91Ó°ÊÓ

${\mathrm{κ}}_T={\mathrm{κ}}_S+\frac{{\mathrm{β}}^{2}VT}{C_P}$

The isothermal and isentropic compressibilities can be calculated as follows:

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

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" $dS={\left(\frac{∂S}{∂P}\right)}_{T}dP+{\left(\frac{∂S}{∂T}\right)}_{P}dT....\left(3\right)$

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

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

Substitute from (3) to get:

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

At constant temperature $dT=0$we get:

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

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

$-V{\mathrm{κ}}_T=-V{\mathrm{κ}}_S+{\left(\frac{∂V}{∂S}\right)}_T{\left(\frac{∂S}{∂P}\right)}_T......\left(4\right)$

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

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

The thermal expansion is determined by:

$\mathrm{β}=\frac{1}{V}{\left(\frac{∂V}{∂T}\right)}_P$

Add these two equations together to get the following result:

${\left(\frac{∂S}{∂P}\right)}_T=-\mathrm{β}V$

substitute into (4) to get:

$-V{\mathrm{κ}}_T=-V{\mathrm{κ}}_S-\mathrm{β}V{\left(\frac{∂V}{∂S}\right)}_T$

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

$dV=\mathrm{β}VdT....\left(6\right)$

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:

${\left(\frac{∂V}{∂S}\right)}_P=\frac{\mathrm{β}VT}{C_P}$

substitute into (5) to get:

$-V{\mathrm{κ}}_T=-V{\mathrm{κ}}_S-\frac{{\mathrm{β}}^2{V}^{2}T}{C_P}$

${\mathrm{κ}}_T={\mathrm{κ}}_S+\frac{{\mathrm{β}}^{2}VT}{C_P}.......\left(8\right)$

For an ideal gas we have:

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

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

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

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

To use $PV=NKT$we get:

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

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

$P{V}^{\mathrm{γ}}=K$

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

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

$\begin{array}{c}{\mathrm{κ}}_S=−\frac{1}{V}\frac{\mathrm{∂}}{\mathrm{∂}P}{\left(\frac{K}{P}\right)}^{1/\mathrm{γ}}\\ {\mathrm{κ}}_S=\frac{1}{V\mathrm{γ}}{\left(\frac{K}{P}\right)}^{(1/\mathrm{γ})−1}\frac{K}{P^2}\\ {\mathrm{κ}}_S=\frac{1}{V\mathrm{γ}}{\left(\frac{K}{P}\right)}^{(1/\mathrm{γ})}\frac{P}{K}\frac{K}{P^2}\\ {\mathrm{κ}}_S=\frac{1}{V\mathrm{γ}}V\frac{P}{K}\frac{K}{P^2}\\ {\mathrm{κ}}_S=\frac{1}{P\mathrm{γ}}\\ {\mathrm{κ}}_S=\frac{1}{P}\left(\frac{f}{2+f}\right)\end{array}$

which is equivalent to (12)

Therefore we proved that${\mathrm{κ}}_T={\mathrm{κ}}_S+\frac{{\mathrm{β}}^{2}VT}{C_P}$