WBJEE · Physics · Kinetic Theory of Gases
The temperature of anideal gas is increased from \(120 \mathrm{~K}\) to \(480 \mathrm{~K}\). If at \(120 \mathrm{~K}\), the root mean square speed of gas molecules is \(v\), then at \(480 \mathrm{~K}\) it will be
- A \(4 v\)
- B \(2 v\)
- C \(\frac{v}{2}\)
- D \(\frac{v}{4}\)
Answer & Solution
Correct Answer
(B) \(2 v\)
Step-by-step Solution
Detailed explanation
\[ \begin{aligned} & \text { Hints: } \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\sqrt{\frac{\mathrm{T}_1}{\mathrm{~T}_2}} \\ & \frac{\mathrm{V}_1}{\mathrm{~V}_2}=\sqrt{\frac{120}{480}}-=\sqrt{\frac{1}{4}}=\frac{1}{2} \\ & \mathrm{~V}_2=2 v \end{aligned} \]
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