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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2024 · 6 April · Evening Shift · Physics · Thermodynamics

A total of 48 ~J heat is given to one mole of helium kept in a...

A total of 48 J48 \mathrm{~J} heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2∘C2^{\circ} \mathrm{C}. The work done by the gas is: Given, R=8.3 J K−1 mol−1\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}.

Options

  1. A24.9 J24.9 \mathrm{~J}
  2. B72.9 J72.9 \mathrm{~J}
  3. C48 J48 \mathrm{~J}
  4. D23.1 J23.1 \mathrm{~J}

Correct answer

D. 23.1 J23.1 \mathrm{~J}

Step-by-step solution

1st 1^{\text{st }} law of thermodynamics ΔQ=ΔU+W⇒+48=nCvΔT+W⇒48=(1)(3R2)(2)+W⇒W=48−3×R⇒W=48−3×(8.3)⇒W=23.1 Joule \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}
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