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Heat and Thermodynamics question

2024 · 6 Apr · Shift 2 · Q74
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Heat and Thermodynamics question

2024 · 6 Apr · Shift 2 · Q74

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A total of 48 J48 \mathrm{~J}48 J heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by 2∘C2^{\circ} \mathrm{C}2∘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}R=8.3 J K−1 mol−1.
  1. A
    23.1 J
  2. B
    48 J
  3. C
    24.9 J
  4. D
    72.9 J
View written solutionFree

Correct answer: A

  1. Use the first law of thermodynamics

For the gas, Q=ΔU+WQ=\Delta U + WQ=ΔU+W where:

  • Q=48 JQ = 48\,\text{J}Q=48J is the heat supplied,
  • ΔU\Delta UΔU is the change in internal energy,
  • WWW is the work done by the gas.
  1. Find change in internal energy for helium

Helium is a monoatomic ideal gas, so for nnn moles: ΔU=nCVΔT\Delta U = n C_V \Delta TΔU=nCV​ΔT with CV=3R2C_V = \frac{3R}{2}CV​=23R​

Given:

  • n=1n=1n=1
  • R=8.3 J mol−1K−1R=8.3\,\text{J mol}^{-1}\text{K}^{-1}R=8.3J mol−1K−1
  • ΔT=2∘C=2 K\Delta T = 2^\circ C = 2\,\text{K}ΔT=2∘C=2K

Thus, ΔU=1×32×8.3×2\Delta U = 1 \times \frac{3}{2} \times 8.3 \times 2ΔU=1×23​×8.3×2 ΔU=3×8.3=24.9 J\Delta U = 3 \times 8.3 = 24.9\,\text{J}ΔU=3×8.3=24.9J

  1. Calculate work done

From W=Q−ΔUW = Q - \Delta UW=Q−ΔU we get W=48−24.9=23.1 JW = 48 - 24.9 = 23.1\,\text{J}W=48−24.9=23.1J

  1. Match with options

W=23.1 JW = 23.1\,\text{J}W=23.1J So the correct option is A.

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