JEE Main 2020 · Physics · Thermodynamics and Kinetic Theory of Gases · easy

A mixture of 1 mole of a monoatomic ideal gas and 1 mole of a diatomic ideal gas (rigid rotor, no vibration) is heated at constant pressure. The…

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A mixture of

1

mole of a monoatomic ideal gas and

1

mole of a diatomic ideal gas (rigid rotor, no vibration) is heated at constant pressure. The fraction of the heat supplied that is converted into work is

A. $\dfrac{2}{7}$
B. $\dfrac{1}{3}$
C. $\dfrac{2}{5}$
D. $\dfrac{1}{4}$

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Answer: (B) $\dfrac{1}{3}$

For the mixture at constant pressure,

C_p = \tfrac52 R + \tfrac72 R = 6R

for the two moles together. The work done is

W = (n_1+n_2)R\,\Delta T = 2R\,\Delta T

while the heat is

Q = 6R\,\Delta T

. Hence

W/Q = 2/6 = 1/3

. The values

2/5

and

2/7

belong to the pure monoatomic and pure diatomic gases respectively.

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