JEE Advanced 2016 Paper 1 · Physics · Thermodynamics and Kinetic Theory of Gases · medium

Two moles of a monatomic ideal gas are taken through a quasi-static process in which the pressure remains directly proportional to the volume…

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Two moles of a monatomic ideal gas are taken through a quasi-static process in which the pressure remains directly proportional to the volume,

P\propto V

. During the process the temperature rises by

100\ \mathrm{K}

. Take

R=8.3\ \mathrm{J\,mol^{-1}K^{-1}}

. The heat absorbed by the gas is

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Answer: (A) $3320\ \mathrm{J}$

For

P=aV

the ideal gas law gives

nRT=aV^2

, so

nR\,\Delta T=a(V_2^2-V_1^2)

. The work done is

W=\int aV\,dV=\tfrac12 a(V_2^2-V_1^2)=\tfrac12 nR\,\Delta T

. The internal energy change of a monatomic gas is

\Delta U=\tfrac32 nR\,\Delta T

. By the first law

Q=\Delta U+W=2nR\,\Delta T

, i.e. the molar heat capacity along this process is

C=2R

(consistent with

C=C_V+R/(1-x)

for

PV^x=

constant with

x=-1

). Numerically

Q=2\times2\times8.3\times100=3320\ \mathrm{J}

. The values

4150\ \mathrm{J}

and

2490\ \mathrm{J}

come from wrongly using

C_P

C_V

for a process that is neither isobaric nor isochoric, and

4980\ \mathrm{J}

from using

3R

Original question written in the JEE Advanced 2016 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake