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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.

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.

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.

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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