Question

Derive an expression for the isothermal compressibility of a substance whose equation of state is $$P=\frac{R T}{v-b}-\frac{a}{v(v+b) T^{1 / 2}}$$.where \(a\) and \(b\) are empirical constants.

Short answer

Answer: The isothermal compressibility, denoted as \(\kappa_T\), for the given substance is $$\kappa_T = -\frac{1}{v}\cdot\frac{\frac{R T}{(P+\frac{a}{v(v+b) T^{1 / 2}})^2}}{1-\frac{a}{T^{1/2}(v+b)^4}(1+v)}$$

Step-by-step solution

Rewrite the given equation for volume

We need to find the partial derivative of volume with respect to pressure. So, let's rewrite the given equation for volume $$v$$ as a function of pressure and temperature: $$v=\frac{R T}{P+\frac{a}{v(v+b) T^{1 / 2}}}-b$$

Implicit differentiation

Now, let's use implicit differentiation with respect to pressure \(P\), keeping the temperature constant. Remember that we need to find \(\frac{\partial v}{\partial P}\). In the implicit differentiation, we treat \(v\) as a function of \(P\) (i.e., \(v(P)\)). After differentiation, we get: $$\frac{\partial v}{\partial P} = \frac{R T}{(P+\frac{a}{v(v+b) T^{1 / 2}})^2} - \frac{a}{T^{1/2}(v+b)^2}\cdot \frac{\partial}{\partial P}\left(\frac{1}{v(v+b)}\right)$$ To find \(\frac{\partial}{\partial P}\left(\frac{1}{v(v+b)}\right)\), we use product rule and chain rule: $$\frac{\partial}{\partial P}\left(\frac{1}{v(v+b)}\right)= -\frac{1}{(v+b)^2}\cdot \frac{\partial v}{\partial P} -\frac{v}{(v+b)^2}\cdot \frac{\partial v}{\partial P}$$

Solve for the partial derivative

Substitute the equation we found in Step 2 back into the main equation: $$\frac{\partial v}{\partial P} = \frac{R T}{(P+\frac{a}{v(v+b) T^{1 / 2}})^2} + \frac{a}{T^{1/2}(v+b)^2}\cdot \left[\frac{1}{(v+b)^2}+\frac{v}{(v+b)^2}\right]\cdot \frac{\partial v}{\partial P}$$ Now, we need to solve for $$\frac{\partial v}{\partial P}$$ : $$\frac{\partial v}{\partial P}\left[1-\frac{a}{T^{1/2}(v+b)^4}(1+v)\right] = \frac{R T}{(P+\frac{a}{v(v+b) T^{1 / 2}})^2}$$ $$\frac{\partial v}{\partial P}= \frac{\frac{R T}{(P+\frac{a}{v(v+b) T^{1 / 2}})^2}}{1-\frac{a}{T^{1/2}(v+b)^4}(1+v)}$$

Substitute into the isothermal compressibility formula

Now, we substitute the expression for $$\frac{\partial v}{\partial P}$$ into the isothermal compressibility formula: $$\kappa_T=-\frac{1}{V}\cdot \frac{\partial V}{\partial P}\Bigg|_{T=\text{const.}} = -\frac{1}{v}\cdot\frac{\frac{R T}{(P+\frac{a}{v(v+b) T^{1 / 2}})^2}}{1-\frac{a}{T^{1/2}(v+b)^4}(1+v)}$$ Therefore, the isothermal compressibility for the given substance is $$\kappa_T = -\frac{1}{v}\cdot\frac{\frac{R T}{(P+\frac{a}{v(v+b) T^{1 / 2}})^2}}{1-\frac{a}{T^{1/2}(v+b)^4}(1+v)}$$