# Consider the cyclic process ABCA, shown in the figure, performed on a sample 2.0 mol of an ideal gas. A total of 1200 J of heat is withdrawn from the sample in the process. Find the work done by the gas during the part BC.

Given

Heat extracted from the system ΔQ=-1200J

(negative sign is because heat is extracted from the system)

Number of moles of the gas n=2.0

From the graph we can write

TA=300K

TB=500K

VA=VC

We know that work done by the gas is given as

ΔW=PΔV

Where ΔV =change in volume

P =pressure

So, work done along line CA will be zero, as VA=VC.

Thus, total work done will be

ΔW=WAB+WBC

WAB = P(VB-VA)

But we know that ideal gas equation is

PV=nRT

Where n= number of moles

R=gas constant

T=temperature

PΔV=nRΔT

Thus, we can write WAB = P(VB-VA) = nR(TB-TA)

ΔW= nR(TB-TA) + WBC

From first law of thermodynamics, we know that,

ΔQ=ΔU+ΔW

Where ΔQ=heat supplied to the system

ΔU=change in internal energy

ΔW=work done by the system

Process ABCA is a cyclic process. The system is brought back to its initial state. Since internal energy is a state function, change in internal energy will be zero.

So, ΔU=0.

So, first law becomes

ΔQ=ΔW

ΔQ= nR(TB-TA) + WBC

We know that R=8.31J/K mol

WBC =ΔQ- nR(TB-TA)

= -1200 - 2×8.31× (500-300)

=-1200 – 3324

=-4524J

Thus, work done along path BC =-4524J

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