Question

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.

	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

(A) P → 4 ; Q → 3 ; R → 1 ; S → 2
(B) P →1 ; Q → 3 ; R → 2 ; S → 4
(C) P → 3 ; Q → 4 ; R → 1 ; S → 2
(D) P → 3 ; Q → 4 ; R  2 ; S → 1

Answer 1

Answer (C)

solution

Process – I is an adiabatic process

ΔQ = ΔU + ΔW ΔQ = 0
W = –ΔU
Volume of gas is decreasing => W < 0
ΔU > 0
=> Temperatuer of gas increases.
=> No heat is exchanged between the gas and surrounding.
Process – II is an isobaric process
(Pressure remain constant)
W = P ΔV = 3P₀[3V₀ – V₀] = 6P₀V₀
Process - III is an isochoric process
(Volume remain constant)
ΔQ = ΔU + W
W = 0
ΔQ = U
Process – IV is an isothermal process
(Temperature remains constant)
ΔQ = ΔU + W
ΔU = 0