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Question

One mole of an ideal gas at 900 K, undergoes two reversible processes, I followed by II, as shown below.
If the work done by the gas in the two process are same, the value of lnV3V2

(U:internal energy, S:entropy, p:pressure, V:volume, R:gas constant)
( Given: molar heat capacity at constant volume, Cv,m of the gas is 52R )
.

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Solution

1st process is adiabatic since entropy is constant.
W1=ΔU
ΔU=450R2250R=1800R
W1=1800R ...(1)

In 2nd process internal energy is constant it means it is a isothermal process.
W2=2303nRTlogV3V2 ...(2)
=nRTlnV3V2 ...(3)
Given, n=1 mole, here temperature is unknown
U=nCvT for process II
450R=1×52RT
T=450×25=180K
Equation (1) = Equation (2)
W1=W2
1800R=1×R×180lnV3V2
lnV3V2=1800180=10

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