JEE Advanced2019PhysicsThermodynamicsActual
One mole of a monoatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature V - T diagram. The correct statement(s) is/are: [ R is the gas constant]
Options
- AThe ratio of heat transfer during processes 1 → 2 and 2 → 3 is Q 1 → 2 Q 2 → 3 = 5 3
- BThe ratio of heat transfer during processes 1 → 2 and 3 → 4 is Q 1 → 2 Q 3 → 4 = 1 2
- CThe above thermodynamic cycle exhibits only isochoric and adiabatic processes.
- DWork done in this thermodynamic cycle 1 → 2 → 3 → 4 → 1 is W = 1 2 R T 0
Correct answer
A. The ratio of heat transfer during processes 1 → 2 and 2 → 3 is Q 1 → 2 Q 2 → 3 = 5 3
Step-by-step solution
A ⇒ Δ Q 1 → 2 Δ Q 2 → 3 = N C P Δ T 1 → 2 N C V Δ T 2 → 3 = C P C V = 5 3 ⇒ correct B ⇒ Δ Q 1 → 2 Δ Q 3 → 4 = T 0 T 0 / 2 = 2 ⇒ Incorrect ( C ) ⇒ No adiabatic process is involved ⇒ incorrect ( D ) ⇒ Work done = area inside P - V diagram = P 0 V 0 = n R T 0 2 [at point 4 in diagram] ⇒ correct. 1 → 2 ⇒ isobaric at pressure 2 P 0 2 → 3 ⇒ isobaric at volume 2 V 0 3 → 4 ⇒ isobaric at pressure P 0 4 → 1 ⇒ isobaric at volume V 0 Given, number of moles, n = 1 and r = C P C V = 5 3 (for mono atomic gas) Now, Δ Q 1 → 2 = n C