heat at temperature and gas is allowed

to expand very slowly by removing weights on the piston. The gas expands and tends to cool down but it

absorbs heat from the source of heat so that its temperature remains constant ( ). This process is

isothermal, so and all the heat added appears as negative of work done on the gas. This process

is shown on PV diagram by the curve AB.

ii. Adiabatic Expansion

The cylinder is now placed on the non-conducting stand so that heat does not enter or leave the gas.

The gas is allowed to expand slowly by removing more weights from the piston. The process is adiabatic and

. The piston does negative work on the gas and the work is done at the cost of internal energy of the gas.

So the temperature of gas decreases to . This process is shown on the PV diagram by the curve BC.

iii. Isothermal Compression

The cylinder is now placed on the sink and the gas is compressed by adding weights on the piston.

During this process the gas rejects heat to sink at . This process is isothermal compression and positive

work is done on gas during compression. The process is shown in the PV diagram by the curve CD.

iv. Adiabatic Compression

The cylinder is finally placed

on the non-conducting stand. The gas

is compressed by adding more

weights on the piston. As heat can’t

leave gas so the process is adiabatic.

The process is such that temperature

of the gas rises to and the gas

goes back to its initial state. The

process is shown on the PV diagram

by curve DA.

So the gas undergoes a cycle

ABCDA. It is a reversible cycle and

is called Carnot cycle.



26.4.1 Efficiency of Carnot Engine

Isothermal Processes

Along the isothermal path AB, temperature remains constant, so . So by the fist law of

thermodynamics, heat transferred from source is equal to magnitude of work.

(

)

Similarly for isothermal compression CD, the heat energy rejected to the sink is

(

)

Dividing equation (1) and (2), we get:

(

)

(

)

(

)

(

)

Adiabatic Processes

For an adiabatic process, we have:

For adiabatic path BC, we can write:

For adiabatic path DA, we have:

Comparing equation (3) and (4), we get:

(

)

(

)

Putting this value in equation (A), we get:

(

)

(

)

So, the efficiency of the Carnot engine is:



From this expression, we find that efficiency of Carnot engine only depend upon temperature of

source and sink between which it operate.

The efficiency is increased by decreasing . The efficiency will be if which is not possible.

Thus, the efficiency of the Carnot engine can’t be .

Sample Problem # 3. The turbine in a steam power plant takes steam from a boiler at and

exhaust it into a condenser at . What is the maximum possible efficiency?

Given Data: Inward Temperature ,

Exhaust Temperature

To Determine: Efficiency

Calculations:

Q # 2: A steam engine has a boiler that operates at 500K.The heat changes water to steam which drives

the piston. The temperature of exhaust is that of outside the air about 300K.What is maximum thermal

efficiency of this system engine.

Given Data: Inward Temperature , Exhaust Temperature


Calculations:

Q # 3. A steam engine operates at 300ºF and exhaust temperature is 150º F . What is maximum

efficiency if Carnot engine.

Given Data: Inward Temperature

Exhaust Temperature


Calculations:

Q # 4. The highest exhaust temperature is 300º C. What is inward temperature if efficiency of engine is

30%.

Given Data: , Efficiency

To Determine: Inward Temperature

Calculations:

Q # 5: The highest theoretical efficiency of a gasoline engine based on the Carnot cycle is 30%.If this

engine expels it gases into the atmosphere which has temperature of 300K.What is temperature in the

cylinder immediately after combustion.

Given Data: , Efficiency



To Determine: Inward Temperature

Calculations:

Q # 6. What are some factors that affect the efficiency of automobile engines?

Ans. First, the efficiency of the automobile engine cannot exceed the Carnot efficiency: it is limited by the

temperature of burning fuel and the temperature of the environment into which the exhaust is dumped.

Second, the engine block cannot be allowed to go over a certain temperature. Third, any practical engine has

friction, incomplete burning of fuel, and limits set by timing and energy transfer by heat.

Q # 7. In practical heat engines, which are we better able to control: the temperature of the hot

reservoir or the temperature of the cold reservoir? Explain.

Ans. It is easier to control the temperature of a hot reservoir. If it cools down, then heat can be added through

some external means, like an exothermic reaction. If it gets too hot, then heat can be allowed to “escape” into

the atmosphere. To maintain the temperature of a cold reservoir, one must remove heat if the reservoir gets

too hot. Doing this requires either an “even colder” reservoir, which you also must maintain, or an

endothermic process.

Q # 8. A steam-driven turbine is one major component of an electric power plant. Why is it

advantageous to have the temperature of the steam as high as possible?

