Here is a first attempt at the entropy of argon atoms (molar mass ) held at in a…

PHYSICS · Multi-part · Fermi University Physics · Thermal, Statistical & Complex Systems · Statistical foundations and the equilibrium ensembles · Statistical mechanics of ideal classical systems

Thermal Physics, Ch The classical ideal gas from the partition function

Problem

Here is a first attempt at the entropy of N=2.00×1023N = 2.00\times10^{23} argon atoms (molar mass 39.95 g mol−139.95\,\mathrm{g\,mol^{-1}}) held at T=250 KT = 250\,\mathrm{K} in a rigid box of volume V=5.00 L=5.00×10−3 m3V = 5.00\,\mathrm{L} = 5.00\times10^{-3}\,\mathrm{m^3}. Argon is monatomic, the gas is dilute enough to be treated as an ideal gas in the classical regime, and only translational states are counted. The attempt writes Z1=V/λT3Z_1 = V/\lambda_T^3 with λT=h/2πmkBT\lambda_T = h/\sqrt{2\pi m k_B T}, then ZN=Z1 NZ_N = Z_1^{\,N}, so that F=−NkBTln⁡(V/λT3)F = -Nk_BT\ln(V/\lambda_T^3) and Strial=NkB[ln⁡ ⁣(VλT3)+32].S_{\rm trial} = Nk_B\left[\ln\!\left(\frac{V}{\lambda_T^3}\right) + \frac{3}{2}\right]. One factor has been left out of ZNZ_N. Name it, then work through the consequences. Use h=6.626×10−34 J sh = 6.626\times10^{-34}\,\mathrm{J\,s}, kB=1.381×10−23 J K−1k_B = 1.381\times10^{-23}\,\mathrm{J\,K^{-1}}, NA=6.022×1023 mol−1N_A = 6.022\times10^{23}\,\mathrm{mol^{-1}}, and Stirling's approximation ln⁡N!≈Nln⁡N−N\ln N! \approx N\ln N - N. ((a)) Evaluate λT\lambda_T and the trial entropy StrialS_{\rm trial}. ((b)) Evaluate the correct entropy from the Sackur-Tetrode formula. ((c)) Show that the difference Strial−SS_{\rm trial} - S is kBln⁡N!k_B\ln N!, and evaluate it. ((d)) The system is now doubled at the same temperature (N→2NN \to 2N, V→2VV \to 2V). By how much does the trial formula exceed twice its original value?

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