91Ó°ÊÓ

Most probable speed of the molecules in a gas at temperature $T_2$is equal to the rms speed of the molecules at temperature, ${\text{T}}_{\text{1}}$, i.e${\text{V}}_P={\text{V}}_{rms}$.,

The expression for the most probable speed is given by,

$V_P=\sqrt{\frac{2RT}{M}}$

Here $V_P$is the most probable speed,R is the universal gas constant, T is the temperature and M is the mass.

The expression for the RMS speed is given by,

${\mathit{V}}_{\mathbf{r}\mathbf{m}\mathbf{s}}\mathbf{=}\sqrt{\frac{\mathbf{3}\mathbf{R}\mathbf{T}}{\mathbf{M}}}$

Here$V_{rms}$ is the RMS speed.

Root-Mean-Square speed of the gas molecules at temperature $T_1$is given as

$V_{rms}=\sqrt{\frac{3R{T}_1}{M}}$

Most probable speed of the gas molecules at temperature${\text{T}}_2$is given as

$V_P=\sqrt{\frac{2R{T}_2}{M}}$

Taking the ratio of ${\text{V}}_P/{\text{V}}_{rms}$, we get

$\begin{array}{c}\frac{{\text{V}}_P}{{\text{V}}_{rms}}=\frac{\sqrt{\frac{2{\text{RT}}_2}{\text{M}}}}{\sqrt{\frac{3{\text{RT}}_1}{\text{M}}}}\\ =\frac{\sqrt{2{\text{T}}_2}}{\sqrt{3{\text{T}}_1}}\end{array}$

Squaring on both the sides, we get

role="math" localid="1661862437312" $\frac{{\text{T}}_2}{{\text{T}}_1}=\frac{3}{2}{\left(\frac{{\text{V}}_P}{{\text{V}}_{rms}}\right)}^2$

But we have the condition that

${\text{V}}_P={\text{V}}_{rms}$

So,

$⇒\frac{T_2}{T_1}=\frac{3}{2}$

Therefore the ratioof $T_2/T_1$ is .$3/2$