JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Heat & Thermodynamics

A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas

(γ = 5/3)

and one mole of an ideal diatomic gas

(γ = 7/5)

. Here,

γ

is the ratio of the specific heats at constant pressure and constant volume of an ideal gas. The gas mixture does a work of 66 Joule when heated at constant pressure. The change in its internal energy is ____Joule.

Answer: 121

Solution

For the mixture:

Monatomic gas: $n_{1} = 2$ mol, $C_{V 1} = \frac{3}{2} R$
Diatomic gas: $n_{2} = 1$ mol, $C_{V 2} = \frac{5}{2} R$

Mixture heat capacity at constant volume:

(C_{V})_{mi x} = \frac{n _{1} C _{V 1} + n _{2} C _{V 2}}{n _{1} + n _{2}} = \frac{2 \times \frac{3}{2} R + 1 \times \frac{5}{2} R}{3} = \frac{11 R}{6}

At constant pressure:

W = n R Δ T = 66

J (for total

n = 3

mol)

Change in internal energy:

Δ U = n (C_{V})_{mi x} Δ T = 3 \times \frac{11 R}{6} \times Δ T = \frac{11}{2} (n R Δ T) = \frac{11}{2} \times 66 = 121

JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Integer answer

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JEE Advanced · 2023 · 04 Jun 2023 · Paper 1 · Heat & Thermodynamics

Solution

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