# Quasistatic isothermal expansion and free expansion

During the quasistatic isothermal expansion of a monoatomic ideal gas, how is the change in entropy related to the heat input Q by the simple formula: ΔS = Q/T?

Why isn't this valid for the free expansion process?

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#### Solution Preview

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 ...

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