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← JEE Advanced 2023 · Paper 1
PhysicsInteger+4Question 28 of 51

JEE Advanced 2023 · Paper 1, Physics question 11

A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas (γ=5/3\gamma = 5/3) and one mole of an ideal diatomic gas (γ=7/5\gamma = 7/5). Here, γ\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 _____ Joule.

Answer and worked solution

Official answer: 121

Work at constant pressure. With 3 moles in total,
W=nRΔT=3RΔT=66 J  ⟹  RΔT=22 J.W=nR\Delta T=3R\Delta T=66\ \text{J}\implies R\Delta T=22\ \text{J}.

Internal energy.
ΔU=(2⋅32+1⋅52)RΔT=5.5×22=121 J.\Delta U=\left(2\cdot\tfrac32+1\cdot\tfrac52\right)R\Delta T=5.5\times22=121\ \text{J}.