Q.

One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV-diagram below. Among these four processes, one is isobaric, one is isochoric, one is isothermal and one is adiabatic. Match the processes mentioned in List-I with the corresponding statements in List-II.        [2018]

  LIST-I   LIST-II
P. In process I 1. Work done by the gas is zero
Q. In process II 2. Temperature of the gas remains unchanged
R. In process III 3. No heat is exchanged between the gas and its surroundings
S. In process IV 4. Work done by the gas is 6P0V0

 

1 P → 4; Q → 3; R → 1; S → 2  
2 P → 1; Q → 3; R → 2; S → 4  
3 P → 3; Q → 4; R → 1; S → 2  
4 P → 3; Q → 4; R → 2; S → 1  

Ans.

(3)

Process I is adiabatic therefore ΔQ=0

Process II is isobaric P=constant therefore W=P(V2-V1)=3P0(3V0-V0)=6P0V0

Process III is isochoric V=constant therefore W=P(V2-V1)=0

Process IV is isothermal, temperature T=constant

  Δu=0