Question

The average speed of the molecules of $He$ in a sample is 5/7 to that of the molecules of $H_2$ in a sample at $0^{o}C$. The temperature of helium sample is:

• A. ${100}^{o}C$
• B. ${278.6}^{o}C$
• C. $0K$
• D. ${5.6}^{o}C$

Answer 1

The correct option is B ${278.6}^{o}C$

As
$u_{avg}=\sqrt{8RT/\pi M}$
so according to given conditions
$u_{avg}$ of $He=\frac{5}{7}\times u_{avg}$ of $H_2$
$u_{He}=\frac{5}{7}\times u_{H_2}$
$\sqrt{\frac{8RT}{\pi \times 4}}=\frac{5}{7}\times \sqrt{\frac{8R\times 273}{\pi \times 2}}$
$\frac{T}{4}=\frac{25}{49}\times \frac{273}{2}$
$T=278.6K$