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Physics · Thermodynamics

JEE Advanced 2023 — Paper 1 — Question 22

One mole of an ideal gas expands adiabatically from an initial state ( TA,V0\mathrm{T}_{\mathrm{A}}, \mathrm{V}_{0} ) to final state (Tf,5 V0)\left(\mathrm{T}_{\mathrm{f}}, 5 \mathrm{~V}_{0}\right). Another mole of the same gas expands isothermally from a different initial state (TB,V0)\left(\mathrm{T}_{\mathrm{B}}, \mathrm{V}_{0}\right) to the same final state (Tf,5 V0)\left(\mathrm{T}_{\mathrm{f}}, 5 \mathrm{~V}_{0}\right). The ratio of the specific heats at constant pressure and constant volume of this ideal gas is γ\gamma. What is the ratio TA/TBT_{A} / T_{B} ?

  1. Option A:

    5γ−15^{\gamma-1}

    Correct
  2. Option B:

    51−γ5^{1-\gamma}

  3. Option C:

    5γ5^{\gamma}

  4. Option D:

    51+γ5^{1+\gamma}

Answer: A

Step-by-step solution

T1 V1γ−1=T2 V2γ−1 TA V0γ−1=Tf(5 V0)γ−1 TA Tf=(5)γ−1 as Tf=TB∴ TA TB=5γ−1\begin{aligned} & \mathrm{T}_{1} \mathrm{~V}_{1}^{\gamma-1}=\mathrm{T}_{2} \mathrm{~V}_{2}^{\gamma-1} \\ & \mathrm{~T}_{\mathrm{A}} \mathrm{~V}_{0}^{\gamma-1}=\mathrm{T}_{\mathrm{f}}\left(5 \mathrm{~V}_{0}\right)^{\gamma-1} \\ & \frac{\mathrm{~T}_{\mathrm{A}}}{\mathrm{~T}_{\mathrm{f}}}=(5)^{\gamma-1} \\ & \text { as } \mathrm{T}_{\mathrm{f}}=\mathrm{T}_{\mathrm{B}} \\ & \therefore \frac{\mathrm{~T}_{\mathrm{A}}}{\mathrm{~T}_{\mathrm{B}}}=5^{\gamma-1} \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2023
Paper
Paper 1
Subject
Physics
Chapter
Thermodynamics
Topic
Different Thermodynamic Processes

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