Question

In a flexible balloon, 2 moles of $S{O}_2$ having initial volume of 1kl at a temperature of ${27}^{o}C$ is filled. ($S{O}_2$ is a linear triatomic gas). The gas is first expanded to thrice its initial volume isobarically and then expanded adiabatically so as to attain its initial temperature. Assuming gas is ideal, $\gamma =\frac{4}{3}$ and $R=\frac{25}{3}Jmo{l}^{-1}{K}^{-1}$. Work done by the gas in the whole process is:

• A. $2kJ$
• B. $3.2kJ$
• C. $40kJ$
• D. $15kJ$

Answer 1

Answer: (D)

$15kJ$

Work Done:
1. Isobaric process:
$\frac{V_1}{T_1}=\frac{V_2}{T_2}$
$\therefore T_2=\frac{V_2}{V_1}{T}_1=3\times (27+273)=3\times 300=900$
Work done $W_{isobaric}=nR\Delta T=2\times 8.31\times (900-300)=9972J$
2. Adiabatic process:
$\gamma =\frac{4}{3}$ for polyatomic gases.
$T{V}^{\gamma -1}=constant$
$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$

$T_2=T_1\times (\frac{V_1}{V_2}{)}^{\gamma -1}$

$T_2=300\times (\frac{1}{3}{)}^{4/3-1}$ $T_2=300\times 0.691=207.3K$ Work done in adiabatic process
$W_{adiabatic}=\frac{nR(T_1-T_2)}{\gamma -1}=\frac{2\times 8.31\times (300-207)}{\frac{4}{3}-1}=-2\times 8.31\times 93\times 3=4637J$

Total work done $W_{total}=9972+4637=14609J=14.6KJ\approx 15KJ$