Internal Energy Physics 202 Professor Lee Carkner Lecture 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) B) C) D) E) T A = TB TA = 2 T B TA = TB TA = 2 TB 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) B) C) D) E) vA = vB vA = 2 v B
v A = vB vA = 2 vB 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) nRT
W dV V 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)nRT Eint = 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 Monatomic No Rotation Polyatomic 2 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 Isothermal
Constant temperature Q=W W = nRTln(Vf/Vi) Isobaric Constant pressure Q=nCp T W=pV Ideal Gas Processes II Adiabatic
No heat (pV = constant, TV-1 = constant) Q= 0 W=-Eint Isochoric Constant volume Q= 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