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 ]

Step-by-step solution

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