The constant-pressure and constant-volume heat capacities appear throughout thermodynamics. For ideal gases, the relationship
\[ C_P - C_V = R \]
is familiar. However, this result is only true for an ideal gas. Our goal is to derive a more general expression that applies to any substance.
The heat capacities are defined by
\[ C_V = \left( \frac{\partial U} {\partial T} \right)_V \]
\[ C_P = \left( \frac{\partial H} {\partial T} \right)_P \]
Therefore,
\[ C_P - C_V = \left( \frac{\partial H} {\partial T} \right)_P - \left( \frac{\partial U} {\partial T} \right)_V \]
Begin with the definition of enthalpy:
\[ H = U + pV \]
Differentiating gives
\[ dH = dU + p\,dV + V\,dp \]
Divide by \(dT\) and constrain to constant pressure:
\[ \left( \frac{\partial H} {\partial T} \right)_P = \left( \frac{\partial U} {\partial T} \right)_P + p \left( \frac{\partial V} {\partial T} \right)_P \]
Thus,
\[ C_P = \left( \frac{\partial U} {\partial T} \right)_P + p \left( \frac{\partial V} {\partial T} \right)_P \]
Consider \(U(V,T)\). Its total differential is
\[ dU = \left( \frac{\partial U} {\partial V} \right)_T dV + \left( \frac{\partial U} {\partial T} \right)_V dT \]
Divide by \(dT\) and constrain to constant pressure:
\[ \left( \frac{\partial U} {\partial T} \right)_P = \left( \frac{\partial U} {\partial V} \right)_T \left( \frac{\partial V} {\partial T} \right)_P + C_V \]
Substituting into the previous expression gives
\[ C_P = C_V + \left[ \left( \frac{\partial U} {\partial V} \right)_T + p \right] \left( \frac{\partial V} {\partial T} \right)_P \]
Therefore,
\[ C_P - C_V = \left[ \left( \frac{\partial U} {\partial V} \right)_T + p \right] \left( \frac{\partial V} {\partial T} \right)_P \]
Earlier in the chapter, we derived
\[ \left( \frac{\partial U} {\partial V} \right)_T = T \left( \frac{\partial p} {\partial T} \right)_V - p \]
Substituting into the previous expression:
\[ C_P - C_V = T \left( \frac{\partial p} {\partial T} \right)_V \left( \frac{\partial V} {\partial T} \right)_P \]
The thermal expansivity is defined as
\[ \alpha = \frac{1}{V} \left( \frac{\partial V} {\partial T} \right)_P \]
so
\[ \left( \frac{\partial V} {\partial T} \right)_P = \alpha V \]
Also, from the definition of the isothermal compressibility,
\[ \kappa_T = -\frac{1}{V} \left( \frac{\partial V} {\partial p} \right)_T \]
it can be shown that
\[ \left( \frac{\partial p} {\partial T} \right)_V = \frac{\alpha}{\kappa_T} \]
Substituting both expressions gives
\[ \boxed{ C_P - C_V = \frac{TV\alpha^2} {\kappa_T} } \]
For an ideal gas,
\[ \alpha = \frac{1}{T} \]
\[ \kappa_T = \frac{1}{p} \]
Substituting into the general expression:
\[ C_P - C_V = \frac{TV(1/T)^2} {(1/p)} \]
\[ C_P - C_V = \frac{pV}{T} \]
Using the ideal gas law,
\[ pV = RT \]
gives
\[ \boxed{ C_P - C_V = R } \]
Big picture: The familiar ideal-gas result is actually a special case of a much more general relationship involving measurable properties: thermal expansivity and compressibility.
Evaluate \(C_P-C_V\) for a gas obeying the van der Waals equation of state:
\[ p = \frac{RT}{V_m-b} - \frac{a}{V_m^2} \]
where \(V_m\) is the molar volume.
Start with the general result:
\[ C_P-C_V = T \left( \frac{\partial p}{\partial T} \right)_{V_m} \left( \frac{\partial V_m}{\partial T} \right)_p \]
First evaluate
\[ \left( \frac{\partial p}{\partial T} \right)_{V_m} = \frac{R}{V_m-b} \]
Next, use the cyclic relationship:
\[ \left( \frac{\partial V_m}{\partial T} \right)_p = - \frac{ \left( \frac{\partial p}{\partial T} \right)_{V_m} } { \left( \frac{\partial p}{\partial V_m} \right)_T } \]
From the van der Waals equation,
\[ \left( \frac{\partial p}{\partial V_m} \right)_T = -\frac{RT}{(V_m-b)^2} + \frac{2a}{V_m^3} \]
Therefore,
\[ \left( \frac{\partial V_m}{\partial T} \right)_p = \frac{ \frac{R}{V_m-b} } { \frac{RT}{(V_m-b)^2} - \frac{2a}{V_m^3} } \]
Substituting into the expression for \(C_P-C_V\):
\[ C_P-C_V = T \left( \frac{R}{V_m-b} \right) \left[ \frac{ \frac{R}{V_m-b} } { \frac{RT}{(V_m-b)^2} - \frac{2a}{V_m^3} } \right] \]
\[ C_P-C_V = \frac{ \frac{TR^2}{(V_m-b)^2} } { \frac{RT}{(V_m-b)^2} - \frac{2a}{V_m^3} } \]
A compact final form is
\[ \boxed{ C_P-C_V = \frac{R} { 1 - \frac{2a(V_m-b)^2}{RTV_m^3} } } \]
Check: If the gas becomes ideal, then \(a=0\) and \(b=0\). The expression reduces to
\[ C_P-C_V = R \]
which is the expected ideal-gas result.
Estimate the difference between \(C_P\) and \(C_V\) for copper at \(298\ \mathrm{K}\).
Use the following room-temperature values:
| Quantity | Value |
|---|---|
| Volumetric thermal expansivity, \(\alpha\) | \(4.95\times10^{-5}\ \mathrm{K^{-1}}\) |
| Isothermal compressibility, \(\kappa_T\) | \(7.02\times10^{-12}\ \mathrm{Pa^{-1}}\) |
| Molar volume, \(V_m\) | \(7.09\times10^{-6}\ \mathrm{m^3\,mol^{-1}}\) |
The general expression for the difference between the constant-pressure and constant-volume heat capacities is
\[ C_P-C_V = \frac{TV_m\alpha^2}{\kappa_T} \]
Substituting the values:
\[ C_P-C_V = \frac{ (298\ \mathrm{K}) (7.09\times10^{-6}\ \mathrm{m^3\,mol^{-1}}) (4.95\times10^{-5}\ \mathrm{K^{-1}})^2 } { 7.02\times10^{-12}\ \mathrm{Pa^{-1}} } \]
\[ C_P-C_V = 0.74\ \mathrm{J\,mol^{-1}\,K^{-1}} \]
\[ \boxed{ C_P-C_V = 0.74\ \mathrm{J\,mol^{-1}\,K^{-1}} } \]
Interpretation: Copper has a room-temperature molar heat capacity of approximately \(24.4\ \mathrm{J\,mol^{-1}\,K^{-1}}\). The difference between \(C_P\) and \(C_V\) is therefore small, only about \(3\%\) of \(C_P\).
This makes physical sense. Solids expand very little when heated and are very difficult to compress, so there is only a small difference between heating at constant pressure and heating at constant volume.
Big picture: The relationship \[ C_P-C_V = \frac{TV\alpha^2}{\kappa_T} \] connects heat capacities to measurable material properties. It serves as an excellent example of how Maxwell Relations and thermodynamic equations of state can be used to transform abstract thermodynamic quantities into forms that can be evaluated experimentally.