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)

(I) is reversible adiabatic expansion.ΔUR=450−2250=−1800∴ΔU=−1800R(II) is an isothermal expansion∴W1=WII nCvT2−T1=−nRT2In⁡V3V2∵w(I)=ΔU=−1800RnCv(ΔT)=−1800R1X52RT2−900=−1800RAlso, 1800 R=−RT2In⁡V3V2−1800R=−180In⁡V3V2∴In⁡V3V2=10="X"32×1040=8

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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 $\frac{V_{3}}{V_{2}}$ is “X”. Find value of $\frac{32 X}{40}$ ? (given U=internal energy, S =entropy, p = pressure, R = gas constant, V = volume, and molar heat capacity of the gas at constant volume $= \frac{5 R}{2})$

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answer is 8.

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Detailed Solution

(I) is reversible adiabatic expansion.
$\begin{array}{l} \frac{ΔU}{R} = 450 - 2250 = - 1800 \\ ∴ ΔU = - 1800 R \end{array}$
(II) is an isothermal expansion
$∴ W_{1} = W_{II} \begin{array}{l} nCv (T_{2} - T_{1}) = - {nRT}_{2} In \frac{V_{3}}{V_{2}} \\ ∵ w (I) = ΔU = - 1800 R \\ nCv (ΔT) = - 1800 R \\ 1 X \frac{5}{2} R (T_{2} - 900) = - 1800 R \end{array}$
Also, 1800 $R=− {RT}_{2} In \frac{V_{3}}{V_{2}}$
$\begin{array}{l} - 1800 R = - 180 In \frac{V_{3}}{V_{2}} \\ ∴ In \frac{V_{3}}{V_{2}} = 10 =" X " \\ \frac{32 \times 10}{40} = 8 \end{array}$