An equimolar mixture of monatomic and diatomic ideal gases undergoes an expansion satisfying dQ = ⅓dU + ½dW. The process obeys PV⁽¹¹⁄ˣ⁾ = constant. Find x. Ignore molecular vibrational degrees of freedom.
Physics · Mixed Chapter
JEE Mains

Question

An equimolar mixture of monatomic and diatomic ideal gases undergoes an expansion satisfying dQ = ⅓dU + ½dW. The process obeys PV⁽¹¹⁄ˣ⁾ = constant. Find x. Ignore molecular vibrational degrees of freedom.

Physics · Mixed Chapter

G1 Physics · Accepted Answer

Step 1 — At ordinary temperatures, molar heat capacities are Cᵥ(mono) = 3R/2 and Cᵥ(dia) = 5R/2. For an equimolar mixture, Cᵥ = (3R/2 + 5R/2)/2 = 2R per mole. Step 2 — First law: dQ = dU + dW. Compare with the given dQ = ⅓dU + ½dW to get (⅔)dU + (½)dW = 0, or 4dU + 3dW = 0. Step 3 — For n moles of ideal mixture, U = 2nRT = 2PV. Consequently dU = 2(PdV + VdP), while dW = PdV. Step 4 — Substitute: 4×2(PdV + VdP) + 3PdV = 0. Hence 11PdV + 8VdP = 0. Step 5 — Divide by PV: 11(dV/V) + 8(dP/P) = 0. In…