Ans. A higher steam temperature means that more energy can be extracted from the steam. For a constant

temperature heat sink at Tc , and steam at Th , the efficiency of the power plant goes as

and is

maximized for a high Th .

Q # 9. Is it possible to construct a heat engine that creates no thermal pollution?

Ans. No. Any heat engine takes in energy by heat and must also put out energy by heat. The energy that is

dumped as exhaust into the low-temperature sink will always be thermal pollution in the outside environment.

Q # 10. Does the second law of thermodynamics contradict or correct the first law? Argue for your

answer.

Ans. No. The first law of thermodynamics is a statement about energy conservation, while the second is a

statement about stable thermal equilibrium. They are by no means mutually exclusive. For the particular case

of a cycling heat engine, the first law implies Qh =Weng + Qc , and the second law implies Qc > 0.

Q # 11. Can a heat pump have a coefficient of performance less than unity? Explain.

Ans. No, because the work done to run the heat pump represents energy transferred into the house by heat.



26.5 Thermodynamic Temperature Scale

The scale of temperature which is independent of the nature of working substance is called

thermodynamic temperature scale or Kelvin temperature scale. Thermodynamic temperature scale has been

derived by Kelvin by the law of thermodynamics.

Consider the operation of three reversible engines 1, 2 and 3. The

engine 1 absorbs energy as heat from the reservoir at , does work

and rejects energy as heat to the reservoir at .

Let the engine 2 absorb energy as heat from the reservoir at

and does work and rejects energy as heat to the reservoir at .

The third reversible engine 3, absorbs energy as heat from the

reservoir at , does work and rejects energy as heat to the

reservoir at .

The efficiency of Carnot reversible heat engine does not depend

on the working substance and depend on the working temperature and

of the source and sink, where and are measured on the perfect gas scale. So the efficiency of Carnot

engine is the function of and :

Now we consider three Carnot engines working between the temperatures , and

, where and . Then

Now



As

is a function of only, so can be omitted

Let

. Therefore,

Where and are called Kelvin temperatures or absolute temperatures.

That is, the ratio of energy absorbed to the energy rejected as heat by a reversible engine is equal to

the ratio of the temperatures of the source and the sink. Such a temperature scale is called thermodynamic (or

Kelvin) Temperature Scale.

The equation can be used to determine the temperature of any reservoir by operating a reversible

engine between that reservoir and another easily reproducible reservoir and by measuring efficiency (heat

interactions). The temperature of easily reproducible thermal reservoir can be arbitrarily assigned a numerical

value (the reproducible reservoir can be at triple point of water and the temperature value assigned 273.16 K).

Then for Carnot engine operating between reservoirs at temperatures and , we have:

We see that on the thermodynamic temperature scale, plays the role of thermodynamic property.

However Q does not depend on the characteristic on any substance because the efficiency of a Carnot engine

is independent of the nature of working substance.

Show that the temperature difference for each Carnot engine in thermodynamic temperature scale is

the same.

Consider a series of Carnot engines coupled together such that each engine does the same amount of work and

also sink of proceeding engine becomes the source of next engine and so on.

It means that

The first engine received heat from the source at temperature , performs the work W and rejects

heat to the second engine.

The second engine receives heat at temperature and rejects heat

to the third engine and so on

As each engine does the same amount of work, therefore we can write:

So,

(

) (

) (

)

(

) (

) (

)



As

It shows that the temperature difference for each Carnot engine is the same. If there are 100 Carnot

engines working between boiling point of water and melting point of ice, then

Where

Temperature corresponding to the boiling point of water

Temperature corresponding to the melting point of ice

26.6 Show that thermodynamic scale of temperature and perfect gas scale are identical.

Ans. Consider a Carnot engine working between the temperatures and on Kelvin scale. Then by Kelvin

scale:

Let these temperatures be and on the perfect gas scale. Therefore

Combining (1) and (2), we get:

As

Similarly

(

)

(

)

Similarly

Or

It shows that temperature of melting point of ice and boiling temperature of water are same on both scales.

Hence the Kelvin scale and perfect gas scale are identical.



26.7 Absolute zero and negative temperature

We can’t have a gas below 1 K and so we can’t measure temperature below 1K using constant volume

gas thermometer. But by the thermodynamic temperature scale, we can measure temperature below 1K.

Consider a system at temperature . We want to measure the temperature of the system. For this we

consider the system around a Carnot cycle. First we do work on the gas adiabatically and temperature of the

gas increases to which is supposed to be known. Then heat is transferred isothermally. Then doing work

adiabatically to decrease the temperature back to Then in the last is rejected isothermally to bring the

system to its initial state.

