Chapter 3.3, Problem 46E

To determine

To calculate: The value of $\frac{dP}{dV}$ .

Answer to Problem 46E

The required value of $\frac{dP}{dV}$ is $-\frac{nRT}{{\left(V-nb\right)}^2}+\frac{2a{n}^2}{V^3}$ .

Explanation of Solution

Given information:

The equation of P: $P=\frac{nRT}{V-nb}-\frac{a{n}^2}{V^2}$

Formula used:

Power rule of differentiation:

  $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$

Calculation:

The given equation of P is $P=\frac{nRT}{V-nb}-\frac{a{n}^2}{V^2}$ and $a$ , $b$ , $n$ and $R$ are constants.

Since, the terms $nRT$ , $nb$ , $n{b}^2$ are constants;

Consider the constants be $C_1=nRT$ , $C_2=nb$ and $C_3=a{n}^2$ .

Rewrite the given equation as:

  $\begin{array}{l}P=\frac{C_1}{V-C_2}-\frac{C_3}{V^2}\\ P=C_1{\left(V-C_2\right)}^{-1}-C_3{V}^{-2}\end{array}$

Differentiate the above expression using the power rule $\frac{d}{dx}\left(x^n\right)=n{x}^{n-1}$ .

  $\begin{array}{c}\frac{dP}{dV}=\frac{d}{dV}\left(C_1{\left(V-C_2\right)}^{-1}\right)-\frac{d}{dV}\left(C_3{V}^{-2}\right)\\ \frac{dP}{dV}=-C_1{\left(V-C_2\right)}^{-2}+2C_3{V}^{-3}\end{array}$

Write the above expression as:

  $\begin{array}{l}\frac{dP}{dV}=-C_1{\left(V-C_2\right)}^{-2}+2C_3{V}^{-3}\\ \frac{dP}{dV}=-\frac{C_1}{{\left(V-C_2\right)}^2}+\frac{2C_3}{V^3}\\ \frac{dP}{dV}=-\frac{nRT}{{\left(V-nb\right)}^2}+\frac{2a{n}^2}{V^3}\end{array}$

Hence, the required value of $\frac{dP}{dV}$ is $-\frac{nRT}{{\left(V-nb\right)}^2}+\frac{2a{n}^2}{V^3}$ .