MHT CET · Physics · Kinetic Theory of Gases
Two gases A and B are at absolute temperatures \(360 \mathrm{~K}\) and \(420 \mathrm{~K}\) respectively. The ratio of average kinetic energy of the molecules of gas B to that of gas A is
- A \(6: 7\)
- B \(\sqrt{7}: \sqrt{6}\)
- C \(7: 6\)
- D \(49: 36\)
Answer & Solution
Correct Answer
(C) \(7: 6\)
Step-by-step Solution
Detailed explanation
The mean kinetic energy \(\frac{1}{2} \mathrm{~m}<\mathrm{v}>^2\) of the gas molecule is proportional to the temperature \(\mathrm{T}\).
\(\frac{\mathrm{K}_{\mathrm{B}}}{\mathrm{K}_{\mathrm{A}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}=\frac{420}{360}=\frac{7}{6}\)
\(\frac{\mathrm{K}_{\mathrm{B}}}{\mathrm{K}_{\mathrm{A}}}=\frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}=\frac{420}{360}=\frac{7}{6}\)
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