A diatomic gas (γ=1.4) does 100 ~J of work in an isobaric expansion. The heat given to the gas is :

Correct answer: C. 350 ~J For Isobaric process aligned & w=P Δ v=nR Δ T=100 ~J & Q=Δ u+w & Δ Q= F 2 nR Δ T+nR Δ T & ( f 2 +1) nR Δ T & ((5)/(2)+1) 100=350 ~J aligned SECTION - B

JEE Main 20248 AprilEvening Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Evening Shift
Subject: Physics
Chapter: Thermodynamics

JEE Main 2024 · 8 April · Evening Shift · Physics · Thermodynamics

A diatomic gas (γ=1.4) does 100 ~J of work in an isobaric...

A diatomic gas

(\gamma=1.4)

does

100 \mathrm{~J}

of work in an isobaric expansion. The heat given to the gas is :

Options

A $250 \mathrm{~J}$
B $150 \mathrm{~J}$
C $350 \mathrm{~J}$
D $490 \mathrm{~J}$

Correct answer

C. $350 \mathrm{~J}$

Step-by-step solution

For Isobaric process

\begin{aligned} & \mathrm{w}=\mathrm{P} \Delta \mathrm{v}=\mathrm{nR} \Delta \mathrm{T}=100 \mathrm{~J} \\ & \mathrm{Q}=\Delta \mathrm{u}+\mathrm{w} \\ & \Delta \mathrm{Q}=\frac{\mathrm{F}}{2} \mathrm{nR} \Delta \mathrm{T}+\mathrm{nR} \Delta \mathrm{T} \\ & \left(\frac{\mathrm{f}}{2}+1\right) \mathrm{nR} \Delta \mathrm{T} \\ & \left(\frac{5}{2}+1\right) 100=350 \mathrm{~J}\end{aligned}

SECTION - B

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