Q.

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 processes are same, the value of ln V3V2 is “X”. Find value of 32X40? (given U=internal energy, S =entropy, p = pressure, R = gas constant, V = volume, and molar heat capacity of the gas at constant volume =5R2)
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answer is 8.

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Detailed Solution

(I) is reversible adiabatic expansion.
ΔUR=4502250=1800ΔU=1800R
(II) is an isothermal expansion
W1=WII   nCvT2T1=nRT2InV3V2w(I)=ΔU=1800RnCv(ΔT)=1800R1X52RT2900=1800R
Also, 1800 R=−RT2InV3V2
1800R=180InV3V2InV3V2=10="X"32×1040=8

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