JEE Advanced 2023 — Paper 1 — PHY-III solution

JEE Advanced2023Paper 1PHY-III

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.

Your answer

Solution

Work at constant pressure. $W = n R Δ T = 66$ J, where total moles $n = 3$ .

Molar specific heat at constant volume of the mixture. $C_{V, mix} = \frac{n _{1} C _{V_{1}} + n _{2} C _{V_{2}}}{n _{1} + n _{2}} = \frac{2 \cdot \frac{3}{2} R + 1 \cdot \frac{5}{2} R}{3} = \frac{( 3 + 5/2 ) R}{3} = \frac{11 R}{6} .$

Change in internal energy. $Δ U = n C_{V, mix} Δ T = \frac{11}{6} \cdot n R Δ T = \frac{11}{6} \cdot 66 = 121$ J.

Answer: $121$ J.

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