Internal Energy

Physics 202Professor Lee

CarknerLecture 14

PAL #13 Kinetic Theory Which process is isothermal?

Since T is constant, nRT is constant and thus pV is constant Initial pV = 20, A: pV = 20, B: pV = 21

3 moles at 2 m3, expand isothermally from 3500 Pa to 2000 Pa For isothermal process: W = nRTln(Vf/Vi) pV = nRT, T = pV/nR = (3500)(2)/(3)(8.31) = 281 K Vf = nRT/pf = (3)(8.31)(281)/(2000) = 3.5 m3

Since T is constant, E = 0, Q = W = 3917 J

Ideal Gas

pV=nRTvrms = (3RT/M)½

Kave =(3/2) k T

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 Strictly true only for monatomic gasses

Note that this is the total internal energy, not the change in internal energy

Molar Specific Heats If we add heat to something, it will change

temperature, depending on the specific heat

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 for an ideal gas

Specific Heat and Internal Energy

If CV = (3/2)R we can find the internal energy in terms of CV

Eint = nCV T

True for any process (assuming monatomic gas)

Specific Heat at Constant Pressure

We can also find the molar specific heat at constant pressure (Cp)

Eint = nCVT

W = pV = nR T

Cp = CV + R Cp is greater than Cv

At constant pressure, you need more heat since you are also doing work

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:

Each degree of freedom (f) has associated with it energy equal to ½RT per mole

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)

It is always true that Cp = CV + R

Oscillation At high temperatures we also have

oscillatory motion

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

Remember also that Eint =-W since Q=0

Ideal Gas Processes I

IsothermalConstant temperature W = nRTln(Vf/Vi)

IsobaricConstant pressure W=pV Eint = nCpT-pV

Ideal Gas Processes II

AdiabaticNo heat (pV = constant,

TV-1 = constant) W=-Eint

IsochoricConstant volume W = 0Eint = Q

Idea Gas Processes III For each type of

process you should know: Path on p-V

diagram Specific

expressions for W, Q and E

Next Time

Read: 21.1-21.4 Note: Test 3 next Friday, Jan 20

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) vA = (1/√2) vB

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