# Isothermal and isentropic compressibilities

Reference: Daniel V. Schroeder, An Introduction to Thermal Physics, (Addison-Wesley, 2000) – Problem 5.16.

An expression similar to that relating the heat capacities can be derived to relate the isothermal and isentropic compressibilities ${\kappa_{T}}$ and ${\kappa_{S}}$, defined as

 $\displaystyle \kappa_{T}$ $\displaystyle \equiv$ $\displaystyle -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T}\ \ \ \ \ (1)$ $\displaystyle \kappa_{S}$ $\displaystyle \equiv$ $\displaystyle -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{S} \ \ \ \ \ (2)$

These quantities measure the fractional change in volume of a substance in response to a change in pressure. To obtain the relation between them, we use a method similar to that for heat capacities ${C_{V}}$ and ${C_{P}}$.

If we write ${S=S\left(P,T\right)}$ then

$\displaystyle dS=\left(\frac{\partial S}{\partial P}\right)_{T}dP+\left(\frac{\partial S}{\partial T}\right)_{P}dT \ \ \ \ \ (3)$

Also, starting with ${V=V\left(P,S\right)}$ we have

$\displaystyle dV=\left(\frac{\partial V}{\partial P}\right)_{S}dP+\left(\frac{\partial V}{\partial S}\right)_{P}dS \ \ \ \ \ (4)$

Substituting 3 into 4 we get

$\displaystyle dV=\left[\left(\frac{\partial V}{\partial S}\right)_{P}\left(\frac{\partial S}{\partial P}\right)_{T}+\left(\frac{\partial V}{\partial P}\right)_{S}\right]dP+\left(\frac{\partial V}{\partial T}\right)_{P}dT \ \ \ \ \ (5)$

At constant temperature ${dT=0}$ and we get

 $\displaystyle \left(\frac{\partial V}{\partial P}\right)_{T}$ $\displaystyle =$ $\displaystyle \left(\frac{\partial V}{\partial S}\right)_{P}\left(\frac{\partial S}{\partial P}\right)_{T}+\left(\frac{\partial V}{\partial P}\right)_{S}\ \ \ \ \ (6)$ $\displaystyle -V\kappa_{T}$ $\displaystyle =$ $\displaystyle \left(\frac{\partial V}{\partial S}\right)_{P}\left(\frac{\partial S}{\partial P}\right)_{T}-V\kappa_{S} \ \ \ \ \ (7)$

From the Maxwell relation from the Gibbs energy

$\displaystyle \left(\frac{\partial S}{\partial P}\right)_{T}=-\left(\frac{\partial V}{\partial T}\right)_{P} \ \ \ \ \ (8)$

Also, from the definition of the thermal expansion coefficient ${\beta}$

$\displaystyle \beta\equiv\frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P} \ \ \ \ \ (9)$

Combining these last two equations gives

$\displaystyle -V\kappa_{T}=-\beta V\left(\frac{\partial V}{\partial S}\right)_{P}-V\kappa_{S} \ \ \ \ \ (10)$

To get rid of the last partial derivative, we observe that the volume change ${dV}$ due to a temperature change ${dT}$ at constant pressure is

$\displaystyle dV=\beta V\;dV \ \ \ \ \ (11)$

The entropy change due to an influx of heat ${dQ}$ at constant pressure at temperature ${T}$ is

 $\displaystyle dS$ $\displaystyle =$ $\displaystyle \frac{dQ}{T}\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle C_{P}\frac{dT}{T} \ \ \ \ \ (13)$

Dividing these two relations gives

$\displaystyle \left(\frac{\partial V}{\partial S}\right)_{P}=\frac{TV\beta}{C_{P}} \ \ \ \ \ (14)$

Inserting this into 10 and cancelling off a factor of ${-V}$ gives the final result

$\displaystyle \kappa_{T}=\kappa_{S}+\frac{TV\beta^{2}}{C_{P}} \ \ \ \ \ (15)$

For an ideal gas, we can use this equation to work out ${\kappa_{S}}$:

 $\displaystyle \beta$ $\displaystyle =$ $\displaystyle \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_{P}=\frac{Nk}{PV}=\frac{1}{T}\ \ \ \ \ (16)$ $\displaystyle \kappa_{T}$ $\displaystyle =$ $\displaystyle -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{T}=\frac{NkT}{P^{2}V}=\frac{1}{P}\ \ \ \ \ (17)$ $\displaystyle C_{P}$ $\displaystyle =$ $\displaystyle C_{V}+Nk\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Nk\left(1+\frac{f}{2}\right)\ \ \ \ \ (19)$ $\displaystyle \kappa_{S}$ $\displaystyle =$ $\displaystyle \frac{1}{P}-\frac{V}{NkT\left(1+\frac{f}{2}\right)}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{P}\frac{f}{\left(f+2\right)} \ \ \ \ \ (21)$

where in the third line, we’ve used Schroeder’s equation 1.48, and ${f}$ is the number of degrees of freedom of each gas molecule.

To check this, recall that for an isentropic (adiabatic) process in an ideal gas

 $\displaystyle PV^{\gamma}$ $\displaystyle =$ $\displaystyle K\ \ \ \ \ (22)$ $\displaystyle V$ $\displaystyle =$ $\displaystyle \left(\frac{K}{P}\right)^{1/\gamma} \ \ \ \ \ (23)$

where ${\gamma=\left(f+2\right)/f}$ and ${K}$ is a constant. So

 $\displaystyle \kappa_{S}$ $\displaystyle =$ $\displaystyle -\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_{S}\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left(\frac{P}{K}\right)^{1/\gamma}\left(-\frac{1}{\gamma}\right)\left(\frac{K}{P}\right)^{1/\gamma}\frac{1}{P}\ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{P\gamma}=\frac{1}{P}\frac{f}{\left(f+2\right)} \ \ \ \ \ (26)$

which is the same as 21, so equation 15 checks out for an ideal gas.