Simple Nature - Light and Matter

where “. . .” indicates terms that may depend on ∆x, ∆p, m, andn, but that have no effect on comparisons of gas samples with thesame number of atoms.self-check CWhy does it make sense that the temperature term has a positive signin the above example, while the pressure term is negative? Why doesit make sense that the whole thing is proportional to n? ⊲ Answer, p.924To show consistency with the macroscopic approach to thermodynamics,we need to show that these results are consistent withthe behavior of an ideal-gas thermometer. Using the new definition1/T = dS/ dQ, we have 1/T = dS/ dE, since transferring anamount of heat dQ into the gas increases its energy by a correspondingamount. Evaluating the derivative, we find 1/T = (3/2)nk/E,or E = (3/2)nkT , which is the correct relation for a monoatomicideal gas.A mixture of molecules example 20⊲ Suppose we have a mixture of two different monoatomic gases,say helium and argon. How would we find the entropy of such amixture (say, in terms of V and E)? How would the energy beshared between the two types of molecules, i.e., would a moremassive argon atom have more energy on the average than aless massive helium atom, the same, or less?⊲ Since entropy is additive, we simply need to add the entropies ofthe two types of atom. However, the expression derived above forthe entropy omitted the dependence on the mass m of the atom,which is different for the two constituents of the gas, so we needto go back and figure out how to put that m-dependence back in.The only place where we threw away m’s was when we identifiedthe radius of the sphere in momentum space with √ 2mE, butthen threw away the constant factor of m. In other words, the finalresult can be generalized merely by replacing E everywhere withthe product mE. Since the log of a product is the sum of the logs,the dependence of the final result on m and E can be brokenapart into two different terms, and we findS = nk ln V + 3 2 nk ln m + 3 nk ln E + . . .2The total entropy of the mixture can then be written asS = n 1 k ln V + n 2 k ln V + 3 2 n 1k ln m 1 + 3 2 n 2k ln m 2+ 3 2 n 1k ln E 1 + 3 2 n 2k ln E 2 + . . .Now what about the energy sharing? If the total energy is E =E 1 + E 2 , then the most ovewhelmingly probable sharing of energySection 5.4 Entropy As a Microscopic Quantity 321