One mole of an ideal monoatomic gas undergoes two reversible processes ( A → B and B → C ) as shown in the given figure: A → B is an adiabatic process. If the total heat absorbed in the entire process ( A → B and B → C ) is RT 2 ln 10 , the value of 2 logV 3 is [Use molar heat capacity of the gas at constant pressure, C p , m = 5 2 R ]
A. A
q A → C = RT 2 ln 10 q A → B = 0  ( ∵  adiabatic)  q A → C = q A → B + q B → C q A → C = q B → C q A → c = nRT 2 ln V 3   V 2 ...(i) For B → C ΔE = q + w ΔE = 0 (since isothermic) q = - w = - - nRT 2 ln V 3   V 2 = nRT 2 ln V 3   V 2 q B → C = nRT 2 ln V 3   V 2 q B → C = RT 2 ln ⁡ V 3 V 2                     [ Since   n   = 1 ] From A → B T 1   V 1 γ - 1 = T 2   V 2 γ - 1 600   V 1 γ - 1 = 60   V 2 γ - 1 10 × 10 5 3 - 1 = V 2 γ - 1 10 5 / 3 = V 2 5 3 - 1 10 5 / 3 = V 2 2 / 3 V 2 = 10 5 3 × 3 2 = 10 5 2 V 2 = 10 5 2         . . . ( 2 ) From equation ( 1 ) q A → C = nRT 2 ln V 3 V 2 Given, q A → C = RT 2 ln 10 RT 2 ln 10 = RT 2 ln V 3   V 2 ln 10 = ln V 3 V 2 ln 10 = ln V 3 10 5 2 10 = V 3 10 5 2 V 3 = 10 1 + 5 2 = 10 7 2 2 logV 3 = 2 log 10 7 / 2 = 7
Related: Chemistry — Thermodynamics (C) · All PYQ Banks