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…

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.

MARKS Content Team · Accepted Answer

Correct answer: 121. Change in internal energy for the mixture can be written as,∆U=n_1Cv_1∆T+n_2Cv_2∆T =n_1Cv_1+n_2Cv_2∆T ...i For isobaric process, work done =P∆V =n_1+n_2R∆T ...ii Divide i by ii, we get (∆U)/W=(n_1Cv_1+n_2Cv_2∆T)/(n_1+n_2R∆T) ⇒∆U=W/R(n_1Cv_1+n_2Cv_2)/(n_1+n_2) =66/R(2×3R2+1×5R2)/(2+1) =121 J

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