A vessel of volume 8 , L contains an ideal ga | vedclass

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(B) Given: Volume $V = 8 \, L = 8 	imes 10^{-3} \, m^3$,Initial pressure $P_1 = 2 \, atm = 2 	imes 1.01325 	imes 10^5 \, Pa \approx 2.0265 	imes 1

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$0.24$

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(B) Given: Volume $V = 8 \, L = 8 	imes 10^{-3} \, m^3$,Initial pressure $P_1 = 2 \, atm = 2 	imes 1.01325 	imes 10^5 \, Pa \approx 2.0265 	imes 10^5 \, Pa$. For simplicity in calculation,using $1 \, atm = 10^5 \, Pa$ as per standard approximation in many textbooks,$P_1 = 2 	imes 10^5 \, Pa$.Initial moles $n_1 = \frac{P_1 V}{RT} = \frac{2 	imes 10^5 	imes 8 	imes 10^{-3}}{8.314 	imes 300} \approx 0.641 \, mol$.Final pressure $P_2 = 125 \, kPa = 1.25 	imes 10^5 \, Pa$.Since temperature $T$ and volume $V$ are constant,$P \propto n$,so $\frac{n_2}{n_1} = \frac{P_2}{P_1}$.$n_2 = n_1 	imes \frac{P_2}{P_1} = 0.641 	imes \frac{1.25 	imes 10^5}{2 	imes 10^5} = 0.641 	imes 0.625 \approx 0.40 \, mol$.Leaked moles $\Delta n = n_1 - n_2 = 0.64 - 0.40 = 0.24 \, mol$.

$A$ vessel of volume $8 \, L$ contains an ideal gas at $300 \, K$ and $2 \, atm$ pressure. The gas is allowed to leak until the pressure becomes $125 \, kPa$. Calculate the number of moles that leaked out if the temperature remains constant.

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