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Internal Energy Physics 202 Professor Lee Carkner Lecture 14

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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 m 3, expand isothermally from 3500 Pa to 2000 Pa For isothermal process: W = nRTln(V f /V i ) pV = nRT, T = pV/nR = (3500)(2)/(3)(8.31) = 281 K V f = nRT/p f = (3)(8.31)(281)/(2000) = 3.5 m 3 Since T is constant, E = 0, Q = W = 3917 J

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Ideal Gas pV=nRT v rms = (3RT/M) ½ K ave =(3/2) k T

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Internal Energy We have looked at the work of an ideal gas, what about the internal energy? E int = (nN A ) K ave = nN A (3/2)kT E int = (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

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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 E int / T = (3/2)nR, so: C V = 3/2 R = 12.5 J/mol K Molar specific heat at constant volume for an ideal gas

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Specific Heat and Internal Energy If C V = (3/2)R we can find the internal energy in terms of C V E int = nC V T True for any process (assuming monatomic gas)

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Specific Heat at Constant Pressure We can also find the molar specific heat at constant pressure (C p ) E int = nC V T W = p V = nR T C p = C V + R C p is greater than C v At constant pressure, you need more heat since you are also doing work

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Degrees of Freedom Our relation C V = (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

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Rotational Motions Monatomic No Rotation Polyatomic 2 Rotational Degrees of Freedom

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Equipartition of Energy Equipartition of Energy: Each degree of freedom (f) has associated with it energy equal to ½RT per mole C V = (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 C p = C V + R

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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

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Internal Energy of H 2

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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 E int =-W since Q=0

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Ideal Gas Processes I Isothermal Constant temperature W = nRTln(V f /V i ) Isobaric Constant pressure W=p V E int = nC p T-p V

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Ideal Gas Processes II Adiabatic No heat (pV = constant, TV -1 = constant) W=- E int Isochoric Constant volume W = 0 E int = Q

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Idea Gas Processes III For each type of process you should know: Path on p-V diagram Specific expressions for W, Q and E

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

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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)T A = T B B)T A = 2 T B C)T A = ½ T B D)T A = √2 T B E)v A = (1/√2) v B

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How does the rms velocity of the molecules in room A compare to the rms velocity of the molecules in room B? A)v A = v B B)v A = 2 v B C)v A = ½ v B D)v A = √2 v B E)v A = (1/√2) v B

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