Two thermodynamical process are shown in the figure. The molar heat capacity for process $A$ and $B$ are ${C}_{A}$ and ${C}_{B}$. The molar heat capacity at constant pressure and constant volume are represented by ${C}_{P}$ and ${C}_{V}$, respectively. Choose the correct statement. ![JEE Main 2024 Physics, Kinetic Theory of Gases — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Kinetic_Theory_of_Gases/6682bd73ba0e94a3c4025cba/question_1__q_6682bd73ba0e94a3c4025cba__cdn-question-pool.getmarks.app__4c085d90-f6f0-4a5e-b68e-ff6a65779362__8c313559a0_final_ppt_sync.png)

![JEE Main 2024 Physics, Kinetic Theory of Gases — solution figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Kinetic_Theory_of_Gases/6682bd73ba0e94a3c4025cba/solution_1__q_6682bd73ba0e94a3c4025cba__cdn-question-pool.getmarks.app__4c085d90-f6f0-4a5e-b68e-ff6a65779362__8c313559a0_final_ppt_sync.png) For process $A$: $\mathrm{log}P=\gamma \mathrm{log}V\Rightarrow P={V}^{\gamma },(\gamma >1)$ $P{V}^{-\gamma }=$ Constant ${C}_{A}={C}_{V}+\frac{R}{1+\gamma }\ldots (i)$ Likewise, for process $B\rightarrow P{V}^{-1}=$ Constant ${C}_{B}={C}_{v}+\frac{R}{1+1}$ ${C}_{B}={C}_{v}+\frac{R}{2}\ldots (\mathrm{ii})$ Also, ${C}_{P}={C}_{v}+R\ldots (\mathrm{iii})$ By $(\mathrm{ii})$ and $(\mathrm{iii})$, we get ${C}_{P}>{C}_{B}>{C}_{v}$

Two thermodynamical process are… - JEE Main 2024 Physics PYQ

JEE Main 2024 — Physics Thermodynamics

2024mcqmedium

Two thermodynamical process are shown in the figure. The molar heat capacity for process $A$ and $B$ are ${C}_{A}$ and ${C}_{B}$ . The molar heat capacity at constant pressure and constant volume are represented by ${C}_{P}$ and ${C}_{V}$ , respectively. Choose the correct statement.

JEE Main 2024 Physics, Kinetic Theory of Gases — question figure

Official previous-year question

Held on 30 Jan 2024 · Verified 6 Jul 2026.

Options

A
${C}_{P}>{C}_{B}>{C}_{V}$
B
${C}_{A}=0$ and ${C}_{B}=\infty$
C
${C}_{P}>{C}_{V}>{C}_{A}={C}_{B}$
D
${C}_{A}>{C}_{P}>{C}_{V}$

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Solution

JEE Main 2024 Physics, Kinetic Theory of Gases — solution figure

For process $A$ :

$\mathrm{log}P=\gamma \mathrm{log}V\Rightarrow P={V}^{\gamma },(\gamma >1)$

$P{V}^{-\gamma }=$ Constant

${C}_{A}={C}_{V}+\frac{R}{1+\gamma }\ldots (i)$

Likewise, for process $B\rightarrow P{V}^{-1}=$ Constant

${C}_{B}={C}_{v}+\frac{R}{1+1}$

${C}_{B}={C}_{v}+\frac{R}{2}\ldots (\mathrm{ii})$

Also, ${C}_{P}={C}_{v}+R\ldots (\mathrm{iii})$

By $(\mathrm{ii})$ and $(\mathrm{iii})$ , we get

${C}_{P}>{C}_{B}>{C}_{v}$

Your note

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Solution

JEE Main 2024 Physics, Kinetic Theory of Gases — solution figure

For process $A$ :

$\mathrm{log}P=\gamma \mathrm{log}V\Rightarrow P={V}^{\gamma },(\gamma >1)$

$P{V}^{-\gamma }=$ Constant

${C}_{A}={C}_{V}+\frac{R}{1+\gamma }\ldots (i)$

Likewise, for process $B\rightarrow P{V}^{-1}=$ Constant

${C}_{B}={C}_{v}+\frac{R}{1+1}$

${C}_{B}={C}_{v}+\frac{R}{2}\ldots (\mathrm{ii})$

Also, ${C}_{P}={C}_{v}+R\ldots (\mathrm{iii})$

By $(\mathrm{ii})$ and $(\mathrm{iii})$ , we get

${C}_{P}>{C}_{B}>{C}_{v}$

Options

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Thermodynamics in other years

More JEE Main Thermodynamics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Modern Physics

Optics

Waves & Oscillations

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