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.
A. A
Change in internal energy for the mixture can be written as, ∆ U = n 1 C v 1 ∆ T + n 2 C v 2 ∆ T = n 1 C v 1 + n 2 C v 2 ∆ T           . . . i For isobaric process, work done = P ∆ V = n 1 + n 2 R ∆ T             . . . ii Divide i by ii , we get ∆ U W = n 1 C v 1 + n 2 C v 2 ∆ T n 1 + n 2 R ∆ T ⇒ ∆ U = W R n 1 C v 1 + n 2 C v 2 n 1 + n 2 = 66 R 2 × 3 R 2 + 1 × 5 R 2 2 + 1 = 121   J
Related: Physics — Thermodynamics · All PYQ Banks