Gases, Heat, and Work

PhysicsMontwood High School

R. Casao

Kinetic Theory of Gas Behavior• A gas consists of atoms (either individually or bound

together as molecules) that fill their container’s volume and exert pressure on the container’s walls.

• The three variables associated with a gas - volume, pressure, and temperature – are all a consequence of the motion of the atoms.– Volume is a result of the freedom atoms have to spread

throughout the container.– Pressure is a result f he collisions of the atoms with the

container’s walls. – Temperature is related to the kinetic energy of the atoms.

• The kinetic theory of gases relates the motion of the atoms to the volume, pressure and temperature of the gas.

• Gas molecules move about randomly, colliding with the walls of the container and with each other.

• Each gas molecule is considered to be a hard sphere that collides elastically with the container’s walls and with other molecules.

• Hard-sphere model assumes that the molecules do not interact with each other except during collisions and that they are not deformed by collisions. – This description works only for

monoatomic gases (a single molecule) for which the energy is entirely linear kinetic energy.

– For diatomic molecules, such as O2, the model must include the internal energy associated with rotations and vibrations of the molecules.

Model of an Ideal Gas• The pressure that a gas exerts on the walls of its

container is a result of the collisions of the gas molecules with the walls of the container.

• The number of molecules is large and the average separation between molecules is great compared to their diameters. The volume of the molecules is negligible when compared with the volume of the container.

• The molecules obey Newton’s laws of motion, but as a whole they move randomly, meaning that any molecule can move in any direction with equal probability.

Model of an Ideal Gas• The molecules undergo elastic collisions with

each other and with the container walls. In collisions, both kinetic energy and momentum are constant.

• The forces between molecules are negligible except during collision. The forces between molecules are short-range, so the molecules interact with each other during collisions.

• The gas under consideration is a pure substance. All of the molecules are identical.

Temperature and Translational Kinetic Energy• A sample of an ideal gas at a given temperature consists of

particles moving with a distribution of velocities; some move fast, some move slowly.

• Root mean square speed (vrms) is a kind of average speed.

– You square each speed, – You find the mean (the average) of all these squared speeds, – You take the square root of the mean.

– Many particles move faster than vrms and other particles move much slower than vrms.

2rms avgv = v

• Equation for vrms:

R = gas constant = 8.3145 J/mol·K T = Kelvin temperature M = molar mass

• The translational kinetic energy of a particle at any instant is

• The average translational kinetic energy is

rms

3 R Tv =

M

21m v

2K

2 2 2avg avg rms

avg

1 1 1m v m v m v

2 2 2K

• Substituting vrms:

• Substituting M/m = NA (Avogadro’s number):

• Boltzmann’s constant , therefore:

where kB = 1.38 x 10-23 J/K

2avg rms

1 1 3 R Tm v m

2 2 MK

avgA

3 R T2 N

K

A

Rk=

N

Bavg

3 k T2

K

• Summary:

• At a given temperature T, all ideal gas molecules – no matter what their mass – have the same average translational kinetic energy.

• RMS particle speed is inversely proportional to the square root of particle mass:

2rms

1 3 k Tm v

2 2

rms

rms

A B

B A

v m

v m

Gases, Heat, and Work

• Consider a gas confined to a cylinder with a movable piston.

• The upward force on the piston due to the pressure of the confined gas is equal to the weight of lead shot loaded on top of the piston.

• The walls of the cylinder are made of insulating material that does not allow any transfer of energy as heat.

• The bottom of the cylinder rests on a reservoir of thermal energy (like a hot plate), with a control knob to adjust the temperature.

• The system ( the gas) starts from an initial state described by a pressure Pi, a volume Vi, and a temperature Ti.

• The system will change to a final state described by a pressure Pf, a volume Vf, and a temperature Tf.

• The process by which the system changes from the initial state to the final state is called a thermodynamic process.

• Energy can be transferred into the system from the thermal reservoir (positive Q) or transferred from the system to the thermal reservoir (negative Q).

