Chapter 8.5, Problem 12E

a.

To determine

To show: Work done $={\displaystyle \underset{\left(p_{1,V_1}\right)}{\overset{p_{2,V_2}}{\int }}p}dV$

Explanation of Solution

Given information:

The gas compressing from state $\left(p_1,V_1\right)$ to $\left(p_2,V_2\right)$

Concept used :

We know that work done for gas = pressure times volume.

We know that work done for gas = pressure times volume.

  $W={\displaystyle \underset{\left(p_{1,V_1}\right)}{\overset{p_{2,V_2}}{\int }}dW}={\displaystyle \underset{\left(p_{1,V_1}\right)}{\overset{p_{2,V_2}}{\int }}pdV}\left[\mathrm{Pr}oved\right]$

b.

To determine

To find: The work done in the given condition.

Answer to Problem 12E

The required work done is $-37968.75$ in.lb

Explanation of Solution

Given information:

  $V_1=243{\text{ in}}^3$ , $V_2=32{\text{ in}}^3$ if $p_1=50{\text{ lb/in}}^3$

Concept Used:

Work done $W={\displaystyle \underset{243}{\overset{32}{\int }}p}dV$

Calculation:

At $p_1=50{\text{ lb/in}}^3$, $V_1=243{\text{ in}}^3$

Therefore, $p_1{V}_1{}^{1.4}=c\Rightarrow c=109350$

  $p{V}^{1.4}=109350\Rightarrow p=\frac{109350}{V^{1.4}}$

  $\begin{array}{c}W={\displaystyle \underset{243}{\overset{32}{\int }}\left(\frac{109350}{V^{1.4}}\right)}dV\\ =109350\times {\left[\frac{V^{-1.4+1}}{-1.4+1}\right]}_{243}^{32}\\ =-37968.75\end{array}$

The required work done is $-37968.75$ in.lb