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Physics Question on Kinetic molecular theory of gases

At what temperature the root mean square velocity of a gas is twice of its root mean velocity at 27∘C27^{\circ} C ?

A

927∘C927^{\circ} C

B

827∘C827^{\circ} C

C

727∘C727^{\circ} C

D

627∘C627^{\circ} C

Answer

927∘C927^{\circ} C

Explanation

Solution

Root mean square velocity υrms=3RTM\upsilon_{rms}=\sqrt{\frac{3RT}{M}} where T is the absolute temperature and M is the molar mass of the gas As the M remains the same υrms∝T\upsilon_{rms} \propto \sqrt{T} As per question (υrms)t∘C=2(υrms)27∘C…(i)\left(\upsilon_{rms}\right)_{t ^{\circ}C} =2\left(\upsilon_{rms}\right)27 ^{\circ}C \ldots\left(i\right) (υrms)t∘C(υrms)27∘C=t+27327+273\frac{\left(\upsilon_{rms}\right)t^{\circ} C}{\left(\upsilon_{rms}\right)27^{\circ} C}=\sqrt{\frac{t+273}{27+273}} 2(υrms)27∘C(υrms)27∘C=t+273300(Using(i))\frac{2\left(\upsilon_{rms}\right)27^{\circ}C}{\left(\upsilon_{rms}\right)27^{\circ}C}=\sqrt{\frac{t+273}{300}} (Using (i)) or 4=t+2733004=\frac{t+273}{300} or t+273=1200t+273=1200 t=1200−273t=1200-273 =927∘C=927^{\circ}C