• Work can be done by the system to raise the loaded piston (positive work) or work can be done on the piston to lower (compress) the system (negative work).

• All changes are performed slowly so that the system can be considered to be in thermal equilibrium at all times.

• Removing a mass of lead shot from the top of the piston will allow the gas to push the piston and remaining shot upward through a small distance with an upward force F.

• Pressure = force/area, so force = pressure·area.

• Work:

ΔVPWΔVΔdA

ΔdAPdFW

Three Possibilities

• If the piston does not move: Fw lead + Fw piston = F gas on piston

• P inside cylinder = P outside cylinder.

• No work is done by the gas or on the gas.

• If the piston moves down: Fw lead + Fw piston > F gas on piston

• P outside cylinder > P inside cylinder.

• Work done by the gas is negative; work done on gas positive.

• Piston stops moving down when P outside cylinder = P inside cylinder.

• If the piston moves up: Fw lead + Fw piston < F gas on piston

• P inside cylinder > P outside cylinder.

• Work done by the gas is positive; work done on gas negative.

• Piston stops moving up when P outside cylinder = P inside cylinder.

• V is the change in volume of the gas due to the movement of the piston. V = Vf – Vi

• When enough shot has been removed to allow the gas to change its volume from Vi to Vf, the total work done by the gas is:

• During the change in volume, the pressure and temperature of the gas may also change.

)T(TRnVVPWifif

Pressure-Volume Diagrams• There are many ways in which a system can move from

its initial state to its final state.• Pressure-volume (PV) diagrams illustrate the changes in

a system as it progresses from its initial state to its final state.

• The work done by the gas as it changes from its initial state to its final state is equal to the area under the curve between points i and f.

• The work done by a gas is positive if the volume increases and negative if the volume decreases.

• A system can be taken from an initial state to a final state in a number of different ways. Heat may or may not be involved and the work W and heat Q will have different values for different processes. Work and heat depend on the path taken from i to f.

The First Law of Thermodynamics• While the work and heat depend on the path

taken from i to f, the quantity Q – W is the same for all paths from i to f.

• The quantity Q – W represents a change in the internal energy U of the system.

• The internal energy U of a system tends to increases if energy is added as heat Q and tends to decrease if energy is lost as work W done by the system.

gasbyWQU

Internal Energy U• Internal energy U: of a system includes the energy in

chemical, thermal, and nuclear forms, and is due to the systems microscopic atomic motion and molecular configurations.– A system's temperature reflects its internal energy;

U = 1.5·n·R·T = 1.5·P·V = 1.5·N·k·T – Change in internal energy:

dU = 1.5·n·R·(Tf – Ti) for a monoatomic gas

dU = 2.5·n·R·(Tf – Ti) for a diatomic gas

– Cannot measure the internal energy directly, but can measure changes in internal energy.

– The kinetic energy of the particles is the only way for an ideal gas to store energy.

• Energy may be transferred through the system boundary either as work W or as heat Q.

• Previously, the term work always meant the work done ON a system. Now we will focus on the work done BY a system.

• The work done ON a system is always the negative of the work done BY the system. In terms of the work done ON a system, the first law of thermodynamics can be expressed mathematically as:

• Internal energy tends to increase if heat is absorbed by the system or if positive work is done on the system.– The system temperature will increase.

• Internal energy tends to decrease if heat is lost by the system or if negative work is done on the system.– The system temperature will decrease.

gasbyWQΔU

A Friend From Chemistry• A mole is the number of atoms in a 12 g

sample of C-12.

• Avogadro’s number: NA = 6.02 x 1023 atoms or molecules per mole

• Number of moles n in a sample of any substance is equal to the ratio of the number of molecules N in the sample to the number of molecules in 1 mole, NA:

AN

Nn

• The moles in a sample can be determined from the mass of the sample Msam and the molar mass M (the mass of 1 mole) or the molecular mass m (the mass of 1 molecule):

• The mass of 1 mol is the product of the mass m of one molecule and the number of molecules NA:

m

M

M

Mn samsam

ANmM

Ideal Gases• Scientists found that the pressures of 1

mole samples of various gases contained in identical volumes and at the same temperature are nearly the same. If lower gas densities are used, then the small differences in pressure disappear.

