2 moles of a monatomic gas requires heat energy $Q$ to be heated from $30^{\circ} \mathrm{C}$ to $40^{\circ}…
2 moles of a monatomic gas requires heat energy $Q$ to be heated from $30^{\circ} \mathrm{C}$ to $40^{\circ} \mathrm{C}$ at constant volume. The heat energy required to raise the temperature of 4 moles of a diatomic gas from $28^{\circ} \mathrm{C}$ to $33^{\circ} \mathrm{C}$ at constant volume is
$2 \mathrm{Q}$
$\frac{7 Q}{2}$
$\frac{4 Q}{3}$
$\frac{5 Q}{3}$
Solution
Heat energy, $\mathrm{Q}$ at constant volume, $\mathrm{Q}=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}$ for monatomic gas $\mathrm{C}_{\mathrm{v}}=\frac{3}{2} \mathrm{R}$ and for diatomic gas, $\mathrm{C}_{\mathrm{v}}=\frac{5}{2} \mathrm{R}$
$\begin{aligned} & \therefore \mathrm{Q}=2 \times \frac{3}{2} \mathrm{R} \times 10 \text { and } \mathrm{Q}^{\prime}=4 \times \frac{5}{2} \mathrm{R} \times 5 \\ & \therefore \frac{\mathrm{Q}^{\prime}}{\mathrm{Q}}=\frac{4 \times \frac{5}{2} \mathrm{R} \times 5}{2 \times \frac{3}{2} \mathrm{R} \times 10}=\frac{5}{3} \therefore \mathrm{Q}^{\prime}=\frac{5}{3} \mathrm{Q}\end{aligned}$