One litre of oxygen at a pressure of 1 ,atm a | vedclass

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(C) The ideal gas equation is given by $pV = nRT$.For oxygen: $p_1 = 1 \,atm$, $V_1 = 1 \,L$. The number of moles is $n_{O_2} = \frac{p_1 V_1}{RT} = \

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(C) The ideal gas equation is given by $pV = nRT$.For oxygen: $p_1 = 1 \,atm$, $V_1 = 1 \,L$. The number of moles is $n_{O_2} = \frac{p_1 V_1}{RT} = \frac{1 	imes 1}{RT} = \frac{1}{RT}$.For nitrogen: $p_2 = 0.5 \,atm$, $V_2 = 2 \,L$. The number of moles is $n_{N_2} = \frac{p_2 V_2}{RT} = \frac{0.5 	imes 2}{RT} = \frac{1}{RT}$.When these gases are introduced into a vessel of volume $V_{mix} = 1 \,L$ at the same temperature $T$, the total number of moles is $n_{mix} = n_{O_2} + n_{N_2} = \frac{1}{RT} + \frac{1}{RT} = \frac{2}{RT}$.The final pressure $p_{mix}$ is given by $p_{mix} = \frac{n_{mix} RT}{V_{mix}} = \frac{(2/RT) 	imes RT}{1} = 2 \,atm$.

One litre of oxygen at a pressure of $1 \,atm$ and two litres of nitrogen at a pressure of $0.5 \,atm$ are introduced into a vessel of volume $1 \,L$. If there is no change in temperature, the final pressure of the mixture of gas (in $atm$) is

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