A monoatomic gas of $n$-moles is heated from temperature $T_1$ to $T_2$ under two different conditions. (i)…

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
  1. More when heated at constant volume
  2. More when heated at constant pressure
  3. Same in both the cases
  4. Zero 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, $\begin{aligned} & (\Delta \mathrm{U})_2=\frac{\mathrm{nfR} \Delta \mathrm{~T}}{2}=\frac{3 \mathrm{nR}\left(\mathrm{~T}_2-\mathrm{T}_1\right)}{2} \\ & \therefore \quad(\Delta \mathrm{U})_1=(\Delta \mathrm{U})_2 \end{aligned}$

Asked in: AP EAMCET 2024 (21 May Shift 1)

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