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Question

One mole of a monatomic ideal gas undergoes four thermodynamic processes as shown schematically in the PV- diagram below.

Answer 1

Answer (C)

Process – I is an adiabatic process

It is a type of process which occurs without transferring of heat or mass between the system and its surrounding. It transfers energy to the surroundings only as work.
Therefore, ΔQ=ΔU+ΔW
Now, ΔQ=0
Therefore, W=−ΔU
Now, the volume of gas is decreasing, then W<0 and ΔU>0
=> Temperature of gas increases.
=> No heat is exchanged between the gas and surrounding.

Process – II is an isobaric process

In this process Pressure remain constant.
W = P ΔV = 3P₀[3V₀ – V₀] = 6P₀V₀

Process - III is an isochoric process
It is a type of process where volume remains constant, so
ΔQ = ΔU + W
W = 0
Therefore, ΔQ=ΔU

Process – IV is an isothermal process
In this process Temperature remains constant.
ΔQ = ΔU + W
ΔU = 0

Answer 2

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