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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Physics
Chapter
Thermodynamics

JEE Main 2025 · 24 January · Morning Shift · Physics · Thermodynamics

The temperature of 1 mole of an ideal monoatomic gas is increased...

The temperature of 1 mole of an ideal monoatomic gas is increased by 50∘C50^{\circ} \mathrm{C} at constant pressure. The total heat added and change in internal energy are E1E_1 and E2E_2, respectively. If E1E2=x9\frac{E_1}{E_2}=\frac{x}{9} then the value of xx is _____

Correct answer

0

Step-by-step solution

Given that process is isobaric ΔT=50∘C\Delta \mathrm{T}=50^{\circ} \mathrm{C} Q in isobaric process =nCpΔ=\mathrm{nC}_{\mathrm{p}} \Delta\mathrm{T}=\mathrm{E}1ΔU_1 \Delta \mathrm{U} in isobaric process =nCvΔ=\mathrm{nC}_{\mathrm{v}} \Delta\mathrm{T}=\mathrm{E}2∴E1E2=CPCv=γ_2 \therefore \frac{\mathrm{E}_1}{\mathrm{E}_2}=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{v}}}=\gamma Given, gas is monoatomic ∴γ=1+2f=1+23=53\begin{aligned} \therefore \gamma & =1+\frac{2}{\mathrm{f}} \\ & =1+\frac{2}{3} \\ & =\frac{5}{3} \end{aligned} Now, as per question. 53=x9x=15\begin{aligned} & \frac{5}{3}=\frac{x}{9} \\ & x=15 \end{aligned}
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