Q137 / 160 · AP EAMCET 2024 (21 May Shift 1) · Physics
A monoatomic gas of $n$-moles is heated from temperature $T_1$ to $T_2$ under two different conditions.
(i) at constant volume and
(ii) at constant pressure.
The change in internal energy of the gas is
The change in internal energy of the gas is
- AMore when heated at constant volume
- BMore when heated at constant pressure
- Correct answerCSame in both the cases
- DZero in both the cases
Solution
For mono atomic gas, $\mathrm{f}=3$
At constant volume, $(\Delta \mathrm{U})_1=\frac{\mathrm{nfR} \Delta \mathrm{~T}}{2}=\frac{3 \mathrm{nR}\left(\mathrm{~T}_2-\mathrm{T}_1\right)}{2}$
At constant pressure, $\begin{aligned} & (\Delta \mathrm{U})_2=\frac{\mathrm{nfR} \Delta \mathrm{~T}}{2}=\frac{3 \mathrm{nR}\left(\mathrm{~T}_2-\mathrm{T}_1\right)}{2} \\ & \therefore \quad(\Delta \mathrm{U})_1=(\Delta \mathrm{U})_2 \end{aligned}$
At constant volume, $(\Delta \mathrm{U})_1=\frac{\mathrm{nfR} \Delta \mathrm{~T}}{2}=\frac{3 \mathrm{nR}\left(\mathrm{~T}_2-\mathrm{T}_1\right)}{2}$
At constant pressure, $\begin{aligned} & (\Delta \mathrm{U})_2=\frac{\mathrm{nfR} \Delta \mathrm{~T}}{2}=\frac{3 \mathrm{nR}\left(\mathrm{~T}_2-\mathrm{T}_1\right)}{2} \\ & \therefore \quad(\Delta \mathrm{U})_1=(\Delta \mathrm{U})_2 \end{aligned}$
Hence, the correct answer is option: 'C'.