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: $p_1 = 1 ~atm$,$V_1 = 1 ~L$. Thus,$n_{O_2} = \frac{p_1 V_1}{RT} = \frac{1 	imes 1}{RT}

Vedclass · Accepted Answer

$2$

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(C) The ideal gas equation is given by $pV = nRT$.For oxygen: $p_1 = 1 ~atm$,$V_1 = 1 ~L$. Thus,$n_{O_2} = \frac{p_1 V_1}{RT} = \frac{1 	imes 1}{RT} = \frac{1}{RT}$.For nitrogen: $p_2 = 0.5 ~atm$,$V_2 = 2 ~L$. Thus,$n_{N_2} = \frac{p_2 V_2}{RT} = \frac{0.5 	imes 2}{RT} = \frac{1}{RT}$.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 $n_{mix} = n_{O_2} + n_{N_2} = \frac{1}{RT} + \frac{1}{RT} = \frac{2}{RT}$.Using the ideal gas equation for the mixture: $p_{mix} V_{mix} = n_{mix} RT$.Substituting the values: $p_{mix} 	imes 1 = \left( \frac{2}{RT} ight) RT$.Therefore,$p_{mix} = 2 ~atm$.

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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