At T(K),a gaseous mixture containing H_2, He | vedclass

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(A) $P_{total} = 1 \ bar$. Let total mass $= 100 \ g$. $W(H_2) = 20 \ g, W(He) = 16 \ g, W(O_2) = 100 - (20 + 16) = 64 \ g$. $n(H_2) = \frac{20}{2} =

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$0.625, 0.250, 0.125$

Vedclass · Answer

(A) $P_{total} = 1 \ bar$. Let total mass $= 100 \ g$. $W(H_2) = 20 \ g, W(He) = 16 \ g, W(O_2) = 100 - (20 + 16) = 64 \ g$. $n(H_2) = \frac{20}{2} = 10 \ mol$. $n(He) = \frac{16}{4} = 4 \ mol$. $n(O_2) = \frac{64}{32} = 2 \ mol$. Total moles $= 10 + 4 + 2 = 16 \ mol$. $x(H_2) = \frac{10}{16} = 0.625$. $x(He) = \frac{4}{16} = 0.250$. $x(O_2) = \frac{2}{16} = 0.125$. Partial pressure $P_i = x_i 	imes P_{total}$. $P(H_2) = 0.625 	imes 1 = 0.625 \ bar$. $P(He) = 0.250 	imes 1 = 0.250 \ bar$. $P(O_2) = 0.125 	imes 1 = 0.125 \ bar$.

At $T(K)$,a gaseous mixture containing $H_2, He$ and $O_2$ exerted a pressure of $1 \ bar$. The weight percentage of $H_2$ and $He$ is $20$ and $16$ respectively. The partial pressure (in bar) of $H_2, He$ and $O_2$ is respectively

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