Most Important Selected Qs for JEE AdvancedPhysicsThermodynamics
An ideal gas has molar heat capacity at constant pressure C_p=5 R / 2 . The gas is kept in a cylindrical vessel fitted with a piston which is free to move. Mass of the frictionless piston is 9 kg . Initial volume of the gas is 0.0027 ~m ^3 and cross-section area of the piston is 0.09 ~m ^2 . The initial temperature of the gas is 300 K . Atmospheric pressure P₀=1.05 10^5 ~N / m ^2 . An amount of 2.5 10^4 ~J of heat en
Options
- AInitial pressure of the gas is 1.06 10^5 ~N / m ^2
- BFinal temperature of the gas is 1000 K
- CFinal pressure of the gas is 1.06 10^5 ~N / m ^2
- DWork done by gas is 9.94 10^3 ~J
Correct answer
D. Work done by gas is 9.94 10^3 ~J
Step-by-step solution
PV = nRT ...(1) Mass of piston is 9 kg & area of piston is 0.09 ~m ^2 So P=P₀+ 9 g 0.09 P=1.06 10^5 ~N / m ^2 As mass of piston is not changed & Piston is frictionless So process is isobaric P_f=P_f=1.06 10^5 ~N / m ^2 dQ = nCpdT ....(2) or P ₂ ~V ₂- P ₁ ~V ₁= nRdT ....(3) (3) (2) P ( V ₂- V ₁ ) dQ = 2 5 aligned & P (V₂-V₁ )= 2 5 2.5 10^4 & W=P V=10^4 ~J & by equation (1) aligned aligned & n= 1.06 10^5 0.0027 R 300 & n= 1.06 270 2500 =0.11448 & or by d Q=n CpdT & T ₂- T ₁= 2.5 10^4 0.11448 5 R 2 & on solving T ₂=10