MHT CET · Physics · Kinetic Theory of Gases
Let the r.m.s. velocity of molecule of a given mass of gas be \(\mathrm{C}_{1}\) at temperature \(27^{\circ} \mathrm{C}\). When the temperature is increased to \(327^{\circ} \mathrm{C}\), the \(\mathrm{r} . \mathrm{m} . \mathrm{s}\). velocity is \(\mathrm{C}_{2}\). Then the ratio \(\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}\) is
- A \(\sqrt{2}\)
- B 2
- C 4
- D \(2 \sqrt{2}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{T}_{1}=27^{\circ} \mathrm{C}=27^{\circ}+273=300 \mathrm{k}\)
\(\mathrm{T}_{2}=327^{\circ} \mathrm{C}=327^{\circ}+273=600 \mathrm{k}\)
\(\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}}=\sqrt{\frac{600}{300}}=\sqrt{2}\)
\(\mathrm{T}_{2}=327^{\circ} \mathrm{C}=327^{\circ}+273=600 \mathrm{k}\)
\(\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}}=\sqrt{\frac{600}{300}}=\sqrt{2}\)
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