Question

Calculate the magnitude of work done when 4 moles of a perfect gas are compressed adiabatically. The initial pressure and volume of the gas are ${10}^5{\text{N/m}}^2$ and 8 litres respectively. The final volume of the gas is 4 litres. Molar specific heat of the gas at constant volume is $3R/2$.

• A. $800\text{J}$
• B. $705\text{J}$
• C. $815\text{J}$
• D. $725\text{J}$

Answer 1

The correct option is B $705\text{J}$

Given that
$P_1={10}^5{\text{N/m}}^2;V_1=8\text{litres}=8\times {10}^{-3}{\text{m}}^3$;
$V_2=4\text{litres}=4\times {10}^{-3}{\text{m}}^3$
$C_v=\frac{3}{2}R$
$\therefore C_p=\frac{5}{2}R[\because C_p-C_v=R]$
$\&\gamma =\frac{C_p}{C_v}=\frac{5}{3}=1.67$

For adiabatic process,
$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$
$\Rightarrow P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }={10}^5{\left(\frac{8}{4}\right)}^{1.67}$
$={10}^5(2{)}^{1.67}=3.18\times {10}^5{\text{N/m}}^2$

Work done in an adiabatic process,
$W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=\frac{({10}^5\times 8\times {10}^{-3}-3.18\times {10}^5\times 4\times {10}^{-3})}{1.67-1}$
$\Rightarrow W=\frac{100[8-12.72]}{0.67}=-704.47\approx -705\text{J}$
[Negative sign indicates compression]