Question

A gas expands from an initial state where the pressure is 340 kPa and volume is 0.0425 m${}^3$ to a final pressure of 136 kPa. The relationship between the pressure and volume is $P{V}^2$ = constant. The work done for the process is________kJ.

• A. $3.21,\text{kJ}$
• B. $7.5,\text{kJ}$
• C. $5.31,\text{kJ}$
• D. $24,\text{kJ}$

Answer 1

The correct option is C $5.31,\text{kJ}$

Given,
$P_1=340kPa:V_1=0.0425m^3;P_2=136kPa$

$Fromtherelation,P{V}^2=C$

$\therefore$ $P_1{V}_1^2=P_2{V}_2^2$

$V_2=\sqrt{\frac{P_1{V}_1^2}{P_2}}=\sqrt{\frac{340\times {0.0425}^2}{136}}$

$=0.0672m^3$



$Now,W_{1-2}={\int }_1^{2}PdV$

$W_{1-2}={\int }_{V_1}^{V_2}\frac{C}{V^2}dV=C{\int }_{V_1}^{V_2}{V}^{-2}dV=-C{\left[\frac{1}{V}\right]}_{V_1}^{V_2}$

=$C[\frac{1}{V_1}-\frac{1}{V_2}]=P_1{V}_1^2[\frac{1}{V_1}-\frac{1}{V_2}]$

$=340\times {0.0425}^2[\frac{1}{0.0425}-\frac{1}{0.0672}]$

$\therefore$ $W_{1-1}=5.31kJ$