PHYSICAL QUANTITIES AND UNITS

h) show an understanding that the Avogadro constant is the number of atoms in 0.012 kg of carbon-12

(i) use molar quantities where one mole of any substance is theamount containing a number of particles equal to the Avogadro

constant

Avogadro’s Constant

• The Avogadro constant, NA, is the number of atoms in 0.012 kg of Carbon-12 or the number of particles in one mole of any substance.

• The substance may be in a solid, liquid or gaseous state, and its basic fundamental unit may either be atomic, molecular or ionic in form.

• NA equals to 6.02 x 10²³ atoms per mole.

• A molar quantity is the amount per mole unit (6.02 x 10²³ particles) e.g.:– molar volume of a gas at s.t.p. is 22.4 dm³.

– Number of particles per mole 6.02 x 1023 atoms/mol

– Molar mass in g/mol

• E.g the number of atoms in 6 g of Carbon-12 is 6/12 x 6.02 x 10²³ or 3.01 x 10²³.

2

MOLAR QUANTITIES

• Avogadro’s constant = no. of particles

mole

• Molar mass = mass

mole

• Molar gas volume = volume

mole

4

Ideal gases• 10. Ideal Gases

• Content

• 10.1 Equation of state

• 10.2 Kinetic theory of gases

• 10.3 Pressure of a gas

• 10.4 Kinetic energy of a molecule

• Learning Outcomes

• (a) recall and solve problems using the equation of state for an ideal gas expressed as

pV = nRT. (n = number of moles)

• (b) infer from a Brownian motion experiment the evidence for the movement of

molecules.

• (c) state the basic assumptions of the kinetic theory of gases.

• (d) explain how molecular movement causes the pressure exerted by a gas and hence

deduce the relationship, p = ⅓ Nm/V < c2> . (N = number of molecules)

[A simple model considering one-dimensional collisions and then extending to three dimensions

using 1

3< 𝑐2 >= < cx2 >

cx is sufficient]

• (e) compare pV = ⅓ Nm< c2> with pV = NkT and hence deduce that the average

translational kinetic energy of a molecule is proportional to T.

5

Microscopic model of a gas

• Physics describes and explains the behaviour of various systems

• In some cases it is impossible to describe what happens to each component of a system e.g properties of a gas in terms of the motion of each of its molecules

• In such cases we describe the average conditions in the gas rather than describing the behaviour of each molecule i.e macroscopic instead of microscopic

• Kinetic theory of an ideal gas is one such example where we make simple assumptions to derive at a law relating kinetic energy to temperature

6

What is the evidence that gas molecules are moving around all the time?

• 1 cm3 of atmospheric air contains approx 3 x 1025 molecules • 1827 biologist Robert Brown observed tiny pollen grains suspended

in water with a microscope with a illuminating beam • Although the water was completely still, the grains were always

moving in a jerky, haphazard manner : we call this Brownian motion

• Hence the assumption that the water molecules are in rapid, random motion under the bombardment from all sides of the water molecules

• We can see the same movement of tiny soot particles in smoke • A century later, Mr. Einstein did a theoretical analysis of Brownian

motion and estimated the diameter of a typical atom in the order of 10-10 m

• It has been found that the average spacing of atoms of liquids is up to 2 times that of solids or still in the order of 10-10 m. For gases it is of the order of 10-9 m

7

The gas laws

• Experiments in the 17th and 18th centuries showed that the volume, pressure and temperature of a given sample of gas are all related

8

Boyle’s Law

Robert Boyle 1627-91

Boyle’s Law states that the pressure of a given mass of gas at constant temperature is inversely proportional to its volume.

p 1/V or pV = constant

If p1,V1 are the initial pressure and volume of the gas, and p2,V2 are the final values after a change of pressure and volume carried out at constant temperature, then

p1V1 = p2V2

9

Charles Law

Jacques Charles 1746-1823

Charles’ Law states that the volume of a given mass of any gas at constant pressure is directly proportional to its thermodynamic (absolute) temperature T

V T or V/T = constant

If V1,T1 are the initial volume and temperature of the gas, and V2,T2 are the final values after a change of volume and temperature carried out at constant pressure, then

V1/T1 = V2/T2

T K = (273 + ) degrees C

10

Pressure Law

Joseph Gay-Lussac 1778-1850

Pressure Law or Gay-Lussac’s Law states that thepressure of a given mass of gas at constant volumeis directly proportional to its absolute temperature.

p T or p/T = constant

If p1,T1 are the initial pressure and temperature of the gas, and p2,T2 are the final values after a change of pressure and temperature carried out at constant volume, then

p1/T1 = p2/T2

11

Ideal Gas Equation In practice, real gases obey the gas laws only at moderate

pressures and at temperatures well above the temperature at which the gas would liquefy.

