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Physics · Thermodynamics

JEE Main 2024 — 8 April, Shift 1 — Question 40

Two different adiabatic paths for the same gas intersect two isothermal curves as shown in P−V\mathrm{P}-\mathrm{V} diagram. The relation between the ratio VaVd\frac{V_{a}}{V_{d}} and the ratio⁡VbVc\operatorname{ratio} \frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}} is: \end{enumerate}

Question figure
  1. Option A:

    VaVd=(VbVc)−1\frac{V_{a}}{V_{d}}=\left(\frac{V_{b}}{V_{c}}\right)^{-1}

  2. Option B:

    VaVd≠VbVc\frac{V_{a}}{V_{d}} \neq \frac{V_{b}}{V_{c}}

  3. Option C:

    VaVd=VbVc\frac{V_{a}}{V_{d}}=\frac{V_{b}}{V_{c}}

    Correct
  4. Option D:

    VaVd=(VbVc)2\frac{V_{a}}{V_{d}}=\left(\frac{V_{b}}{V_{c}}\right)^{2}

Answer: C

Step-by-step solution

adiabatic process

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

Ta⋅Vaγ−1=Td⋅Vdγ−1\mathrm{T}_{\mathrm{a}} \cdot \mathrm{V}_{\mathrm{a}}^{\gamma-1}=\mathrm{T}_{\mathrm{d}} \cdot \mathrm{V}_{\mathrm{d}}^{\gamma-1}

(VaVd)γ−1=TdTa\left(\frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{d}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{d}}}{\mathrm{T}_{\mathrm{a}}}

Tb⋅Vbγ−1=Tc⋅Vcγ−1\mathrm{T}_{\mathrm{b}} \cdot \mathrm{V}_{\mathrm{b}}^{\gamma-1}=\mathrm{T}_{\mathrm{c}} \cdot \mathrm{V}_{\mathrm{c}}^{\gamma-1}

(VbVc)γ−1=TcTb\left(\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}}\right)^{\gamma-1}=\frac{\mathrm{T}_{\mathrm{c}}}{\mathrm{T}_{\mathrm{b}}}

VaVd=VbVc(∵Td=TcTa=Tb)\frac{\mathrm{V}_{\mathrm{a}}}{\mathrm{V}_{\mathrm{d}}}=\frac{\mathrm{V}_{\mathrm{b}}}{\mathrm{V}_{\mathrm{c}}} \quad\binom{\because \mathrm{T}_{\mathrm{d}}=\mathrm{T}_{\mathrm{c}}}{\mathrm{T}_{\mathrm{a}}=\mathrm{T}_{\mathrm{b}}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
8 April, Shift 1
Subject
Physics
Chapter
Thermodynamics
Topic
Different Thermodynamic Processes