One mole of an ideal diatomic gas at 900 K, undergoes two reversible processes, I followed by II, as shown below. If work done by the gas in the two processes are same, then which of the given option(s) is/are correct?

(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 $\frac{5}{2}$ R)

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For process II, $Δ U = w$ .

${(Δ U)}_{I I} = 0$

Temperature after process I is 180 K.

I is a reversible adiabatic process.

answer is A, B, C.

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Detailed Solution

(B)
Sol: I is a reversible adiabatic process and II is isothermal reversible process.
T₁ = 900 K
Calculation of T₂ after reversible adiabatic process (I)
   ${(Δ U)}_{I} = n C_{v} d T$
$- 1800 R = 1 \times \frac{5}{2} R (T_{2} - 900);$ T2 = 180 K
   ${(Δ U)}_{I} = w_{I}$
   $W_{I I} = - n R T_{2} \ln \frac{V_{3}}{V_{2}} = W_{I}$
$- 1 \times R \times 180 \ln \frac{V_{3}}{V_{2}} = - 1800 R;$ $\ln \frac{V_{3}}{V_{2}} = 10$