So by the relation:

So, knowing and measuring and , we can measure . Similarly we can take the system

around another Carnot cycle to a still lower temperature . We can continue this process to the absolute zero

temperature.

But smaller the temperature, small is heat transferred in the isothermal process between two adiabatic

process. Near absolute zero, the system will undergo an isothermal process without the transfer of heat.

It is found experimentally that lower the temperature, the more difficult it is to go still lower. So we

can state 3rd

law of thermodynamics as follows:

It is impossible by any procedure, no matter how idealized, to reduce any system to

the absolute zero of temperature in a finite number of operations.

Hence we can’t have a sink at absolute zero or a heat engine with 100% efficiency is impossible.

26.8 Entropy Function

Entropy is a Greek terms means change. It is a measure of disorder or randomness of molecular

motion of the system. It is a thermal property of a system which remains constant as long as no heat enters or

leaves the system. It is a real physical quantity. Entropy of a system increases if heat flows into the system at

constant temperature and decreases if leaves the system at constant temperature.

It is a state function and depends on the state of system. The absolute value of entropy can’t be

determined however change in entropy can be determined by the relation:

This expression shows that entropy of a system increases if heat flows into the system at constant

temperature and decreases if leaves the system at constant temperature. The SI unit of entropy is J/K.

26.9 Entropy in Reversible Process

As we know that for a Carnot cycle:

As and always have opposite signs, so the above expression can be written as:



This equation shows that the sum of algebraic quantity

is zero for a Carnot cycle.

For a finite number of cycles, equation (1) becomes:

∑

For infinite number of cycles, equation (1) becomes:

∮

Here ∮ indicates that the integral is calculated over a close path and quantity is the small quantity

of heat which enters or leave the system. Equation (3) is called CLAUSIUS THEOREM.

The Clausius theorem gives the definition of entropy function. Consider a reversible cycle represented

by the close curve as shown in the figure.

Let and denote the initial and final states of a system. Let the system undergo a change from to

along path-1 and back to along path-2. These two path form a reversible cycle. By Clausius theorem,

∮

This integral can be written as the sum of two integrals:

∮

∫

∫

∫

∫

∫

∫

This shows that the integrals on two reversible paths

are equal. It means that the integral ∫

is same along the

reversible paths from to . In the language of mathematics, the integrand

is an exact differential of some

function S i.e.,

∫

∫

This quantity S is called entropy of the system. It should be noted that is an inexact differential but

is

an exact differential. Change in entropy is independent of path and only depends on the initial and final

states.



26.10 Entropy in Irreversible Process

In nature there is no reversible process due to friction and heat transfer. So every thermodynamic

process is irreversible.

To find the entropy change for an irreversible process, we choose a path by connecting the initial and

final state and calculate the entropy change by the equation:

∫

Examples of Irreversible Process

(i) Free Expansion

Consider an ideal gas enclosed in an insulated container. When the ideal gas rushes into evacuated

chamber, its temperature and internal energy remains constant i.e., and by 1st law of

thermodynamics:

The free expansion is an irreversible process. There is an entropy change between initial and final states. To

find the entropy change, we choose a path from to . Let us consider an isothermal expansion that takes the

gas from initial state to ( ). Now

∫

∫

Putting the value of Q, we get:

In an isothermal process, (

). Therefore,

(

)

This is equal to the entropy change irreversible free expansion. It should

be noted that entropy change is positive for an irreversible process. As

there is no heat transfer to environment in free expansion, so entropy

change for environment is zero. Thus the total entropy of

increases during free expansion.

(ii) Irreversible Heat Transfer

Consider two block at temperatures and . Both the blocks

have same mass and specific heat. We bring the blocks into thermal

contact. After some time, they reach common temperature . Like free

expansion, this is also irreversible process.

To find the entropy change in this irreversible process, we

consider block 1 at lower temperature and imagine a series of thermal

reservoirs at temperatures . We fist start with block 1 in contact



with the 1st reservoir and move it one step at a time along a sequence. In each step, a small amount of heat

enters the block. Entropy change for block 1 is given by:

∫

∫

( | | ) (

)

Similarly, for block 2 between temperatures and , the entropy change is given by:

∫

∫

( | | ) (

)

The total entropy change is given by:

(

) (

)

(

)

[ (

) ]

[ (

) ]

Now, we can show that the total entropy change is positive. For this purpose, we have to prove that

.

We first find by considering the total heat flow equal to zero.

Now

(

)

This expression shows that

. So (

) is greater than zero and so the entropy change is positive.

