Physics · Thermodynamics

JEE Advanced 2023 — Paper 1 — Question 27

A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas $(\gamma=5 / 3)$ and one mole of an ideal diatomic gas $(\gamma=7 / 5)$ . Here, $\gamma$ is the ratio of the specific heats at constant pressure and constant volume of an ideal gas. The gas mixture does a work of 66 Joule when heated at constant pressure. The change in its internal energy is $\qquad$ Joule.

Answer: 121

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Step-by-step solution

for isobaric process, work done $\mathrm{W}=\left(\mathrm{n}_{1}+\mathrm{n}_{2}\right) \mathrm{R} \Delta \mathrm{T}$ $W=3 R \Delta T$ $66=3 R \Delta T$ $\mathrm{R} \Delta \mathrm{T}=22$ Change in internal energy $\Delta u=\frac{f_{1}}{2} n_{1} R \Delta T+\frac{f_{2}}{2} n_{2} R \Delta T$ [Degree of freedom, $\left.\mathrm{f}=\frac{2}{\gamma-1}\right]$ $\frac{3}{2} \times 2 \times \mathrm{R} \Delta \mathrm{T}+\frac{5}{2} \times 1 \times \mathrm{R} \Delta \mathrm{T}$ $\frac{11}{2} \times R \Delta T=121$ Joule

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Exam: JEE Advanced 2023
Paper: Paper 1
Subject: Physics
Chapter: Thermodynamics
Topic: Calculation of Work, Heat and Internal Energy

One mole of an ideal gas expands adiabatically from an initial state ( T A, V 0 ) to final state ( T f, 5 V 0 ) . Another mole of the same…

Answer: 121

Step-by-step solution

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