Problem Set 5 Solutions - Thermal Physics | PHYS 224, Assignments of Thermal Physics

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224 problem set 5 solutions

224 problem set 5 solutions Problem 2.31. Starting from equation 2.40 for the multiplicity, we have for the entropy of an ideal gas ca 1 VN qansa ana . Ommtt Ve"! ; S= 1a (asm GNOME) ) = inv" + in( 3 ) — InN! tn[(3NV/2)! Iam \? | BN 73Ny\ 3N = Nin +Nin( i ) ~NinN +N -*n() > vo Qnml! \/? BN\W? 5 V ¢4nml! 3/2 5 * [in 5 +1n( ne ) -n(>) +3] ~ wlin( 5 (3 ) ) * 3: In the second line I've used Stirling’s approximation twice, in the form of equation 2.16 which omits the merely “large” factor of ¥20N. The final expression is the Sackur-Tetrode result, equation 2.49. Problem 2.34. The increase in entropy during quasistatic isothermal expansion of an ideal gas is computed in equation 2.51 as where V, and V, are the initial and final volumes. But the heat input during this process was computed in equation 1.31 as vo V NEP ty =k in“, Q=-Wwat] o> V Dividing this expression by T gives the preceding expression for AS, so indeed, AS = Q/T. For the free expansion process, however, AS is still given by the same expression but Q = 0; therefore AS is most definitely not equal to Q/T. Problem 2.35. Writing 5/2 as In e°/?, the Sackur-Tetrade equation becomes we") | $= Neto] We want to know when this quantity is negative, that is, when the argument of the logarithm is less than 1, So set it equal to 1 and use the equipartition theorem to write U in terms of T: Vv 4nmU 32 QamkT 3/2 — 5/2 — © 8/2 '= ne (Ga) We ( ie y Solving for T gives Nyt h? T=(5) say Vi Qre’/Smk We're to assume that N/V is the same as at room temperature (Jy) and atmospheric pressure (P)), so we can use the ideal gas law to write it as Py/kTy, then plug in Fy = 10° Pa and Ty = 300 K. The mass of a helium atom is 4 u, where ] u = 1.66 x 10-*7 kg. Plugging in all these numbers, I get 7’ = 0.01 K. Below this temperature, the methods of Chapter 7 become necessary.