Question

Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are ${10}^5$ $N/m^2$ and $6$ litres respectively. The final volume of the gas is $2$ litres. Molar specific heat of the gas at constant volume is $3R/2$.

• A. $872$J.
• B. $972$J.
• C. $772$J.
• D. $672$J.

Answer 1

Answer: (B)

$972$J.

Given : Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are 105105 N/m2N/m2 and 66 litres respectively. The final volume of the gas is 22 litres. Molar specific heat of the gas at constant volume is 3R/23R/2.

Solution :

Work done in an adiabatic process is

$W=\frac{nR\Delta T}{\gamma -1}$ = $\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}$

$P_1={10}^{5}N/m^2$ , $V_1=6l$ , $V_2=2l$

$P_2=P_1(\frac{V_1}{V_2}{)}^{\gamma }$

$\gamma =1.67$

On solving we get $P_2=6.26\times {10}^{5}N/m^2$

$W=\frac{6.26\times {10}^5\times 2-{10}^5\times 6}{1.67-1}$

On solving we get W=973.1 J $\approx$ 972 J

The correct Opt = B