JEE Main 2019 · Physics · Thermodynamics and Kinetic Theory of Gases · hard

Two moles of a diatomic ideal gas (rigid molecules, CV=5/2R) undergo a process in which the pressure is directly proportional to the volume. During…

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Two moles of a diatomic ideal gas (rigid molecules,

C_V=\tfrac52R

) undergo a process in which the pressure is directly proportional to the volume. During this process the temperature rises from

300\ \mathrm{K}

400\ \mathrm{K}

. Take

R=8.31\ \mathrm{J\,mol^{-1}K^{-1}}

. The heat absorbed by the gas is nearly

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Answer: (C) $4.99\ \mathrm{kJ}$

With

P = kV

, the work is

W=\int P\,dV=\tfrac12 k(V_2^2-V_1^2)=\tfrac12(P_2V_2-P_1V_1)=\tfrac12 nR\,\Delta T = \tfrac12(2)(8.31)(100)=831\ \mathrm{J}

. The change in internal energy is

\Delta U = nC_V\Delta T = 2\cdot\tfrac52(8.31)(100)=4155\ \mathrm{J}

. By the first law,

Q=\Delta U + W = 4986\ \mathrm{J}\approx 4.99\ \mathrm{kJ}

. Equivalently the molar heat capacity of this process is

C=C_V+R/2=3R

. Using

C_V

alone gives

4.16\ \mathrm{kJ}

and

C_P

gives

5.82\ \mathrm{kJ}

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