Thermodynamics

JEE Main level, single correct, medium

15 of 190

Question 15 of 190

Initial pressure and volume of a monoatomic ideal gas are PP and VV. The change in internal energy of this gas in adiabatic expansion to volume Vfinal=27VV_{final}=27V is ________ J.

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Answer: (C) −43PV-\dfrac{4}{3}PV

For a monoatomic ideal gas, the ratio of specific heats is γ=53\gamma = \dfrac{5}{3}.

In an adiabatic process, PVγ=constantPV^{\gamma} = \text{constant}.

P1V1γ=P2V2γP_1 V_1^{\gamma} = P_2 V_2^{\gamma}

PV5/3=P2(27V)5/3P V^{5/3} = P_2 (27V)^{5/3}

P2=P(127)5/3=P243P_2 = P \left(\dfrac{1}{27}\right)^{5/3} = \dfrac{P}{243}

The change in internal energy is given by:

ΔU=P2V2−P1V1γ−1\Delta U = \dfrac{P_2 V_2 - P_1 V_1}{\gamma - 1}

Substituting the values:

ΔU=(P243)(27V)−PV53−1\Delta U = \dfrac{\left(\dfrac{P}{243}\right)(27V) - PV}{\dfrac{5}{3} - 1}

ΔU=PV9−PV23\Delta U = \dfrac{\dfrac{PV}{9} - PV}{\dfrac{2}{3}}

ΔU=−89PV23=−43PV\Delta U = \dfrac{-\dfrac{8}{9}PV}{\dfrac{2}{3}} = -\dfrac{4}{3}PV

Answer: −43PV-\dfrac{4}{3}PV

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