Q139 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Physics
Work done on heating one mole of monoatomic gas adiabatically through $20^{\circ} \mathrm{C}$ is $W$. Then, the work done on heating 6 moles of rigid diatomic gas through the same change in temperature
- A$9 W$
- Correct answerB$10 W$
- C$12 W$
- D$8 W$
Solution
Work done in an adiabatic expansion is
$\Delta W=\frac{\mu R \Delta T}{\gamma-1}$
For 1 mole of monoatomic gas,
$\Delta T=20^{\circ} \mathrm{C}, \gamma=\frac{5}{3}$
For 6 moles of rigid diatomic gas, $\begin{aligned} \Delta T & =20^{\circ} \mathrm{C}, \gamma=\frac{7}{5} \\ \Delta W^{\prime} & =\frac{6 \times R \times 20}{\left(\frac{7}{5}-1\right)}=\frac{5 \times 6 \times R \times 20}{2} \end{aligned}$ $=10 \times 30 R=10 \mathrm{~W} \quad$ [from Eq. (i)]
For 6 moles of rigid diatomic gas, $\begin{aligned} \Delta T & =20^{\circ} \mathrm{C}, \gamma=\frac{7}{5} \\ \Delta W^{\prime} & =\frac{6 \times R \times 20}{\left(\frac{7}{5}-1\right)}=\frac{5 \times 6 \times R \times 20}{2} \end{aligned}$ $=10 \times 30 R=10 \mathrm{~W} \quad$ [from Eq. (i)]
Hence, the correct answer is option: 'B'.