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Internal Energy

In this lecture, we explore the fundamentals of isothermal processes and the internal energy of gases. We discuss the relationship between pressure, volume, and temperature in such processes, emphasizing that internal energy depends solely on temperature for ideal gases, particularly monatomic ones. We cover molar specific heats at constant volume and pressure, and the roles of degrees of freedom and equipartition of energy. We also analyze adiabatic expansions along with key formulas. This foundational knowledge is crucial for thermodynamics applications.

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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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