• Ideal gas law: P·V = n·R·T– P = gas pressure; V = gas volume– n = number of moles of gas present– R = gas constant = 8.3145 J/mol·K– T = Kelvin temperature

• The ideal gas law can also be expressed in terms of the number of gas molecules present: P·V = N·k·T– P = gas pressure; V = gas volume– N = number of molecules of gas present– k = gas constant = 1.38 x 10-23 J/K– T = Kelvin temperature

• Because n·R or N·k are constants for a gas that undergoes changes in volume, pressure, and temperature:

f

ff

i

ii

T

PV

T

PV

Work Done by Ideal Gas at Constant Temperature

• When an ideal gas in a piston-cylinder arrangement is allowed to expand from an initial volume Vi to a final volume Vf while keeping the temperature constant, this is called an isothermal expansion.

• An isothermal compression would require doing work on the piston-cylinder system to cause the volume to decrease.

• On a P-V diagram, an isotherm is a curve that connects points having the same temperature.

• An isotherm is a curve that connects points that have the same temperature.

• The figure shows three isotherms, each corresponding to a constant value of T.

• The values of T for the isotherms increase upward to the right. T1<T2<T3.

• The red line on the 310 K isotherm is the path followed by a gas during an isothermal expansion from state i to state f at a constant temperature of 310 K.

• Work:

• Unit: Joule, J

• Isothermal Expansion

i

f

V

VlnTRnW

• An isothermal compression occurs when the volume of the gas decreases at constant temperature.

• Expansion: Vf > Vi; ratio of Vf /Vi >1 and W done by the gas is positive.

• Compression: Vf < Vi; ratio of Vf / Vi < 1 and W done by the gas is negative.

• Isothermal Compression

Work Done at Constant Volume/Constant Pressure

• If the volume of the gas is constant (the vertical path in the figure), then W = P·(Vf – Vi) = 0 J because Vf – Vi = 0.

• If the volume changes from Vi to Vf (Path A) while the pressure stays constant, then:

W = P·(Vf – Vi).

Path A: isobaric process (constant pressure)

Vertical path: isochoric process (constant volume)

Problem Example 1• A cylinder contains 12 L of oxygen at 20° C and 15

atm. The temperature is raised to 35° C and the volume is reduced to 8.5 L. What is the final pressure of the gas in Pascals (N/m2)?

• Variables: Vi = 12 L; Ti = 20° C; Pi = 15 atm; Tf = 35° C; Vf = 8.5 L

• Conversions:– Ti = 20° C + 273 = 293 K

– Tf = 35° C + 273 = 308 K

– Pi = 15 atm·101300 Pa/atm = 1519500 Pa

Pa2254997.82490.5

Pa5616072000P

K293L8.5

K308Pa1519500L12P

TV

TPVP

T

PV

T

PV

f

f

if

fii

f

f

ff

i

ii

Problem Example 2• 2 moles of oxygen

expands at a constant temperature of 310 K from an initial volume of 12 L to a final volume of 19 L. How much work is done by the gas during the expansion?

• The shaded area is equal to the work done by the gas in expanding from Vi to Vf.

• Variables: Vi = 12 L; Vf = 19 L; T = 310 K; n = 2 mol

• Constants: R = 8.3145 J/mol·K

• Equation:

J2368.88W

L12

L19lnK310

Kmol

J8.3145mol2W

V

VlnTRnW

i

f

Adiabatic Processes• An adiabatic process occurs so fast or

happens in a system that is so well insulated that no transfer of energy occurs between the system and the environment; Q = 0 J.

