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Internal Energy

Internal Energy. Physics 202 Professor Lee Carkner Lecture 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 m 3 , expand isothermally from 3500 Pa to 2000 Pa

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Internal Energy

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  1. Internal Energy Physics 202 Professor Lee Carkner Lecture 14

  2. 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, DE = 0, Q = W = 3917 J

  3. Ideal Gas pV=nRT vrms = (3RT/M)½ Kave =(3/2) k T

  4. 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

  5. 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 DEint/DT = (3/2)nR, so: CV = 3/2 R = 12.5 J/mol K • Molar specific heat at constant volume for an ideal gas

  6. Specific Heat and Internal Energy • If CV = (3/2)R we can find the internal energy in terms of CV DEint = nCVDT • True for any process (assuming monatomic gas)

  7. Specific Heat at Constant Pressure • We can also find the molar specific heat at constant pressure (Cp) DEint = nCVDT W = pDV = nR DT Cp = CV + R • Cp is greater than Cv • At constant pressure, you need more heat since you are also doing work

  8. 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

  9. Rotational Motions Polyatomic 2 Rotational Degrees of Freedom Monatomic No Rotation

  10. 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

  11. 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

  12. Internal Energy of H2

  13. Adiabatic Expansion • It can be shown that the pressure and temperature are related by: pVg = constant • You can also write: TVg-1 = constant • Remember also that DEint =-W since Q=0

  14. Ideal Gas Processes I • Isothermal Constant temperature W = nRTln(Vf/Vi) • Isobaric Constant pressure W=pDV DEint = nCpDT-pDV

  15. Ideal Gas Processes II • Adiabatic No heat (pVg = constant, TVg-1 = constant) W=-DEint • Isochoric Constant volume W = 0 DEint = Q

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

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

  18. 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? • TA = TB • TA = 2 TB • TA = ½ TB • TA = √2 TB • vA = (1/√2) vB

  19. How does the rms velocity of the molecules in room A compare to the rms velocity of the molecules in room B? • vA = vB • vA = 2 vB • vA = ½ vB • vA = √2 vB • vA = (1/√2) vB

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