A total of $48 \mathrm{~J}$ heat is given to one mole of helium kept in a cylinder. The temperature of…

A total of $48 \mathrm{~J}$ heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $2^{\circ} \mathrm{C}$. The work done by the gas is: Given, $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.
  1. $24.9 \mathrm{~J}$
  2. $72.9 \mathrm{~J}$
  3. $48 \mathrm{~J}$
  4. $23.1 \mathrm{~J}$

Solution

$1^{\text {st }}$ law of thermodynamics $\begin{aligned} & \Delta \mathrm{Q}=\Delta \mathrm{U}+\mathrm{W} \\ & \Rightarrow+48=\mathrm{nC}_{\mathrm{v}} \Delta \mathrm{T}+\mathrm{W} \\ & \Rightarrow 48=(1)\left(\frac{3 \mathrm{R}}{2}\right)(2)+\mathrm{W} \\ & \Rightarrow \mathrm{W}=48-3 \times \mathrm{R} \\ & \Rightarrow \mathrm{W}=48-3 \times(8.3) \\ & \Rightarrow \mathrm{W}=23.1 \text { Joule } \end{aligned}$

Asked in: JEE Main 2024 (06 Apr Shift 2)

Practice more Thermodynamics questions on Aicharya