Internal Energy

Physics 202Professor Lee

CarknerLecture 16

Consider two rooms of a house, room A and room B. If the (otherwise identical) molecules in room B have twice as much average kinetic energy than the ones in A, how does the temperature of room A compare to the temperature of room B?

A) TA = TB

B) TA = 2 TB

C) TA = ½ TB

D) TA = √2 TB

E) TA = (3/2) TB

How does the rms velocity of the molecules in room A compare to the rms velocity of the molecules in room B?

A) vA = vB

B) vA = 2 vB

C) vA = ½ vB

D) vA = √2 vB

E) vA = (1/√2) vB

PAL #15 Kinetic Theory Which process is isothermal?

Start with p =4 and V = 5 Since T is constant, nRT is constant and thus pV is constant Initial pV = 20, A: pV = 20, B: pV = 21 A is isothermal

3 moles at 2 m3, expand isothermally from 3500 Pa to 2000 Pa For isothermal process: W = nRTln(Vf/Vi) Need T, Vf

pV = nRT, T = pV/nR = (3500)(2)/(3)(8.31) = 281 K Vf = nRT/pf = (3)(8.31)(281)/(2000) = 3.5 m3

W = (3)(8.31)(281)ln (3.5/2) = 3917 J Since T is constant, E = 0, Q = W = 3917 J

Ideal Gas We will approximate most gases as

ideal gases which can be represented by:

vrms = (3RT/M)½

dVVnRT

W Vf

Vi

Internal Energy We have looked at the work of an ideal gas,

what about the internal energy?

Eint = (nNA) Kave = nNA(3/2)kT

Eint = (3/2) nRT

Internal energy depends only on temperature

Since monatomic gasses can only have energy of motion

Molar Specific Heats How does heat affect an ideal gas?

The equation for specific heat is:

From the first law of thermodynamics:

Consider a gas with constant V (W=0),

But Eint/T = (3/2)nR, so:

CV = 3/2 R = 12.5 J/mol K Molar specific heat at constant volume

Specific Heat and Internal Energy

Eint = (3/2)nRTEint = nCVT

Eint = nCV T

True for any process (assuming monatomic gas)

Specific Heat at Constant Pressure

We can also find the specific heat at constant pressure:

Eint = nCVT

W = pV = nR T Solving for Cp we find:

Cp = CV + R

For a constant pressure or constant volume situation (assuming a monatomic ideal gas) we can find how much heat is required to produce any temperature change

Degrees of Freedom Our relation CV = (3/2)R = 12.5 agrees

with experiment only for monatomic gases

We assumed that energy is stored only in translational motion

For polyatomic gasses energy can also be stored in modes of rotational motion

Each possible way the molecule can store energy is called a degree of freedom

Rotational Motions

MonatomicNo Rotation

Polyatomic2 Rotational Degrees of Freedom

Equipartition of Energy Equipartition of Energy:

We can now write CV as

CV = (f/2) R = 4.16f J/mol K Where f = 3 for monatomic gasses (x,y and z

translational motion and f=5 for diatomic gases (3 trans. + 2 rotation)

Oscillation

The atoms oscillate back and forth as if the bonds were springs

So there are 3 types of microscopic motion a molecule can experience: translational -- rotational -- oscillatory --

If the gas gets too hot the molecules will disassociate

Internal Energy of H2

Adiabatic Expansion

It can be shown that the pressure and temperature are related by:

pV = constant

You can also write:TV-1 = constant

Ideal Gas Processes I

IsothermalConstant temperatureQ=WW = nRTln(Vf/Vi)

IsobaricConstant pressureQ=nCp T W=pV

Ideal Gas Processes II

AdiabaticNo heat (pV = constant,

TV-1 = constant)Q = 0W=-Eint

IsochoricConstant volumeQ= nCVT W = 0

Idea Gas Processes III For each type of

process you should know: Path on p-V

diagram What is constant Specific

expressions for W, Q and E

Next Time

Read: 21.1-21.4 Homework: Ch 19, P: 44, 46, 53,

Ch 20, P: 2, 4 Note: Test 2 next Friday, Feb 1