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(C) Since the temperature $T$ and volume $V$ of the vessel remain constant,we use the ideal gas law $PV = nRT$.From this,$n = \frac{PV}{RT}$. Since $V

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$20\%$ of the gas escaped due to the leakage

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(C) Since the temperature $T$ and volume $V$ of the vessel remain constant,we use the ideal gas law $PV = nRT$.From this,$n = \frac{PV}{RT}$. Since $V, R,$ and $T$ are constant,$n \propto P$.Initial state: $P_1 = 5 	ext{ atm}$,$n_1 = n$.Final state: $P_2 = 4 	ext{ atm}$,$n_2 = n'$.Therefore,$\frac{n'}{n} = \frac{P_2}{P_1} = \frac{4}{5} = 0.8$.This means $n' = 0.8n$.The amount of gas that escaped is $\Delta n = n - n' = n - 0.8n = 0.2n$.Percentage of gas escaped = $\frac{0.2n}{n} 	imes 100\% = 20\%$.

$A$ gas is enclosed in a vessel at a constant temperature at a pressure of $5 \text{ atm}$ and volume $4 \text{ L}$. Due to a leakage in the vessel,after some time,the pressure is reduced to $4 \text{ atm}$. As a result,the:

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