Why isn't this valid for the free expansion process?© BrainMass Inc. brainmass.com October 9, 2019, 6:49 pm ad1c9bdddf
The entropy is given by:
S = Nk Log[V/N (E/N)^(3/2)] + 5/2 Nk + 3/2NkLog[4 pi m/(3 h^(2))]
If you keep the gas at a constant temperature and let it expand then obviously, N stays the same. The total energy for an ideal mono atomic gas is 3/2 N kT and this also stays the same. So, the entropy increases by N k Log(V_2/V_1), where V_2 is the new volume and V_1 is the old volume.
Because the total internal energy is constant during the expansion, the heat Q that is supplied to the gas must equal the work done by the gas. The work that is performed by the gas follows from the ideal gas law:
P = N k T/V
Integral P dv from V_1 to V_2 is N k T ...
A detailed solution is given.