Question

A formula analogous to that for $C_P-C_V$relates the isothermal and isentropic compressibilities of a material:

${\kappa }_T={\kappa }_S+\frac{TV{\beta }^2}{C_P}.$

(Here ${\kappa }_S=-(1/V)(\partial V/\partial P{)}_S$is the reciprocal of the adiabatic bulk modulus considered in Problem 1.39.) Derive this formula. Also check that it is true for an ideal gas.

Short answer

${\kappa }_T={\kappa }_S+\frac{{\beta }^{2}VT}{C_P}$

Step-by-step solution

To  prove

${\kappa }_T={\kappa }_S+\frac{{\beta }^{2}VT}{C_P}$

Explanation

The isothermal and isentropic compressibilities can be calculated as follows:

$$\begin{gathered} {\kappa }_T=-\frac{1}{V}{\left(\frac{\partial V}{\partial P}\right)}_T.....\left(1\right) \\ {\kappa }_S=-\frac{1}{V}{\left(\frac{\partial V}{\partial P}\right)}_S.....\left(2\right) \end{gathered}$$

If S is a function of P and T, then $V=V(P,T)$is obtained from the definition of the derivative:

role="math" localid="1648438079130" $dS={\left(\frac{\partial S}{\partial P}\right)}_{T}dP+{\left(\frac{\partial S}{\partial T}\right)}_{P}dT....\left(3\right)$

If V is a function of P and S, then $V=V(P,S)$is obtained from the definition of the derivative:

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

Substitute from (3) to get:

$$\begin{gathered} dV={\left(\frac{\partial S}{\partial P}\right)}_{T}dP+{\left(\frac{\partial V}{\partial S}\right)}_T\left[{\left(\frac{\partial S}{\partial P}\right)}_{T}dP+{\left(\frac{\partial S}{\partial T}\right)}_{P}dT\right] \\ dV=\left[{\left(\frac{\partial S}{\partial P}\right)}_T+{\left(\frac{\partial V}{\partial S}\right)}_T{\left(\frac{\partial S}{\partial P}\right)}_T\right]dP+{\left(\frac{\partial S}{\partial T}\right)}_{P}dT \end{gathered}$$

At constant temperature $dT=0$we get:

$$\begin{gathered} (dV{)}_T=\left[{\left(\frac{\partial S}{\partial P}\right)}_T+{\left(\frac{\partial V}{\partial S}\right)}_T{\left(\frac{\partial S}{\partial P}\right)}_T\right]dP \\ {\left(\frac{\partial V}{\partial P}\right)}_T={\left(\frac{\partial S}{\partial P}\right)}_T+{\left(\frac{\partial V}{\partial S}\right)}_T{\left(\frac{\partial S}{\partial P}\right)}_T \end{gathered}$$

Further continuation for the proof

substitute from (1) and (2) to get:

$-V{\kappa }_T=-V{\kappa }_S+{\left(\frac{\partial V}{\partial S}\right)}_T{\left(\frac{\partial S}{\partial P}\right)}_T......\left(4\right)$

From the Gibbs energy and the Maxwell relation, we can conclude:

${\left(\frac{\partial S}{\partial P}\right)}_T={\left(\frac{\partial V}{\partial T}\right)}_P$

The thermal expansion is determined by:

$\beta =\frac{1}{V}{\left(\frac{\partial V}{\partial T}\right)}_P$

Add these two equations together to get the following result:

${\left(\frac{\partial S}{\partial P}\right)}_T=-\beta V$

substitute into (4) to get:

$-V{\kappa }_T=-V{\kappa }_S-\beta V{\left(\frac{\partial V}{\partial S}\right)}_T$

Further continuation for the proof 

At constant pressure, the volume changes owing to a temperature change, and it is given by:

$dV=\beta VdT....\left(6\right)$

and the entropy change is given by:

$dS=\frac{dQ}{T}=C_P\frac{dT}{T}....\left(7\right)$

divide (6) over (7) to get:

${\left(\frac{\partial V}{\partial S}\right)}_P=\frac{\beta VT}{C_P}$

substitute into (5) to get:

$-V{\kappa }_T=-V{\kappa }_S-\frac{{\beta }^2{V}^{2}T}{C_P}$

${\kappa }_T={\kappa }_S+\frac{{\beta }^{2}VT}{C_P}.......\left(8\right)$

Further continuation for the proof 

For an ideal gas we have:

$\beta =\frac{1}{V}{\left(\frac{\partial V}{\partial T}\right)}_P=\frac{1}{V}\frac{\partial }{\partial T}{\left(\frac{NkT}{P}\right)}_P=\frac{Nk}{PV}=\frac{1}{T}......\left(9\right)$

${\kappa }_T=-\frac{1}{V}{\left(\frac{\partial V}{\partial P}\right)}_T=\frac{1}{V}\frac{\partial }{\partial P}{\left(\frac{NkT}{P}\right)}_T=\frac{NkT}{P^{2}V}....\left(10\right)$

substitute from (9), (10) and (11) into (8) to get:

$$\begin{gathered} {\kappa }_S={\kappa }_T-\frac{{\beta }^{2}VT}{C_P} \\ {\kappa }_S=\frac{NkT}{P^{2}V}-{\beta }^{2}VT{\left[Nk\left(1+\frac{f}{2}\right)\right]}^{-1} \\ {\kappa }_S=\frac{NkT}{P^{2}V}-{\beta }^{2}VT{\left[Nk\left(\frac{2+f}{2}\right)\right]}^{-1} \\ {\kappa }_S=\frac{NkT}{P^{2}V}-\frac{V}{NkT}\left(\frac{2}{2+f}\right) \end{gathered}$$

To use $PV=NKT$we get:

$$\begin{gathered} {\kappa }_S=\frac{PV}{P^{2}V}-\frac{V}{PV}\left(\frac{2}{2+f}\right) \\ {\kappa }_S=\frac{1}{P}\left(\frac{f}{2+f}\right)......\left(12\right) \end{gathered}$$

Further continuation for the proof 

Now we must verify this relationship: in an ideal gas, in an isentropic (adiabatic) process, we have:

$P{V}^{\gamma }=K$

$\text{where}\gamma =(f+2)/f\text{and}K\text{is constant, write for}V\text{to get:}$

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

$\begin{array}{c}{\kappa }_S=-\frac{1}{V}\frac{\mathrm{\partial }}{\mathrm{\partial }P}{\left(\frac{K}{P}\right)}^{1/\gamma }\\ {\kappa }_S=\frac{1}{V\gamma }{\left(\frac{K}{P}\right)}^{(1/\gamma )-1}\frac{K}{P^2}\\ {\kappa }_S=\frac{1}{V\gamma }{\left(\frac{K}{P}\right)}^{(1/\gamma )}\frac{P}{K}\frac{K}{P^2}\\ {\kappa }_S=\frac{1}{V\gamma }V\frac{P}{K}\frac{K}{P^2}\\ {\kappa }_S=\frac{1}{P\gamma }\\ {\kappa }_S=\frac{1}{P}\left(\frac{f}{2+f}\right)\end{array}$

which is equivalent to (12)

Therefore we proved that${\kappa }_T={\kappa }_S+\frac{{\beta }^{2}VT}{C_P}$