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Two moles of a monoatomic ideal gas $(C_{v} = 1.5 R)$ initially at 400 K in an isolated, 1.0 L, piston is allowed to expand against a constant pressure of a 1.0 atm till the final volume reaches to 10 L. Which of the following conclusions regarding the above change(s) is (are) true?

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The final temperature of the gas is 364K.

If the same process were carried out to the same final volume but under reversible conditions, final temperature would have been less than 364 K.

In the above process, the initial and final temperatures and volumes are related as $(\frac{T_{1}}{T_{2}}) = {(\frac{V_{1}}{V_{2}})}^{γ - 1}$

Entropy change of a system $(Δ S_{s y s t})$ is zero.

answer is A, B.

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Detailed Solution

a) $Δ U = Q + W$

$\begin{array}{l} n C_{v} (T_{2} - T_{1}) = - 1 [V_{2} - V_{1}] \\ 2 \times 1.5 R [T_{2} - 400] = - 9 \end{array}$

$\begin{array}{l} T_{2} - 400 = \frac{- 3}{R} - 36 \\ T_{2} = 400 - 36 \\ T_{2} = 364 K \end{array}$

b) If Reversible:

$8 = \frac{C_{p}}{C_{v}} = \frac{2.5}{1.5} = \frac{5}{3}$

$T_{1} {V_{1}}^{8 - 1} = T_{2} {V_{2}}^{8 - 1}$

$400 \times 1^{2 / 3} = T_{2} {(10)}^{2 / 3}$

$T_{2} = \frac{400}{{(10)}^{2 / 3}} = 86.20 K$

c) $T_{1} {V_{1}}^{8 - 1} = T_{2} V_{2}^{8 - 1}$ is applicable only for reversible adiabatic expansion

d) $Δ S_{s y s} = n C_{v} l n T_{2} / T_{1} + n R l n \frac{V_{2}}{V_{1}} \neq 0$

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