MHT CET · Physics · Kinetic Theory of Gases
The temperature of an ideal gas is increased from \(140 \mathrm{~K}\) to \(560 \mathrm{~K}\). If the r.m.s. speed of gas molecules is \(v\) at \(140 \mathrm{~K}\), then at \(560 \mathrm{~K}\), r.m.s. speed becomes
- A \(4 v\)
- B \(\frac{v}{4}\)
- C \(\frac{v}{2}\)
- D \(2 v\)
Answer & Solution
Correct Answer
(D) \(2 v\)
Step-by-step Solution
Detailed explanation
The r.m.s. speed \(v\) is directly proportional to the square root of temperature \(T\) in kelvin.
\(v \propto \sqrt{T}\)
The new temperature is \(T^{\prime}=4 \mathrm{~T}\)
Hence, \(v^{\prime}=2 v\)
\(v \propto \sqrt{T}\)
The new temperature is \(T^{\prime}=4 \mathrm{~T}\)
Hence, \(v^{\prime}=2 v\)
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