Hence by placing two block in thermal contact produces no change at all in the environment and so

for environment. But the total entropy of increases in this irreversible heat

transfer.



Q # 12. Give various examples of irreversible processes that occur in nature.

Ans. A slice of hot pizza cools off. Road friction brings a skidding car to a stop. A cup falls to the floor and

shatters. Your cat dies.

26.11 Entropy and the 2nd

law of thermodynamics

The 2nd

law of thermodynamics can be stated in terms of entropy as:

In any thermodynamic process, that proceeds from one equilibrium state to another,

the entropy of the system + environment either remains unchanged or increases.

For reversible process, the entropy does not change. For irreversible processes i.e., for all natural processes,

the total entropy of the must increase. It is possible that the entropy of system

might decrease, but entropy of environment shows increase of greater magnitude, so that the total change in

entropy is always positive.

No natural process can ever shown decrease in the total entropy of the .

This is another statement of 2nd

law of thermodynamics. Let us consider this statement of 2nd

law for the

following cases.

Example

The reverse process of free expansion is called free compression.

The change in entropy in free expansion is given by:

(

)

In free compression, , so (

) is negative. So is negative. Which means entropy decreases. But

this violates the 2nd

law according to which entropy of is always

positive.

In free compression, there is no change in entropy of environment like free expansion.

So the 2nd

law of thermodynamics in terms of entropy denies the process of free compression.

The Kelvin-Planck Form of 2nd

Law of Thermodynamic

Because all engines operates in cycle, so entropy change for the system in complete cycle is zero. In a

perfect engine, the environment (source) releases heat at temperature and its entropy change is

, a

negative quantity. The total entropy change of is therefore

negative in a perfect heat engine which violates the 2nd

law in terms of entropy. So the 2nd

law of

thermodynamics denies the possibility of perfect engine.

Clausius form of 2nd

law

In a perfect refrigerator, there is no change entropy of the system in one complete cycle. But the

environment (cold body) releases heat ( ) at temperature and absorbs heat at temperature . The total

change entropy of the environment is:

(

)

Because , so is negative which violates the second law in terms of entropy. So the 2nd

law denies

the possibility of perfect refrigerator.



The Arrow of Time

All the natural processes are taking place in such direction that the change in entropy of

is always positive i.e., if in any process the entropy of the system decreases then there is large

increases in the environment such that total entropy change is positive.

Q # 12. Discuss three common examples of natural processes that involve an increase in entropy.

Ans. (1) Energy flows by heat from a hot bowl of chili into the cooler surrounding air. Heat lost by the hot

stuff is equal to heat gained by the cold stuff, but the entropy decrease of the hot stuff is less than the entropy

increase of the cold stuff. (2) As you inflate a soft car tire at a service station, air from a tank at high pressure

expands to fill a larger volume. That air increases in entropy and the surrounding atmosphere undergoes no

significant entropy change. (3) The brakes of your car get warm as you come to a stop. The shoes and drums

increase in entropy and nothing loses energy by heat, so nothing decreases in entropy.

Q # 13. How could you increase the entropy of 1 mol of a metal that is at room temperature? How could

you decrease its entropy?

Ans. To increase its entropy, raise its temperature. To decrease its entropy, lower its temperature.

Q # 28. Calculate charge in entropy when 300 g of head melts at 600 K. Head has latent heat of fusion

J/Kg.

Given Data: Mass , Temperature , Latent Heat

To Determine: Change in Entropy

Calculations: As

Now, Change in Entropy

Q#20: A large cold object is at 273 K and a large hot object is at 373K. Show that it is impossible for a

small amount of heat energy say 8J to be transferred from cold object to hot object without decreasing

the entropy of universe and hence violating the second law.

Given Data: Heat Transferred Temperature of Cold Body

Temperature of Hot Body

To Determine: Net Change in Entropy

Calculations:

Now the Net Change in Entropy

As the net entropy change is negative, which means that this process couldn’t occur in nature.

UNIVERSITY OF SARGODHA

B.A. / B.Sc. 1st Annual Exam 2017

Physics Paper: B

Time Allowed: 3 Hours Maximum Marks: 50

Note: Attempt any five questions in all, selecting one question from section I and four questions from section-II. All

questions carry equal marks.

SECTION-I

Q # 1. (a) List at least five properties of ideal gas and then determine the rms speed of ideal gas molecules.

(b) Calculate the most probable speed and average speed for oxygen molecules atT = 300K .

(c) How the speed of sound related to the gas variables in the kinetic theory model?

Q # 2. (a) What is meant by thermal Isolation? Then calculate the work done during adiabatic expansion.

