# Relations between specific heats

The laws of thermodynamics imply the following relations between the heat capacity at constant volume, $C_{V}$, and the heat capacity at constant pressure, $C_{P}$:

$C_{P} - C_{V}= V T\frac{\alpha^{2}}{\beta_{T}}\,$
$\frac{C_{P}}{C_{V}}=\frac{\beta_{T}}{\beta_{S}}\,$

Here $\alpha$ is the thermal expansion coefficient:

$\alpha=\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P}\,$

$\beta_{T}$ is the isothermal compressibility:

$\beta_{T}=-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T}\,$

and $\beta_{S}$ is the isentropic compressibility:

$\beta_{S}=-\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{S}\,$

The first relation allows one to obtain the heat capacity for solids at constant volume which is not readily measured in terms of quantities that are more easily measured. The second relation allows one to express the isentropic compressibility in terms of the heat capacity ratio.

## Derivation

If an infinitesimal small amount of heat $\delta Q$ is supplied to a system in a quasistatic way then, according to the second law of thermodynamics, the entropy change of the system is given by:

$dS = \frac{\delta Q}{T}\,$

Since

$\delta Q = C dT\,$

where C is the heat capacity, it follows that:

$T dS = C dT\,$

The heat capacity depends on how the external variables of the system are changed when the heat is supplied. If the only external variable of the system is the volume, then we can write:

$dS = \left(\frac{\partial S}{\partial T}\right)_{V}dT+\left(\frac{\partial S}{\partial V}\right)_{T}dV$

From this we see that:

$C_{V}=T\left(\frac{\partial S}{\partial T}\right)_{V}\,$

Expressing dS in terms of dT and dP similarly as above leads to the expression:

$C_{P}=T\left(\frac{\partial S}{\partial T}\right)_{P}\,$

We can find the above expression for $C_{P}-C_{V}$ by expressing dV in terms of dP and dT in the above expression for dS. We have

$dV = \left(\frac{\partial V}{\partial T}\right)_{P}dT+\left(\frac{\partial V}{\partial P}\right)_{T}dP\,$

which gives

$dS = \left[\left(\frac{\partial S}{\partial T}\right)_{V}+ \left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}\right]dT+\left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial P}\right)_{T}dP$

and we see that:

$C_{P} - C_{V} = T\left(\frac{\partial S}{\partial V}\right)_{T}\left(\frac{\partial V}{\partial T}\right)_{P}=VT\alpha\left(\frac{\partial S}{\partial V}\right)_{T}\,$

The partial derivative $\left(\frac{\partial S}{\partial V}\right)_{T}$ can be rewritten in terms of variables that do not involve the entropy using a suitable Maxwell relation. These relations follow from the fundamental thermodynamic relation:

$dE = T dS - P dV\,$

It follows from this that the differential of the Helmholtz free energy $F = E - T S$ is:

$dF = -S dT - P dV\,$

This means that

$-S = \left(\frac{\partial F}{\partial T}\right)_{V}\,$

and

$-P = \left(\frac{\partial F}{\partial V}\right)_{T}\,$

The symmetry of second derivatives of F w.r.t. T and V then implies

$\left(\frac{\partial S}{\partial V}\right)_{T} =\left(\frac{\partial P}{\partial T}\right)_{V}\,$

allowing us to write:

$C_{P} - C_{V} = VT\alpha\left(\frac{\partial P}{\partial T}\right)_{V}\,$

The r.h.s. contains a derivative at constant volume, which can be difficult to measure. We can rewrite it as follows. In general, we have:

$dV= \left(\frac{\partial V}{\partial P}\right)_{T}dP+\left(\frac{\partial V}{\partial T}\right)_{P}dT\,$

Since the partial derivative $\left(\frac{\partial P}{\partial T}\right)_{V}$ is just the ratio of dP and dT for dV = 0, we can obtain this by putting dV = 0 in the above equation and solving for this ratio. We then obtain:

$\left(\frac{\partial P}{\partial T}\right)_{V}=-\frac{\left(\frac{\partial V}{\partial T}\right)_{P}}{\left(\frac{\partial V}{\partial P}\right)_{T}}=\frac{\alpha}{\beta_{T}}\,$

which yields the expression:

$C_{P} - C_{V}= V T\frac{\alpha^{2}}{\beta_{T}}\,$

The expression for the ratio of the heat capacities can be obtained as follows. We have:

$\frac{C_{P}}{C_{V}} = \frac{\left(\frac{\partial S}{\partial T}\right)_{P}}{\left(\frac{\partial S}{\partial T}\right)_{V}}\,$

The partial derivative in the numerator can be expressed as a ratio of partial derivatives of the pressure w.r.t. temperature and entropy. If in the relation

$dP = \left(\frac{\partial P}{\partial S}_{T}\right)dS+\left(\frac{\partial P}{\partial T}\right)_{S}dT\,$

we put $dP = 0$ and solve for the ratio $\frac{dS}{dT}$ we obtain $\left(\frac{\partial S}{\partial T}\right)_{P}$. Doing so gives:

$\left(\frac{\partial S}{\partial T}\right)_{P}=-\frac{\left(\frac{\partial P}{\partial T}\right)_{S}}{\left(\frac{\partial P}{\partial S}\right)_{T}}\,$

We can similarly rewrite the partial derivative $\left(\frac{\partial S}{\partial T}\right)_{V}$ by expressing dV in terms of dS and and dT, puting dV equal to zero and solving for the ratio $\frac{dS}{dT}$. If we substitute that expression in the heat capacity ratio expressed as the ratio of the partial derivatives of the entropy above, we obtain:

$\frac{C_{P}}{C_{V}}=\frac{\left(\frac{\partial P}{\partial T}\right)_{S}}{\left(\frac{\partial P}{\partial S}\right)_{T}} \frac{\left(\frac{\partial V}{\partial S}\right)_{T}}{\left(\frac{\partial V}{\partial T}\right)_{S}}\,$

Taking together the two derivatives at constant S:

$\frac{\left(\frac{\partial P}{\partial T}\right)_{S}}{\left(\frac{\partial V}{\partial T}\right)_{S}} = \left(\frac{\partial P}{\partial T}\right)_{S}\left(\frac{\partial T}{\partial V}\right)_{S}=\left(\frac{\partial P}{\partial V}\right)_{S}\,$

Taking together the two derivatives at constant T:

$\frac{\left(\frac{\partial V}{\partial S}\right)_{T}}{\left(\frac{\partial P}{\partial S}\right)_{T}}=\left(\frac{\partial V}{\partial S}\right)_{T}\left(\frac{\partial S}{\partial P}\right)_{T}=\left(\frac{\partial V}{\partial P}\right)_{T}\,$

We can thus write

$\frac{C_{P}}{C_{V}}=\left(\frac{\partial P}{\partial V}\right)_{S}\left(\frac{\partial V}{\partial P}\right)_{T}=\frac{\beta_{T}}{\beta_{S}}\,$