JEE Main 2016 · Physics · Thermodynamics and Kinetic Theory of Gases · hard

A mixture contains 1 mole of helium (monatomic) and 2 moles of oxygen (diatomic, rigid), both treated as ideal gases.

Single correct option

A mixture contains

1

mole of helium (monatomic) and

2

moles of oxygen (diatomic, rigid), both treated as ideal gases. The ratio

C_P/C_V

of the mixture is

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Answer: (D) $\dfrac{19}{13}$

Internal energy per kelvin adds for non-reacting gases, so the molar

C_V

of the mixture is the mole-weighted average:

C_V=\dfrac{1\times\frac32R+2\times\frac52R}{3}=\dfrac{6.5R}{3}

. For any ideal-gas mixture

C_P-C_V=R

per mole, so

C_P=\dfrac{6.5R}{3}+R=\dfrac{9.5R}{3}

. Therefore

\gamma=\dfrac{C_P}{C_V}=\dfrac{9.5}{6.5}=\dfrac{19}{13}\approx1.46

. Averaging the

\gamma

values directly,

\dfrac{1.67+2(1.4)}{3}=1.49

, is wrong because

\gamma

itself is not additive.

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