(b) A Carnot engine whose high-temperature reservoir is at 400K has an efficiency of 30%. By how much should the

temperature of the low temperature reservoir be changed to increase the efficiency to 40%.

(c) What is the difference between heat of fusion and heat of vaporization?

SECTION-II

Q # 3. (a) Determine the electric field due to a charged disk of radius R.

(b) How much is required to turn an electric dipole 1800 in a uniform electric field of magnitude if

and the initial angle of 64°.

(c) State two Shell theorems of electrostatics.

Q # 4. (a) Define Capacitance and its unit, also determine capacitance due to cylindrical capacitor.

(b) Four 18 Ω are connected in parallel across 25 V ideal battery. What is the current through the battery?

(c) State macroscopic and microscopic forms of Ohm‘s law.

Q # 5. (a) Define electric potential and its unit. Calculate the electric potential due to electric dipole.

(b) An infinite line of charge produces the field of magnitude at a distance of 2 m. Calculate the linear

charge density.

(c) What is an analogy between electrical and mechanical oscillator.

Q # 6. (a) Discuss the growth of charge of a capacitor in RC circuit and also find the current.

(b) A particle undergoes uniform circular motion of radius 26.1 µm in a uniform magnetic field. The magnetic force on the

particle has the magnitude of . What is the kinetic energy of the particle?

(c) How does doubling the frequency alter the capacitive and inductive reactance?

Q # 7. (a) State and explain Ampere’s law and write its integral and differential form.

(b) A solenoid has length 1.23 m and inner diameter 3.55 cm and it carries the current 5.57A. It consists of five close-

packed layers, each with 450 tums along length. What is magnetic field at its center?

(c) What is meant by capacitive reactance? Also write its unit.

Q # 8. (a) Define Inductance and its unit. Derive the expression for Inductance for a Solenoid.

(b) What is magnetic energy per unit length of a coaxial cable of inner radius l.2 mm and outer radius 3.5 mm? If 2.7 current

flows.

(c) What is the rms value of secondary voltage if its peak value is 100 V?

Q # 9. (a) An ac emf is applied to RLC series circuit. Calculate the expression for average power dissipated.

(b) A series RLC circuit has inductance L = 12 mH, capacitance C = 1.6 µF and resistance R = 1.5Ω. At what time t will the

amplitude of the charge oscillations in the circuit be 50% of its initial value’?

What is difference between acceptor and rejecter circuit?

Q # 10. Write the note on any two of the following

a) Carnot heat Engine

b) Dielectrics and Gauss Law

c) Magnetic Materials

d) Maxwell Equations



Physics Paper: B




SECTION-I

Q # 1. (a) What is Carnot Heat Engine? Calculate the work done and efficiency of Carnot Heat Engine.

(b) In an interstellar gas cloud. at 50K, the pressure is Pa. Assuming that the molecular diameter of the gases in

the cloud are all 20 nm, what is their mean free path?

Q # 2. (a) Discuss entropy and calculate the entropy change for irreversible free expansion of a gas.

(b) A Carnot engine whose high-temperature reservoir is at 400K has an efficiency of 30%. By how much should the

temperature of the low temperature reservoir be changed to increase the efficiency to 40%.

(c) What are the two statements of Second Law of Thermodynamic?

SECTION-II

Q # 3: (a) Determine the electric field due to a ring of uniform positive charge distribution.

(b) An electron is in the vacuum near Earth’s surface and located at y=0 on the vertical y-axis. At what value of y should the

second electron is placed such that its electrostatic force on the first electron balance the gravitational force on the first

electron?

(c) Electric charge is Conserved. Comment

Q # 4: (a) Define Capacitance of a capacitor and determine capacitance due to cylindrical capacitor.

(b) A neutral water molecule in its vapor state has an electric dipole moment of magnitude . How far apart

are the molecules centers of positive and negative charges?

(c) How Coulomb’s Law can be derived from Gauss’s Law?

Q # 5: (a) Define electric potential and its unit. Calculate the electric potential due to electric dipole.

(b) A point charge causes an electric flux of 750 N-m2/C to pass through a spherical Gaussian surface of l0 cm radius

centered on the charge. If the radius of the Gaussian’s surface is doubled, how much flux is passed through the surface?

(c) How the energy density and electric field are related.

Q # 6: (a) Discuss the discharging of capacitor in RC circuit. Also calculate the current.

(b) A capacitor discharge through a resistor. After how many time constant does the stored energy fall to one half of its

initial value?

(c) Define Ampere’s Law.

