 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2010/paper-1/section-c/c10/solution

d G = - S d T + V d p

gives

S = - (\partial G / \partial T)_{p}

and

V = (\partial G / \partial p)_{T}

. Equality of mixed derivatives yields

(\partial S / \partial p)_{T} = - (\partial V / \partial T)_{p}

. Therefore

d S = - (\frac{\partial V}{\partial T})_{p} d p + (\frac{\partial S}{\partial T})_{p} d T .

For one mole of ideal gas,

V_{m} = RT / p

, so

(\partial V_{m} / \partial T)_{p} = R / p

. Reversible heat at fixed pressure gives

(\partial S_{m} / \partial T)_{p} = C_{m p} / T

, and consequently

d S_{m} = - \frac{R}{p} d p + \frac{C _{m p}}{T} d T .

With constant positive molar heat capacity, integrate from a reference state

(p_{0}, T_{0})

S_{m} = S_{0} - R ln \frac{p}{p _{0}} + C_{m p} ln \frac{T}{T _{0}} .

If heat capacity varies, the last term must instead be

\int_{T_{0}}^{T} C_{m p} (T^{'}) d T^{'} / T^{'}

. The reference ratios make the logarithms dimensionless. The following classical ideal-gas entropy limits test where the constant-heat-capacity expression remains applicable.

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