Question

Most probable speed of a gas at 150 K is equal to the root mean square speed of $O_2$ gas at 400 K. Molecular mass of the gas is.

Answer 1

$V_{mp}=\sqrt{\frac{2RT}{M}}\&V_{rms}=\sqrt{\frac{3RT}{M}}$

According to the question

$\sqrt{\frac{2R\times 150}{M}}=\sqrt{\frac{3R\times 400}{32}}$

$\Rightarrow \frac{300}{M}=\frac{3\times 400}{32}$

$\Rightarrow M=8units$