At T K,equal weights of H_2, D_2,and T_2 are | vedclass

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(B) Let the equal weight of each gas be $w \ g$. The molar masses are $M(H_2) = 2 \ g/mol$,$M(D_2) = 4 \ g/mol$,and $M(T_2) = 6 \ g/mol$.Number of mol

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$0.18 : 0.27 : 0.54$

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(B) Let the equal weight of each gas be $w \ g$. The molar masses are $M(H_2) = 2 \ g/mol$,$M(D_2) = 4 \ g/mol$,and $M(T_2) = 6 \ g/mol$.Number of moles: $n(H_2) = \frac{w}{2}$,$n(D_2) = \frac{w}{4}$,$n(T_2) = \frac{w}{6}$.Total moles $n_{	ext{total}} = w(\frac{1}{2} + \frac{1}{4} + \frac{1}{6}) = w(\frac{6+3+2}{12}) = \frac{11w}{12}$.Mole fractions: $x(H_2) = \frac{w/2}{11w/12} = \frac{6}{11} \approx 0.545$,$x(D_2) = \frac{w/4}{11w/12} = \frac{3}{11} \approx 0.273$,$x(T_2) = \frac{w/6}{11w/12} = \frac{2}{11} \approx 0.182$.Since partial pressure $P_i = x_i P_{	ext{total}}$,the ratio of partial pressures $P(T_2) : P(D_2) : P(H_2)$ is equal to the ratio of their mole fractions $x(T_2) : x(D_2) : x(H_2) = 0.18 : 0.27 : 0.54$.

At $T \ K$,equal weights of $H_2, D_2$,and $T_2$ are present in a closed vessel. The pressure exerted by this gaseous mixture is $P \ atm$. The ratio of partial pressures of $T_2, D_2$,and $H_2$ is approximately. ($H, D$ and $T$ are isotopes of hydrogen).

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