Q # 7: (a) Define magnetic force and calculate the magnetic force between two parallel current carrying wires

(b) A coil has an inductance of 53mH and a resistance of 0.35 ohms. If a l2V emf is applied across the coil, how much

energy is stored in the magnetic field after the current has built up to its equilibrium value?

(c) What is meant be capacitive reactance? Also describe its unit.

Q # 8: (a) What is solenoid? Determine magnetic field due to solenoid by applying Ampere’s law.

(b) What is intensity of a travelling electromagnetic wave if Bm is T?

(c) What is the maximum value of an ac voltage whose rms value is 100 V?

Q # 9: (a) An ac emf is applied to RLC series circuit. Calculate the expression for average power dissipated.

(b) A 45.2 mH inductor has the resistance of l.2 kΩ. What is the frequency and find the capacitance of a capacitor with the

same reactance.

(c) Define Curie’s Law.

Q # 10: Write the note on any two of the following:

(i) Mean free path

(ii) Dielectrics and Gauss Law

(iii) Transformer

(iv) Semiconductors and Superconductors



Physics Paper: B


Note: Attempt any five questions in all, selecting one question from section I and four questions from section-II. All questions carry

equal marks.

SECTION-I

Q # 1. (a) The Maxwell Speed Distribution function is given as:

( ) (

)

⁄

⁄ √

(b) One mole of oxygen gas expands isothermally at 310 K from initial volume of 12 L to the final volume of 19 L. What would be the

final temperature if the gas had expanded adiabatically to the same final volume?

(c) There are five number 5, 11, 32, 67 and 89. What is the rms value of these numbers.

Q # 2. (a) What is Carnot engine? Calculate the work done and efficiency of Carnot Heat Engine.

(b) The temperature of one mole of mono-atomic gas is raised reversibly 300 K to 400 K with the volume kept constant. What is the

entropy change of the gas?

(c) What are the two statements of Second Law of Thermodynamics?

SECTION-II

Q # 3: (a) What is electric dipole? Calculate the electric field at the bisector point at the distance x due to dipole.

(b) The nucleus of iron atom has a radius of about m and contains 26 protons. Calculate the magnitude repulsive force

between two of the protons.

(c) What will be the value of flux if the surface encloses negative charge?

Q # 4: (a) Define and prove Gauss’s law and write its integral and differential form.

(b) A Gaussian surface in the form of cylinder of radius R immersed in a uniform electric field E, with the cylinder axis parallel to the

field. What is the flux of the electric field through this closed surface?

(c) How does the gravitational and electric force resemble each other? How do they differ?

Q # 5: (a) Define electric potential and its unit. Calculate the electric potential from electric field.

(b) A wire 4 m long and 6 mm in diameter has a resistance of . A potential difference of 23 V is applied between the ends. What

is current density?

Q # 6: (a) Discuss the growth of charge in RC circuit also calculate the current.

(b) A capacitor discharge through resistor. After how many time constant does the stored energy fall to one half of its initial value?

(c) Show that the product RC has the dimension of time.

Q # 7: (a) Calculate the magnetic force acting on a moving charge in a magnetic field.

(b) A proton traveling at 230 with respect to magnetic field

2.3 mT experiences a magnetic force of .

Calculate the speed and KE of proton in eV.-

(c) What do you mean by strength of magnetic field and what is its unit?

Q # 8: (a) State Biot-Savart law? And calculate magnetic field due to long straight wire using Biot-Savart law.

(b) At a certain location, the magnetic field has the magnitude 42 µT and point downward at 57° to the vertical. Calculate the magnetic

flux through a horizontal surface of area 2.5 m2.

(c) Does the induced emf depend upon the resistance of the circuit?

Q # 9: (a) An AC emf is applied to RLC series circuit. Calculate the expression for average power dissipated.

(b) A 45.2 mH inductor has the resistance of 1.2 kΩ. What is the frequency and find the capacitance of capacitor with the same

resistance?

(c) Why the electromagnetic waves are not deflected by magnetic field.


i. Van Der Waal equation of state

ii. Faraday’s law of electromagnetic induction

iii. Transformer

iv. Electromagnetic wave spectrum



Physics Paper: B




SECTION-I

Q # 1. (a) Discuss the distribution of molecular energies. Hence derive the Maxwell Boltzmann energy distribution formula.

(b) A cylindrical container of length 56 cm and diameter 12.5 cm holds 0.35 moles of nitrogen gas at the pressure of 2.05

atm. Calculate rms speed of the nitrogen molecules.

(c) Show that for adiabatic process

Q # 2. (a) What are the two statements of second law of thermodynamics? Show that the statements are equivalent.

