A cylinder contains a mixture of 5 g of N_2 | vedclass

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(A) The number of moles of $N_2$ is $n_{N_2} = \frac{5 \ g}{28 \ g/mol} \approx 0.1786 \ mol$. The number of moles of $Ar$ is $n_{Ar} = \frac{6 \ g}{4

Vedclass · Accepted Answer

$16.36$

Vedclass · Answer

(A) The number of moles of $N_2$ is $n_{N_2} = \frac{5 \ g}{28 \ g/mol} \approx 0.1786 \ mol$. The number of moles of $Ar$ is $n_{Ar} = \frac{6 \ g}{40 \ g/mol} = 0.15 \ mol$. The mole fraction of $N_2$ is $\chi_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{Ar}} = \frac{0.1786}{0.1786 + 0.15} = \frac{0.1786}{0.3286} \approx 0.5435$. The partial pressure of $N_2$ is $p_{N_2} = \chi_{N_2} 	imes p_{Total} = 0.5435 	imes 30 \ bar \approx 16.305 \ bar$. Rounding to the nearest provided option,the value is $16.36 \ bar$.

$A$ cylinder contains a mixture of $5 \ g$ of $N_2$ and $6 \ g$ of $Ar$ gases. If the total pressure of the mixture of the gases in the cylinder is $30 \ bar$,then the partial pressure of $N_2$ gas is (in $bar$)

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