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One mole of an ideal gas,initially at 200 K at 4 atm is expanded adiabatically to 1 atm in such a way that the temperature of the gas falls to 160 K. $C_{p}$ of the gas is 28 $J K^{- 1} m o l^{- 1}$ and is constant over the temperature range of this process. Choose the correct option(s): [Given: R = 8.3 $J K^{- 1} m o l^{- 1},$ ln2=0.7, ln5 = 1.6]

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$| w | = 788 k J$

$q = 0$

$17.22 J K^{- 1} m o l^{- 1}$

1120 J

answer is A, C, D.

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Detailed Solution

$Δ U = n C_{v} Δ T = (28 - 8.3) \times (- 40) = - 788 J$

$∵ q = 0, w = Δ U = - 788 J$

$Δ H = n C_{P} Δ T = (28) \times (- 40) = - 1120 J$

$Δ S_{s y s} = n C_{P} l n \frac{T_{2}}{T_{1}} + n R l n \frac{P_{1}}{P_{2}} = 28 l n \frac{5}{4} + 8.3 l n \frac{4}{1} = 28 (1.6 - 1.4) + 8.3 \times 1.4$

$= 17.22 J K^{- 1} m o l^{- 1}$

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