A mixture of N_2 and Ar gases in a cylinder c | vedclass

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(D) Step $1$: Calculate the number of moles of each gas.$n_{N_2} = \frac{7 \ g}{28 \ g \ mol^{-1}} = 0.25 \ mol = \frac{1}{4} \ mol$$n_{Ar} = \frac{8

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$15$

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(D) Step $1$: Calculate the number of moles of each gas.$n_{N_2} = \frac{7 \ g}{28 \ g \ mol^{-1}} = 0.25 \ mol = \frac{1}{4} \ mol$$n_{Ar} = \frac{8 \ g}{40 \ g \ mol^{-1}} = 0.2 \ mol = \frac{1}{5} \ mol$Step $2$: Calculate the mole fraction of $N_2$ $(x_{N_2})$.$x_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{Ar}} = \frac{1/4}{1/4 + 1/5} = \frac{0.25}{0.25 + 0.20} = \frac{0.25}{0.45} = \frac{5}{9}$Step $3$: Calculate the partial pressure of $N_2$ $(p_{N_2})$.$p_{N_2} = x_{N_2} 	imes P_{total} = \frac{5}{9} 	imes 27 \ bar = 5 	imes 3 \ bar = 15 \ bar$.

$A$ mixture of $N_2$ and $Ar$ gases in a cylinder contains $7 \ g$ of $N_2$ and $8 \ g$ of $Ar$. If the total pressure of the mixture of gases in the cylinder is $27 \ bar$,the partial pressure of $N_2$ is $......... \ bar$.
[Use atomic masses (in $g \ mol^{-1}$) : $N=14, Ar=40$]

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$A$ mixture of $N_2$ and $Ar$ gases in a cylinder contains $7 \ g$ of $N_2$ and $8 \ g$ of $Ar$. If the total pressure of the mixture of gases in the cylinder is $27 \ bar$,the partial pressure of $N_2$ is $......... \ bar$.[Use atomic masses (in $g \ mol^{-1}$) : $N=14, Ar=40$]