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Internal Energy

Internal Energy. Physics 202 Professor Lee Carkner Lecture 16.

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Internal Energy

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  1. Internal Energy Physics 202 Professor Lee Carkner Lecture 16

  2. 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 • TA = (3/2) TB

  3. 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

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

  5. Ideal Gas • We will approximate most gases as ideal gases which can be represented by: vrms = (3RT/M)½

  6. 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

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

  8. Specific Heat and Internal Energy Eint = (3/2)nRT Eint = nCVT DEint = nCVDT • True for any process (assuming monatomic gas)

  9. Specific Heat at Constant Pressure • We can also find the specific heat at constant pressure: DEint = nCVDT W = pDV = nR DT • 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

  10. 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

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

  12. 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)

  13. 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

  14. Internal Energy of H2

  15. Adiabatic Expansion • It can be shown that the pressure and temperature are related by: pVg = constant • You can also write: TVg-1 = constant

  16. Ideal Gas Processes I • Isothermal Constant temperature Q=W W = nRTln(Vf/Vi) • Isobaric Constant pressure Q=nCp DT W=pDV

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

  18. 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 DE

  19. 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

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