(b) A refregerator whose COP is 4.7, extract heat from the food chamber at the rate of 250 J/cycle. How much work per

cycle is required to operate the refrigerator?

(c) What is the connection between entropy and the second law of thermodynamics.

SECTION-II

Q # 3: (a) What is electric dipole? Find the electric field at a point on the bisector plane due to dipole.

(b) An infinite line of charge produces the field at a distance of 1.92 m. Calculate the linear charge density.

(c) Do electrons tend to go to region of high potential or low potential?

Q # 4: (a) Derive the formula to calculate the electric potential due to a point charge and collection of point charges.

(b) What is the electric potential at the surface of the gold nucleus? The radius of the surface of the gold nucleus is 7 fm and

atomic number is 79.

(c) Can two equipotential surfaces intersect each other?

Q # 5: (a) Define capacitance. Show that capacitance of a cylindrical capacitor depends upon the radii of inner and outer

spheres.

(b) What is the capacitance of the earth viewed as an isolating conducting sphere of radius 6370 km?

(c) What type of energy is stored in the capacitor?

Q # 6: (a) Discuss the growth of charge in RC series circuit and the expression of current which is used to charge the

capacitor.

(b) The capacitor C discharges through the resistance R. After how many time constants does the charge drop to half of its

initial value?

(c) Under what circumstances can the terminal potential difference of a battery exceed its emf?

Q # 7: (a) State and explain Ampere’s law and write its integral and differential form.

(b) The earth's magnetic field has a value of 39 µT and is horizontal and due north. The net field is zero 8.13 cm above the

long straight horizontal wire that carries a steady current. Calculate current and find its direction of flow.

(c) What type of magnetic field exists around a current carrying straight conductor?

Q # 8: (a) What is atomic magnetism? Find the relation between total orbital magnetic dipole moment and total angular

momentum.

(b) What must be the magnitude of uniform electric field if it is to have the same energy density as possessed by a 0.5 T

magnetic field?

(c) How does doubling the frequency alter the capacitive and inductive reactance?

Q # 9: (a) Explain the phase lag or lead between current and voltage in a purely inductive AC circuit. Also draw phasor

diagram.

(b) In RLC series circuit R= 160 Ω, C=15 µF, L= 230 mH and f=60 Hz. Find the phase constant.


i. Brownian motion

ii. Lenz’s law

iii. Maxwell’s equations

iv. Electromagnetic wave spectrum

UNIVERSITY OF SARGODHA B.A. / B.Sc. 1st Annual Exam 2013

Physics Paper: B




SECTION-I

Q # 1. (a) State and explain the first law of thermodynamics and discuss its applications for constant temperature.

(b) Apparatus that liquefies helium is in a room at 300 K. If the helium is in the apparatus is at 4.0 K, what is the minimum

ratio of heat delivered to the room to the heat removed from the helium?

(c) What factors reduce the efficiency of a heat engine from its ideal value.

Q # 2. (a) What is ideal gas? Describe the work done on ideal gas at constant pressure and temperature.

(b) An ideal gas undergoes a reversible isothermal expansion at . The entropy of the gas increases by 46 J/K. How

much heat was absorbed?

(c) Explain why temperature of a gas drops in an adiabatic expansion.

SECTION-II

Q # 3: (a) Define dipole and dipole moment. Calculate the relation for torque and potential energy of an electric dipole

when it is placed in electric field.

(b) An electric dipole consists of charges +2e and -2e separated be 0.78 nm. It is in electric field strength .

Calculate the magnitude of Torque on the dipole, when the dipole moment is (i) parallel to (ii) antiparallel to the electric

field.

(c) A point charge is moving in an electric field at right angle to the lines of force. Does any force act on it.

Q # 4: (a) What is electric potential? Find the electric potential at a point P on the axis of charge disc.

(b) The potential at the center of a uniformly charged circular disc of radius is . What is the total

charge on the disc?

(c) Define equipotential surfaces. What is the potential difference between two points on an equipotential surface.

Q # 5: (a) Derive the relation for energy stored in an electric field of charged capacitor, also calculate energy density.

(b) An isolated conducting sphere whose radius is 685 cm has a charge 1.25 nC. What is energy density at the surface of

sphere?

(c) What is dielectric and what is the effect of dielectric on capacitance of capacitor?

Q # 6: (a) Discuss decay and growth of charge in RC circuit.

(b) A capacitor of capacitance C is discharging through a resistance R. In terms of time constant , when will its

charge be one half of its initial value?

(c) What do you mean by a node in DC circuit and state Kirchhoff’s node rule?

