Chemistry - Thermodynamics - I Question with Solution | TestHub

Let p = P/P_A, u = V/V_A. The line through (1, 1) and (0.5, 2) is p = 3 - 2u. Work on the gas along the line = 1/2(1 + 2)(0.5) PV = 0.75 PV; isothermal work on the gas = PVln 2 = 0.693 PV, so (a) is true. T ∝ pu = 3u - 2u², a parabola in u: (b) true. With u = (3 - p)/2, T ∝ p(3 - p) is also quadratic, so P-T is a parabola, not a hyperbola: (c) false. d(3u - 2u²)/du = 0 gives u = 0.75, a maximum (T_max = 1.125 T_A): (d) true. Ans: (a), (b), (d).