MHT CET · Physics · Kinetic Theory of Gases
The root mean square velocity of molecules of a gas is \(200 \mathrm{~m} / \mathrm{s}\). What will be the
root mean square velocity of the molecules, if the molecular weight is doubled and
the absolute temperature is halved?
- A \(50 \mathrm{~m} / \mathrm{s}\)
- B \(100 \mathrm{~m} / \mathrm{s}\)
- C \(200 \mathrm{~m} / \mathrm{s}\)
- D \(\frac{100}{\sqrt{2}} \mathrm{~m} / \mathrm{s}\)
Answer & Solution
Correct Answer
(B) \(100 \mathrm{~m} / \mathrm{s}\)
Step-by-step Solution
Detailed explanation
(B)
R.M.S. velocity \(C=\sqrt{\frac{3 R T}{M}}\)
\(\therefore \quad C \quad \propto \sqrt{\frac{T}{M}}\)
\(\frac{C^{\prime}}{C}=\sqrt{\frac{T^{\prime}}{T} \frac{M}{M^{\prime}}}=\sqrt{\frac{1}{2} \times \frac{1}{2}}=\frac{1}{\sqrt{4}}=\frac{1}{2}\)
\(\therefore C^{\prime}=\frac{C}{2}=\frac{200}{2}=100 \mathrm{~m} / \mathrm{s}\)
R.M.S. velocity \(C=\sqrt{\frac{3 R T}{M}}\)
\(\therefore \quad C \quad \propto \sqrt{\frac{T}{M}}\)
\(\frac{C^{\prime}}{C}=\sqrt{\frac{T^{\prime}}{T} \frac{M}{M^{\prime}}}=\sqrt{\frac{1}{2} \times \frac{1}{2}}=\frac{1}{\sqrt{4}}=\frac{1}{2}\)
\(\therefore C^{\prime}=\frac{C}{2}=\frac{200}{2}=100 \mathrm{~m} / \mathrm{s}\)
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