Nitrogen gas N_2 of mass 28 ,g is kept in a v | vedclass

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(B) Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$ (mass/molar mass).Since the volume $V$ and molar mass $M$ are constant,we have $P

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$63/5$

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(B) Using the ideal gas equation $PV = nRT$,where $n = \frac{m}{M}$ (mass/molar mass).Since the volume $V$ and molar mass $M$ are constant,we have $P = \frac{m}{M} \frac{RT}{V}$,which implies $P \propto m \cdot T$ or $m \propto \frac{P}{T}$.Initial state: $m_i = 28 \,g$,$P_i = 10 \,atm$,$T_i = 57 + 273 = 330 \,K$.Final state: $m_f$,$P_f = 5 \,atm$,$T_f = 27 + 273 = 300 \,K$.Taking the ratio: $\frac{m_f}{m_i} = \frac{P_f}{P_i} 	imes \frac{T_i}{T_f}$.$\frac{m_f}{28} = \frac{5}{10} 	imes \frac{330}{300} = \frac{1}{2} 	imes \frac{11}{10} = \frac{11}{20}$.$m_f = 28 	imes \frac{11}{20} = 15.4 \,g$.The amount of gas leaked out is $\Delta m = m_i - m_f = 28 - 15.4 = 12.6 \,g$.$12.6 \,g = \frac{126}{10} = \frac{63}{5} \,g$.

Nitrogen gas $N_2$ of mass $28 \,g$ is kept in a vessel at a pressure of $10 \,atm$ and temperature $57^{\circ} C$. Due to leakage of $N_2$ gas,its pressure falls to $5 \,atm$ and temperature to $27^{\circ} C$. The amount of $N_2$ gas leaked out is .......... $g$.

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