3.II.34E

Statistical Physics | Part II, 2008

Derive the following two relations:

TdS=CpdTTVTpdpT d S=C_{p} d T-\left.T \frac{\partial V}{\partial T}\right|_{p} d p

and

TdS=CVdT+TpTVdVT d S=C_{V} d T+\left.T \frac{\partial p}{\partial T}\right|_{V} d V

[You may use any standard Maxwell relation without proving it.]

Experimentalists very seldom measure CVC_{V} directly; they measure CpC_{p} and use thermodynamics to extract CVC_{V}. Use your results from the first part of this question to find a formula for CpCVC_{p}-C_{V} in terms of the easily measured quantities

α=1VVTp\alpha=\left.\frac{1}{V} \frac{\partial V}{\partial T}\right|_{p}

(the volume coefficient of expansion) and

κ=1VVpT\kappa=-\left.\frac{1}{V} \frac{\partial V}{\partial p}\right|_{T}

(the isothermal compressibility).

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