Question

We have two vessels of equal volume, one filled with hydrogen and the other with an equal mass of Helium. The common temperature is ${27}^{\circ }C$.
If the temperature of Helium is kept at ${27}^{\circ }C$ and that of hydrogen is changed, at what temperature will its pressure become equal to that of helium? The molecular weights of hydrogen and helium are $2$ and $4$ respectively.

Answer 1

Again, we have
$pV=nkT$
$H_2$ and $He$ have equal volumes. Since pressure are equal, we must have
$n_H{T}_H=n_{He}{T}_{He}$
or, $\frac{T_{He}}{T_H}=\frac{n_H}{n_{He}}=2$
Here, $T_{He}=27+273=300K$
$\therefore T_H=\frac{1}{2}{T}_{He}=150K=(150-273{)}^{\circ }C=-{123}^{\circ }C$.