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Physics · Thermodynamics

JEE Main 2024 — 9 April, Shift 1 — Question 49

A sample of 1 mole gas at temperature TT is adiabatically expanded to double its volume. If adiabatic constant for the gas is γ=32\gamma=\frac{3}{2}, then the work done by the gas in the process is:

  1. Option A:

    RT[2−2]\mathrm{RT}[2-\sqrt{2}]

    Correct
  2. Option B:

    RT[2−2]\frac{\mathrm{R}}{\mathrm{T}}[2-\sqrt{2}]

  3. Option C:

    RT[2+2]\mathrm{RT}[2+\sqrt{2}]

  4. Option D:

    TR[2+2]\frac{\mathrm{T}}{\mathrm{R}}[2+\sqrt{2}]

Answer: A

Step-by-step solution

TVγ−1=\mathrm{TV}^{\gamma-1}= constant

⇒T(V)32−1=Tf(2 V)32−1\Rightarrow \mathrm{T}(\mathrm{V})^{\frac{3}{2}-1}=\mathrm{T}_{\mathrm{f}}(2 \mathrm{~V})^{\frac{3}{2}-1}

⇒TV12=Tf(2)12( V)12\Rightarrow \mathrm{TV}^{\frac{1}{2}}=\mathrm{T}_{\mathrm{f}}(2)^{\frac{1}{2}}(\mathrm{~V})^{\frac{1}{2}}

⇒Tf=(T2)\Rightarrow \mathrm{T}_{\mathrm{f}}=\left(\frac{\mathrm{T}}{\sqrt{2}}\right) Now, W.D. =nRΔT1−γ=1⋅R[T2−T]1−32=\frac{\mathrm{nR} \Delta \mathrm{T}}{1-\gamma}=\frac{1 \cdot \mathrm{R}\left[\frac{\mathrm{T}}{\sqrt{2}}-\mathrm{T}\right]}{1-\frac{3}{2}}

⇒\Rightarrow W.D. =2RT[1−12]=2 \mathrm{RT}\left[1-\frac{1}{\sqrt{2}}\right]

⇒\Rightarrow W.D. == RT [2−2][2-\sqrt{2}]

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 1
Subject
Physics
Chapter
Thermodynamics
Topic
Calculation of Work, Heat and Internal Energy

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