JEE Advanced2013PhysicsThermodynamicsActual
One mole of a monatomic ideal gas is taken along two cyclic processes E → F → G → E and E → F → H → E as shown in the PV diagram. The processes involved are purely isochoric, isobaric, isothermal or adiabatic. Match the paths in List I with the magnitudes of the work done in List II and select the correct answer using the codes given below the lists. List I List II A. G → E P
Options
- Aa-q;b-s;c-r;d-p;
- Ba-s;b-r;c-q;d-p;
- Ca-q;b-s;c-r;d-p;
- Da-q;b-r;c-s;d-p;
Correct answer
B. a-s;b-r;c-q;d-p;
Step-by-step solution
F → G  work done in isothermal process is nRT ln V f V i = 3 2 P 0 V 0 ln 3 2 V 0 V 0 = 3 2 P 0 V 0 ln 2 5 = 1 0  P 0 V 0 ln 2 ln  G ⟶ E , Δ W = P 0 3 1 V 0 = 3 1  P 0 V 0 ln  G ⟶ H work done is less than 3 1  P 0 V 0 i.e., 2 4 P 0 V 0 ln  F ⟶ H work done is 3 6  P 0 V 0