The partition function for an ideal gas in general is 1 3N/2 zideal gas 2Tm ZN N! VN(Zint ( T) ) N h2 B At room temperature (300 K), consider two ideal gases. The first is monatomic and its chemical potential is /monatomic = -0.6 eV. The second is diatomic, has the same mass per particle as the first (for simplicity) and at this temperature, its Zint = 200. What is the chemical potential of this diatomic gas, in ev? evConsider a monoatomic ideal gas, with Zint = 1. The partition function is then 3N 2 Zmonnatomic ideal gas 1 ( 2m ) f VN N : I125 Use F = len Z, the Stirling approximation In N! : NlnN Nand the appropriate partial derivative to derive the chemical potential of the monoatomic ideal gas as a function of T, N and V. You may want to compare your result with what you got in Weekly Practice 9. (a) Take the atomic mass of Xenon to be 131 (Xenon has 8 different stable isotopes and many more metastable ones). What is the chemical potential for pure Xe gas at 1 atm (101,325 Pa) and T = 300 K? Use the ideal gas law and give your answer in eV. Iv (b) How would your answer to part (a) change if the gas was pure Krypton at the same temperature and pressure? Take the atomic mass of Krypton to be 84. Ivv (b) Repeat the computation from part (a) if Xe is only 1% (by number density or, equivalently, partial pressure) of a mixture of different gases. Note: if pure Xenon is allowed to come in contact with the gas in part (b), the net flow of Xenon atoms should be into the mixture. This tells you that your answer to part (b) should be smaller than your answer to part (a). Iv