An ideal gas is one which obeys the 3 gas laws and is given by:

pV T or pV/T = constant

i.e. p1V1/T1 = p2V2/T2

All the above laws relate to a fixed mass of gas

It is also found by experiments that the volume of a gas is proportional to its mass giving a combined equation of

pV mT or pV = AmT

where m is the mass of the gas and A is a constant of proportionality

As A is different for different gases we express the fixed mass of gas in terms of the number of moles of gas present

12

Avogadro’s Hypothesis

Avogadro’s hypothesis states that equal volume of all gases, at the same temperature and pressure, contain equal number of molecules.

As such, gases like O2, N2, CH4, CO2, CO, H2 etc all contain the equal number of molecules at the same temperature and pressure.

13

Mole (abbreviated as mol)

A mole is the number of elementary units (atoms or molecules) of any substance which is equal to the amount of atoms in 12 g of carbon-12.

The number of molecules per mole for all substances is the same and is called the Avogadro’s constant or number NA

NA is 6.02 x 1023/mol

14

Relative atomic mass & atomic mass unit

The relative atomic mass Ar is the ratio of the mass of an atomto one-twelfth of the mass of an atom of carbon-12

The relative atomic mass is numerically equal to the mass in grams of a mole of atoms

The relative molecular mass Mr is the ratio of the mass of a molecule to one-twelfth of the mass of an atom of carbon-12, and is numerically equal to the mass in grams of a mole of molecules

The atomic mass unit u is one-twelfth of the mass of an atom of carbon-12, and has the value 1.66 x 10-27 kg (also called the unified atomic mass constant)

Simple way to find the mass of 1 mole of an element is to take its nucleon number expressed in grams

e.g nucleon number of argon 40Ar is 40. Therefore, 1 mole of argon then has a mass of 40 g

15

Gas laws in terms of moles & molar gas constant • Hence in terms of moles n,

pV nTand putting in a new constant of proportionality R,

pV = nRT• R is called the molar gas constant (or sometimes universal

gas constant as it has the same value for all gases)• R has a value of 8.3 J K-1mol-1

• The above equation is expressed in the form pVm = RTwhere Vm is the volume occupied by 1 mole of the gas

• Another version is pV = NkT where N is the number of molecules in the gas and k is a constant called the Boltzmann constant which has a value of 1.38 x 10-23 J K-1

• The molar gas constant R and the Boltzmann constant k are connected through the Avogadro constant NA

nR = kNA

16

Experiment has confirmed that one mole of all gases at s.t.p. ( 0 degrees C and 760mm Hg or 1.01 x 105 Pa ) occupies 22.4 litres (1 litre is 1000 cm3).

From pV/T = k , substituting the above values, the constant for one mole R, called the molar gas constant, is therefore the same for all gases and has a value of 8.31 J/K/mol

At s.t.p.:pressure = 1.01 x 105Patemperature = 273.15 Kmolar gas volume = 22.4dm3

17

Equation of state

• The precise relation between volume, pressure, temperature and the mass of the given gas in a sample is called the equation of state of the gas

• An ideal gas is one which obeys the equation of state:

pV = nRT

• For approximate calculations the ideal gas equation can be used with real gases

18

Exercise

• Find the volume occupied by 1 mole of air at stp (273K and 1.01 x 105 Pa), taking R as 8.3 J K-1 mol-1 for air.

• Solution:

pV = nRT, so V = nRT/p

= 2.24 x 10-2 m3

19

Exercise

• Find the number of molecules per cubic metre of air at stp.

Solution:from above example, 1 mole of air at stp is 2.24 x 10-2

m3

but 1 mole of air contains NA molecules where NA is the Avogadro constant.

thus the number of molecules per cubic metre of air is 6.02 x 1023/2.24 x 10-2 = 2.69 x 1025 m-3

20

Exercise

• A syringe contains 25 x 10-6 m3 of helium gas at a temperature of 20° C and a pressure of 5.0 x 104 Pa. The temperature is increased to 400° C and the pressure on the syringe is increased to 2.4 x 105 Pa.

