Question

Consider, the two identical containers, one with 1 mole of $H_2$ and the other with 1 mole of $He$. If the root mean square (RMS) velocities of the two gases are the same, then the ratio of the temperatures, $\frac{T_{\left(H_2\right)}}{T_{\left(He\right)}}$ is $a:b$. Find the value of $a+b$.

Answer 1

We know,
$v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$
$R=\text{Universal gas constant}$
$M=\text{Molar mass}$
Here, $(v_{\text{rms}}{)}_{H_2}=(v_{\text{rms}}{)}_{He}$
$\Rightarrow \sqrt{\frac{3R{T}_{H_2}}{M_{H_2}}}=\sqrt{\frac{3R{T}_{He}}{M_{He}}}\Rightarrow \frac{T_{H_2}}{2}=\frac{T_{He}}{4}$
$\Rightarrow \frac{T_{H_2}}{T_{He}}=\frac{a}{b}=\frac{1}{2}$
Hence, $a+b=1+2=3$.