StreakPeaked· Practice

ExamsJEE AdvancedChemistry

An ideal gas with molar heat capacity at constant volume Cv,m = (5/2)R undergoes a reversible adiabatic expansion from 1 litre to 32 litres. Its initial temperature is 327 degrees C. The molar enthalpy change (in J/mol) for the process is

  1. -1125 R
  2. -575 R
  3. -1575 R
  4. None of these

Correct answer: -1575 R

Solution

Cv,m = (5/2)R, so Cp,m = Cv,m + R = (7/2)R, and gamma = Cp/Cv = 7/5. Initial temperature T1 = 327 + 273 = 600 K. For reversible adiabatic: T1*V1^(gamma-1) = T2*V2^(gamma-1). (gamma-1) = 2/5. T2 = T1*(V1/V2)^(2/5) = 600*(1/32)^(2/5) = 600*(1/4) = 150 K. Delta_H = Cp,m * delta_T = (7/2)*R*(150 - 600) = (7/2)*R*(-450) = -1575 R.

Related JEE Advanced Chemistry questions

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →