JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Heat & Thermodynamics

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Two moles of helium gas is mixed with three moles of hydrogen molecules (taken to be rigid). What is the molar specific heat of mixture at constant volume? (

R = 8.3 J / mol K

)

(A) $17.4 J / mol K$ ✓ Correct
(B) $15.7 J / mol K$
(C) $19.7 J / mol K$
(D) $21.6 J / mol K$

Answer: (A)

Solution

For a mixture, the effective degrees of freedom are the mole-weighted average,

f_{mix} = \frac{n _{1} f _{1} + n _{2} f _{2}}{n _{1} + n _{2}}

. Helium is monatomic with

f_{1} = 3

and

n_{1} = 2

, while rigid-molecule hydrogen is diatomic with

f_{2} = 5

and

n_{2} = 3

, so

f_{mix} = \frac{2 \times 3 + 3 \times 5}{5} = \frac{21}{5}

. The molar specific heat at constant volume is

C_{v} = \frac{f}{2} R = \frac{21}{5} \times \frac{8.3}{2} \approx 17.4 J mol^{- 1} K^{- 1}

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Heat & Thermodynamics

Solution

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