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(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 52 R)

ΔuR=2250 - 450=1800⇒Δu=1800×R=W1800R=nRTln⁡V3V2For process IQ = 0Δu=W =1800RFor process IIΔu=0u=W=nRTln⁡V3V2T1=900KΔu=nCvΔT −1800R=1×5R2(T−900)T=180K1×180ln⁡V3V2=1800 ln⁡V3V2=10

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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
(U : internal energy, S: entropy, p : pressure, V: volume, R : gas constant)
(Given: molar heat capacity at constant volume, C_V,m of the gas is $\frac{5}{2}$ R)

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answer is 10.

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Detailed Solution

$\begin{array}{l} \frac{Δu}{R} = 2250 - 450 = 1800 \Rightarrow Δu = 1800 \times R = W \\ 1800 R = nRTln \frac{V_{3}}{V_{2}} \end{array}$

For process I

Q = 0

$Δu = W = 1800 R$

For process II

$\begin{array}{l} Δu = 0 \\ u = W = nRTln \frac{V_{3}}{V_{2}} \\ T_{1} = 900 K \end{array}$

$\begin{array}{l} Δu = {nC}_{v} ΔT - 1800 R = 1 \times \frac{5 R}{2} (T - 900) \\ T = 180 K \\ 1 \times 180 \ln \frac{V_{3}}{V_{2}} = 1800 \ln \frac{V_{3}}{V_{2}} = 10 \end{array}$