At $\mathrm{T}(\mathrm{K})$, equal weights of $\mathrm{H}_2, \mathrm{D}_2$, and $\mathrm{T}_2$ are present…
At $\mathrm{T}(\mathrm{K})$, equal weights of $\mathrm{H}_2, \mathrm{D}_2$, and $\mathrm{T}_2$ are present in closed vessel. The pressure exerted by this gaseous mixture is $\mathrm{P}$ atm. The ratio of partial pressures of $\mathrm{T}_2, \mathrm{D}_2$ and $\mathrm{H}_2$ is approximately
(H, D and $T$ are isotopes of hydrogen).
$0.33: 0.33: 0.33$
$0.18: 0.27: 0.54$
$0.25: 0.50: 0.25$
$0.54: 0.27: 0.18$
Solution
Let's take the equal weights of the three isotopes as $2 g$.
$\Rightarrow \mathrm{n}_1=1 \mathrm{~mol}, \mathrm{n}_2=\frac{1}{2} \mathrm{~mol}, \mathrm{n}_3=\frac{1}{3} \mathrm{~mol}$
$\mathrm{n}_{\text {total }}=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}$
$\Rightarrow x_1=\frac{1}{\frac{11}{6}}=0.54, x_2=\frac{\frac{1}{2}}{\frac{11}{6}}=0.27, x_3=\frac{\frac{1}{3}}{\frac{11}{6}}=0.18$
From Dalton's law of partial pressures:-
$\begin{aligned}
& P_i=x_i P_{\text {total }} \\
& \Rightarrow \frac{P_3}{P_2}=\frac{x_3 P}{x_2 P}=\frac{0.18}{0.27} \text { and } \frac{P_2}{P_1}=\frac{x_2}{x_1}=\frac{0.27}{0.54} \\
& \Rightarrow P_3: P_2: P_1=0.18: 0.27: 0.54
\end{aligned}$