• If work is done by the system (gas), W is positive and U decreases by the amount of work.

gasby

gasbygasby

ΔWΔU

ΔWJ0ΔWΔQΔU

• If work is done on the system (gas), W is negative and U increases by the amount of work.

• In an idealized adiabatic process, heat cannot enter or leave because of the insulation.

• The only way energy can be transferred between the system and the environment is by work.

• If you remove shot from the piston and allow the gas to expand, the work done by the system (gas) is positive and U decreases by the amount of work.

• If you add shot to the piston, the system (gas) is compressed. The work done by the system is negative and U increases by the amount of work.

• Adiabatic processes are common in the cycle of a gasoline engine.

• At any time in the adiabatic process, we assume that the gas is in an equilibrium state, so that P·V = n·R·T is valid.

• The pressure and volume at any time during an adiabatic process is given by:

v

p

γ

c

cγ

constantVP

Adiabatic Expansion • Cp is the specific heat at constant pressure; Cv is the specific heat at constant volume.

• = 1.4 for diatomic gases and 1.67 for monoatomic gases.

• For the adiabatic expansion shown, the gas cools as it expands and does work on the surroundings (W is negative); if T is negative, U is negative.

• For an adiabatic compression, the gas heats as the surroundings do work on the gas and compress it (W is positive); if T is positive, U is positive.

• From the initial and final states of the expansion or compression:

1-γ

ff

1-γ

ii

γ

ff

γ

ii

VTVT

VPVP

Work in an Adiabatic Process• U = -Wby gas

• If the number of moles and the initial and final temperatures are known:

W = n·CV·(Ti – Tf)CV is the specific heat at constant volume

• Using P·V = n·R·T and CV =R/( – 1):

– Expansion: temperature drops, Ti > Tf; PiVi >PfVf; W positive.

– Compression: temperature increases, Tf > Ti; PfVf >PiVi; W negative.

1γVPVP

W ffii

Problem Example 3• Air in the cylinder of a diesel engine at 20° C is

compressed from an initial pressure of 1 atm and volume of 800 cm3 to a volume of 60 cm3. Assuming that air behaves as an ideal gas ( = 1.4) and that the compression is adiabatic, find the final pressure and temperature.

• Variables: Vi = 800 cm3; Pi = 1 atm; Vf = 60 cm3

atm37.6308.611

11596.5atm1P

cm60

cm800atm1

V

VPP

f

1.43

1.43

γ

f

γ

ii

f

• For the temperature, convert ° C to K:Ti = 20° C + 273 = 293 K

Could also have used:

K56.2685.14

4248.5T

5.14

14.5K293

60

800K293T

cm60

cm800K293

V

VTT

f

0.4

0.4

f

11.43

11.43

1γ

f

1γ

ii

f

f

ff

i

ii

T

PV

T

PV

Summary of the First Law of Thermodynamics

For all Paths: U = Q – W by gasdU = 1.5·n·R·(Tf – Ti) for a monoatomic gasdU = 2.5·n·R·(Tf – Ti) for a diatomic gas

Path Constant Process Equations

1 P Isobaric W = P·(Vf – Vi)

2 T Isothermal U = Q = W =n·R·T·ln(Vf/Vi)

3 P·V, T·V Adiabatic Q = 0; U = W

4 V Isochoric W = 0; U = Q

1γVPVP

W ffii

Internal Energy and Cyclic Processes

• The change in internal energy, dU, for any processes that take a gas from any initial conditions to any intermediate conditions and then returns the gas to the original conditions is 0 J; dU = 0 J.

• Example: as the gas in the figure is taken thru the cycle ABCDA, the change in internal energy is 0 J.

• When using the R = 8.3145 J/mol·K gas constant , the pressure must be in Pa and the volume must be in m3.

• Conversions: 1 atm = 1.013 x 105 Pa 1 m3 = 1000 L

Thermodynamic Equilibrium Virtual Laboratory

• http://jersey.uoregon.edu/vlab/Thermodynamics/