MHT CET · Physics · Kinetic Theory of Gases
An ideal gas of molar mass ' \(\mathrm{M}_0\) ' has r.m.s. velocity ' \(\mathrm{V}\) ' at temperature ' \(\mathrm{T}\) '. Then
- A \(\mathrm{VT}^2=\) constant
- B \(\frac{\mathrm{V}^2}{\mathrm{~T}}=\) constant
- C \(\mathrm{V}^2 \mathrm{~T}=\) constant
- D \(\mathrm{V}\) is independent of \(\mathrm{T}\)
Answer & Solution
Correct Answer
(B) \(\frac{\mathrm{V}^2}{\mathrm{~T}}=\) constant
Step-by-step Solution
Detailed explanation
R.M.S. velocity is given by
\(
\begin{aligned}
& \mathrm{V}=\sqrt{\frac{3 R T}{\mathrm{M}_0}} \\
& \therefore \mathrm{V}^2=\frac{3 \mathrm{RT}}{\mathrm{M}_0} \\
& \therefore \frac{\mathrm{V}^2}{\mathrm{~T}}=\frac{3 \mathrm{R}}{\mathrm{M}_0}=\text { constant }
\end{aligned}
\)
\(
\begin{aligned}
& \mathrm{V}=\sqrt{\frac{3 R T}{\mathrm{M}_0}} \\
& \therefore \mathrm{V}^2=\frac{3 \mathrm{RT}}{\mathrm{M}_0} \\
& \therefore \frac{\mathrm{V}^2}{\mathrm{~T}}=\frac{3 \mathrm{R}}{\mathrm{M}_0}=\text { constant }
\end{aligned}
\)
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