Q # 7: (a) State Ampere's law and derive the relation for magnetic field due to a toroid.

(b) A 200 turns solenoid having a length of 25cm carries a current of 0.30A. Find magnetic field B inside the solenoid.

(c) Discuss analogies and differences between Coulomb’s law and Biot-savrat law.

Q # 8: (a) Discuss RLC series circuit and derive the relation for impedance and resonance frequency.

(b) At what frequency would a 6.0 mH inductor and 10 µF capacitor have the same resonance?

(c) Define rms value and power factor in AC circuit.

Q # 9: (a) What is inductor? Derive the relation for energy stored in magnetic field due to solenoid?

(b) A coil has an inductance of 53mH and resistance of 0.35 Ω. If a 12 V emf is applied across coil, how much energy

stored in magnetic field after current has been built up to its equilibrium value.

(c) Does the induced emf always act to decrease the magnetic flux through a circuit?

Q # 10: Write a note on any two of the following:

i. Hysteresis Loop

ii. Gauss Law in Dielectric

iii. Carnot cycle

UNIVERSITY OF SARGODHA B.A. / B.Sc. 1st Annual Exam 2012

Physics Paper: B




SECTION-I

Q # 1. (a) What is an ideal gas? Discuss work done on an ideal gas during thermal isolation.

(b) Calculate work done by the external agent in compressing 1.12 mole of oxygen from a volume of 22.4 L at 1.53 atm

pressure to 15.3 L at the same temperature.

(c) How is the speed of sound related to gas variables in kinetic theory model?

Q # 2. (a) Define mean free path of a molecule travelling through a gas, also calculate mean free path on microscopic level.

(b) Calculate the mean free path for 35 spherical jelly beans in a jar that is vigorously shaken, the volume of jar is 1.0 L and

the diameter of jelly beans is 1.0 cm.

(c) List three examples of Brownian motion in physical phenomenon.

SECTION-II

Q # 3: (a). What is an electric dipole? Discuss the motion of dipole in the external electric field. Also calculate work done

and energy stored in the dipole.

(b) A proton is placed in a uniform electric field D. What must be the magnitude and direction of the field if the electrostatic

force acting on the proton is just to balance its weight?

(c) A point charge is moving in an electric field at right angle to the lines of force. Does any force act on it.

Q # 4: (a). Define electric potential and calculate electric potential due to a dipole.

(b)What must be the magnitude of an isolated positive charge for the electric potential at 15 cm from the charge to be 120

V?

(c) How can you ensure that the electric potential in a given region of space has constant value?

Q # 5: (a) Define capacitance? Calculate the capacitance of a cylindrical capacitor.

(b) A parallel plate capacitor has circular plate of 8.22 cm radius and 1.31 mm separation. Calculate the capacitance and

what charge will appear on each plate if a potential difference of 1116 V is applied.

(c) Water has high value of dielectric constant, why is it not used as a dielectric material in capacitor?

Q # 6: (a) Discuss decay of charge in RC series circuit. Calculate current and define time constant of the circuit.

(b) When 115 V is applied across 9.66 m long wire, the current density is 1.44 . Calculate the conductivity of the

wire material.

(c) Under what circumstances would you want to connect batteries in parallel and in series?

Q # 7: (a) State Biot Savart Law. Determine the magnetic field at point, distance R away from the long straight wire

carrying current.

(b) In the Bohr Model of hydrogen atom the electron circulates around the nucleus in a path of radius

at a frequency of Hertz. What value of B is set up at the center of the orbit?

(c) Is B uniform for all points within a circular loop of wire carrying current?

Q # 8: (a) Derive an expression for the average power dissipated in RLC series circuit containing an AC source.

(b) In an RLC series circuit R = 160 Ω, C = 15 µF, = 230 mH, f = 60 Hz and ε=36V (AC supply). Find: (a) the rms emf

(b) the rms current (c) the power factor and (d) average power dissipated.

(c) Why it is useful lo use rms notation for alternating current and voltage. '

Q # 9: (a) Discuss the process of generation of electromagnetic waves. How Maxwell equations lead towards the existence

of electromagnetic waves.

(b) A beam of light with an intensity I of l2 W cm-1

falls perpendicularly on a perfectly reflecting plane mirror of l.5 cm2

area. What force acts on the mirror.

(c) How does a microwave oven cook food‘?

Q # 10. Write note on any two of the following:

i. Faradays law of electromagnetic induction.

ii. Maxwell Equations

iii. Second Law of thermodynamics and entropy.

b.sc. physics complete notes of thermodynamics

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