Find the new volume of gas in the syringe.

Solution:

Using p1V1/T1 = p2V2/T2 with T1 = 293 K & T2 = 673 K

V2 = 12 x 10-6 m3

21

Exercise

Oxygen gas contained in a cylinder of volume 1 x 10-2 m3 has a temperature of 300 K and a pressure 2.5 x 105 Pa. Calculate the mass of the oxygen used when the oxygen pressure has fallen to 1.3 x 105 Pa.

22

Exercise

Oxygen gas contained in a cylinder of volume 1 x 10-2 m3 has a temperature of 300 K and a pressure 2.5 x 105 Pa. Calculate the mass of the oxygen used when the oxygen pressure has fallen to 1.3 x 105 Pa.

SolutionLet the initial and final number of moles of oxygen be n1 and n2 respectively. n1 = p1V/(RT), n2 = p2V/(RT)

no. of moles used = n1 – n2 = (p1 – p2)V/(RT)= [(2.5 – 1.3) x 105 x 10-2]/(8.31 x 300)

= 0.48 moles= 0.48 x 32 x 10-3 kg = 0.015 kg.

23

The kinetic theory• Robert Boyle in 17th century developed an explanation

of how a gas exerts pressure• Daniel Bernoulli in 18th century explained it in greater

detail• The basic idea was that gases consist of atoms or

molecules moving about at great speed (later visualised by Robert Brown) and that the gas exerts a pressure on the walls of its container because of the continued impacts of the molecules with the walls

It’s value is the total rate of the momentum change of the molecules per unit area of the wall.

• The assumptions of the kinetic theory of an ideal gas are: All molecules behave as identical, hard, perfectly elastic spheres The volume of the molecules is negligible compared with the volume of the containing vessel There are no forces of attraction or repulsion between molecules There are many molecules, all moving randomly so that average behaviour can be considered

Derivation of pressure of gas

1. Derive the equation:

p V= 1/3Nm<c2>

p = 1/3ρ<c2>

2. Define the following:a) the mean speed <c>

b) the mean speed squared <c>2

c) the mean square speed <c2>

d) the rms speed.

c

25

Exercise

• The speed of 7 molecules in a gas are numerically equal to 2,4,6,8,10,12 and 14 units. Find the numerical values of:

a) the mean speed <c>b) the mean speed squared <c>2

c) the mean square speed <c2>d) the rms speed.

a) <c> = 8 unitsb) <c>2 = 64 units2

c) <c2> = 80 units2

d) rms speed = 8.94 units

26

Exercise

• Calculate the rms speed of hydrogen gas at s.t.p. where its density is 0.09 kg m-3.

• Solution

At s.t.p., p = 1.013 x 105 Pa

27

Mean Energy of Molecules

The kinetic energy of a molecule moving at an instant with a speed v is ½ mv2

The average kinetic energy <Ek> of translation of the random motion of the molecule of a gas is therefore ½ m<c2>.

Show that:i. 3/2kT = 1/2m<c2>

ii. <c> = √(3kT/m)

28

Conclusion

• The r.m.s. speed or velocity of the molecules of a gas T.

• The r.m.s. speed of the molecules of different gases at the same temperature 1/M, so gases of higher molecular mass have smaller r.m.s. speeds

• Hence the rms speed is proportional to the square root of the thermodynamic temperature of the gas, and inversely proportional to the square root of the mass of the molecule.

• Thus at a given temperature, less massive molecules move faster, on average, than more massive molecules i.e the higher the temperature, the faster the molecules move.

• The molecules of air at normal temperatures and pressures have an average velocity of the order of 480 m/s

29

Exercise

• The r.m.s. speed of hydrogen gas at s.t.p. is 1840 m/s. Calculate its new r.m.s. speed at 1000 C and the same pressure. What is the r.m.s. speed of oxygen at s.t.p.?

• Let cr = r.m.s. speed of hydrogen at 1000 Ccs = r.m.s. speed of oxygen at s.t.p.H2 is 2, O2 is 32

cr = 2150 m/s

cs = 460 m/s

30

Exercise

• Find the total kinetic energy of the molecules in one mole of an ideal gas at standard temperature.

Solution:

Ek = 3400 J mol-1

31

Exercise

• Find the rms speed of the molecules in nitrogen gas at 27° C. The mass of a nitrogen molecule is 4.6 x 10-26 kg.

• Solution:

crms

= 520 ms-1