A vessel of volume 24.6 L contains 1.5 mole | vedclass

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(C) The partial pressure of a gas in a mixture is the pressure it would exert if it alone occupied the entire volume of the container at the same temp

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$2.5$

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(C) The partial pressure of a gas in a mixture is the pressure it would exert if it alone occupied the entire volume of the container at the same temperature.Using the ideal gas equation, $PV = nRT$, the partial pressure of $N_2$ $(p_{N_2})$ is given by $p_{N_2} = \frac{n_{N_2} R T}{V}$.Given:$n_{N_2} = 2.5 \ mol$$V = 24.6 \ L$$T = 300 \ K$$R = 0.0821 \ L \ atm \ K^{-1} \ mol^{-1}$Substituting the values:$p_{N_2} = \frac{2.5 \ mol 	imes 0.0821 \ L \ atm \ K^{-1} \ mol^{-1} 	imes 300 \ K}{24.6 \ L}$$p_{N_2} = \frac{61.575}{24.6} \ atm$$p_{N_2} = 2.5 \ atm$.

$A$ vessel of volume $24.6 \ L$ contains $1.5 \ moles$ of $H_2$ and $2.5 \ moles$ of $N_2$ at $300 \ K$. Calculate the partial pressure of $N_2$ in the vessel. (in $atm$)

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