A closed container contains a homogeneous mixture of two moles of an ideal monatomic gas and one mole of an ideal diatomic gas . Here, is the ratio of the specific heats at constant pressure and constant volume of an ideal gas. The gas mixture does a wo

Question

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. [2023]

Smart Achievers · Accepted Answer

(121)

$At constant pressure, work done$

$W = P Δ V = n R Δ T = 66$

$Δ U = n {(C_{V})}_{mix} Δ T$

${(C_{V})}_{mix} = \frac{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}}{n_{1} + n_{2}}$

$\Rightarrow {(C_{V})}_{mix} = \frac{2 \times \frac{3}{2} R + 1 \times \frac{5}{2} R}{3} = \frac{11}{6} R$

$Δ U = \frac{11}{6} (n R Δ T) = \frac{11}{6} \times 66 = 121 J$

Solution

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Solution

Ans. (121) At constant pressure, work done W=PΔV=nRΔT=66 ΔU=n(CV)mixΔT (CV)mix=n1CV1+n2CV2n1+n2 ⇒(CV)mix=2×32R+1×52R3=116R ΔU=116(nRΔT)=116×66=121 J

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