Q. 18

# Figure shows the variation in the internal energy U with the volume V of 2.0 mol of an ideal gas in a cyclic process abcda. The temperatures of the gas at b and c are 500 K and 300 K respectively. Calculate the heat absorbed by the gas during the process.

Given

Number of moles n=2

Temperature at b Tb=500K

Temperature at c Tc=300K

From the graph it is clear that

Tb=Ta and Td=Tc

Thus, path ab and cd are isothermal paths.

We know that work done by the gas is given as

ΔW=PΔV

Where ΔV =change in volume

P =pressure

Again, from the graph we can see that ΔV=0 for path bc and da.

Therefore, work done along path bc and da are zero.

So, total work done ΔW=Wab+Wcd

We know that work done in an isothermal process is given as

Where n=number of moles

R=gas constant =8.31J/Kmol

T=temperature

Vf=final volume

Vi=initial volume

Wab = 8310×ln2

Similarly,

Wcd = -4986 ×ln2

So, total work done ΔW=Wab+Wcd

=8310×ln2 - 4986×ln2

=3324×ln2

ΔW =3324×0.693=2304.02J

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=2304.02J

Thus, heat absorbed by the system is 2304.02J.

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