JEE Advanced2016PhysicsThermodynamicsActual
A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure P i = 10 5 P a a n d v o l u m e V i = 10 - 3 m 3 changes to a final state at P f = 1 32 × 10 5 P a a n d V f = 8 × 10 - 3 m 3 in an adiabatic quasi - static process, such that P 3 V 5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same
Options
- A112 J
- B294 J
- C588 J
- D813 J
Correct answer
C. 588 J
Step-by-step solution
In adiabatic process P 3 V 5 = constant ⇒ P V 5 3 = constant ⇒ γ = 5 3 ⇒ C V = 3 2 R a n d C P = 5 2 R In another process ∆ Q = n C P ∆ T + n C v ∆ T = 5 2 n R T B - T A + 3 2 n R T C - T B ∆ Q = 5 2 P B V B - P A V A + 3 2 P C V C - P B V B Putting values ∆ Q = 587.5 J ≈ 588 J