28, g of N_2 gas is contained in a flask at a | vedclass

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(D) The ideal gas law $PV = nRT$ must hold both before and after the leakage of the gas.Since the volume $V$ of the flask remains constant,we have $\f

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$63/5\, g$

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(D) The ideal gas law $PV = nRT$ must hold both before and after the leakage of the gas.Since the volume $V$ of the flask remains constant,we have $\frac{P_i}{n_i T_i} = \frac{P_f}{n_f T_f}$.Given initial conditions: $P_i = 10\, atm$,$T_i = 57 + 273 = 330\, K$,$m_i = 28\, g$,$n_i = \frac{28}{28} = 1\, mol$.Given final conditions: $P_f = \frac{10}{2} = 5\, atm$,$T_f = 27 + 273 = 300\, K$.Using the relation $n_f = \frac{P_f}{P_i} \cdot \frac{T_i}{T_f} \cdot n_i$,we get:$n_f = \frac{5}{10} \cdot \frac{330}{300} \cdot 1 = \frac{1}{2} \cdot \frac{11}{10} = \frac{11}{20}\, mol$.The final mass $m_f = n_f 	imes M = \frac{11}{20} 	imes 28 = 15.4\, g$.The mass of $N_2$ gas that leaked out is $\Delta m = m_i - m_f = 28 - 15.4 = 12.6\, g$.Converting $12.6\, g$ to fraction: $12.6 = \frac{126}{10} = \frac{63}{5}\, g$.

$28\, g$ of $N_2$ gas is contained in a flask at a pressure of $10\, atm$ and at a temperature of $57^\circ C$. It is found that due to leakage in the flask,the pressure is reduced to half and the temperature is reduced to $27^\circ C$. The quantity of $N_2$ gas that leaked out is

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