An ideal gas of molar mass ' $\mathrm{M}_0$ ' has r.m.s. velocity ' $\mathrm{V}$ ' at temperature '…
An ideal gas of molar mass ' $\mathrm{M}_0$ ' has r.m.s. velocity ' $\mathrm{V}$ ' at temperature ' $\mathrm{T}$ '. Then
- $\mathrm{VT}^2=$ constant
- $\frac{\mathrm{V}^2}{\mathrm{~T}}=$ constant
- $\mathrm{V}^2 \mathrm{~T}=$ constant
- $\mathrm{V}$ is independent of $\mathrm{T}$
Solution
R.M.S. velocity is given by
$\begin{aligned}
& \mathrm{V}=\sqrt{\frac{3 R T}{\mathrm{M}_0}} \\
& \therefore \mathrm{V}^2=\frac{3 \mathrm{RT}}{\mathrm{M}_0} \\
& \therefore \frac{\mathrm{V}^2}{\mathrm{~T}}=\frac{3 \mathrm{R}}{\mathrm{M}_0}=\text { constant }
\end{aligned}$
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Asked in: MHT CET 2021 (22 Sep Shift 2)
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