Physics · Thermodynamics

JEE Main 2024 — 8 April, Shift 2 — Question 47

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

Answer: A

For Isobaric process

$\mathrm{w}=\mathrm{P} \Delta \mathrm{v}=\mathrm{nR} \Delta \mathrm{T}=100 \mathrm{~J}$

$\mathrm{Q}=\Delta \mathrm{u}+\mathrm{w}$

$\Delta \mathrm{Q}=\frac{\mathrm{F}}{2} \mathrm{nR} \Delta \mathrm{T}+\mathrm{nR} \Delta \mathrm{T}$

$\left(\frac{\mathrm{f}}{2}+1\right) \mathrm{nR} \Delta \mathrm{T}$

$\left(\frac{5}{2}+1\right) 100=350 \mathrm{~J}$

Answer key and solution verified before publishing.

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