Let the r.m.s. velocity of molecule of a given mass of gas be $\mathrm{C}_{1}$ at temperature $27^{\circ}…

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
  1. $\sqrt{2}$
  2. 2
  3. 4
  4. $2 \sqrt{2}$

Solution

$\mathrm{T}_{1}=27^{\circ} \mathrm{C}=27^{\circ}+273=300 \mathrm{k}$ $\mathrm{T}_{2}=327^{\circ} \mathrm{C}=327^{\circ}+273=600 \mathrm{k}$ $\frac{\mathrm{C}_{2}}{\mathrm{C}_{1}}=\sqrt{\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}}=\sqrt{\frac{600}{300}}=\sqrt{2}$ :

Asked in: MHT CET 2020 (14 Oct Shift 1)

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