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

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(A) Initial state: $PV = n_1RT$. Given $P_1 = 2\, atm = 2.026 	imes 10^5\, Pa$,$V = 8 	imes 10^{-3}\, m^3$,$T = 300\, K$,$R = 8.314\, J/(mol\cdot K)

Vedclass · Accepted Answer

$0.24$

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(A) Initial state: $PV = n_1RT$. Given $P_1 = 2\, atm = 2.026 	imes 10^5\, Pa$,$V = 8 	imes 10^{-3}\, m^3$,$T = 300\, K$,$R = 8.314\, J/(mol\cdot K)$.$n_1 = \frac{P_1V}{RT} = \frac{2.026 	imes 10^5 	imes 8 	imes 10^{-3}}{8.314 	imes 300} \approx 0.65\, mol$.Using the provided approximation $1\, atm = 10^5\, Pa$ as per the solution logic: $n_1 = \frac{2 	imes 10^5 	imes 8 	imes 10^{-3}}{8.31 	imes 300} = 0.64\, mol$.Since temperature $T$ and volume $V$ are constant,$P \propto n$,so $\frac{P_2}{P_1} = \frac{n_2}{n_1}$.$P_2 = 125\, kPa = 1.25 	imes 10^5\, Pa$.$n_2 = n_1 	imes \frac{P_2}{P_1} = 0.64 	imes \frac{1.25 	imes 10^5}{2 	imes 10^5} = 0.64 	imes 0.625 = 0.40\, mol$.Leaked moles $= n_1 - n_2 = 0.64 - 0.40 = 0.24\, mol$.

$A$ 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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