internal energy physics 202 professor lee carkner lecture 16
Post on 20-Dec-2015
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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
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
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