Thermodynamics

JEE Main level, single correct, medium

112 of 190

Question 112 of 190

77 mole of certain monoatomic ideal gas undergoes a temperature increase of 40K40K at constant pressure. The increase in the internal energy of the gas in this process is
(Given R=8.3JK−1mol−1R=8.3J{K}^{-1}{mol}^{-1})

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Answer: (B) 3486J3486J

For the given process pressure is constant therefore, it is an isobaric process.
For a quasi-static process the change in internal energy of an ideal gas is independent of the nature of the process and is given by,
ΔU=nCvΔT\Delta U=n{C}_{v}\Delta T
=n×3R2×ΔT=n\times \dfrac{3R}{2}\times \Delta T
[molar heat capacity at constant volume for monatomic gas =3R2=\dfrac{3R}{2}]
ΔU=7×32×8.3×40=3486J\Delta U=7\times \dfrac{3}{2}\times 8.3\times 40=3486J

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