At what temperature the root mean square velocity of a gas is twice of its root mean velocity at $27^{\circ} C $?
Updated On: Jul 2, 2022
$927^{\circ} C $
$827^{\circ} C $
$727^{\circ} C $
$627^{\circ} C $
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The Correct Option isA
Solution and Explanation
Root mean square velocity
$\upsilon_{rms}=\sqrt{\frac{3RT}{M}}$
where T is the absolute temperature and M is the molar mass of the gas
As the M remains the same
$\upsilon_{rms} \propto \sqrt{T}$
As per question
$\left(\upsilon_{rms}\right)_{t ^{\circ}C} =2\left(\upsilon_{rms}\right)27 ^{\circ}C \ldots\left(i\right)$$\frac{\left(\upsilon_{rms}\right)t^{\circ} C}{\left(\upsilon_{rms}\right)27^{\circ} C}=\sqrt{\frac{t+273}{27+273}}$$\frac{2\left(\upsilon_{rms}\right)27^{\circ}C}{\left(\upsilon_{rms}\right)27^{\circ}C}=\sqrt{\frac{t+273}{300}} (Using (i))$
or $4=\frac{t+273}{300}$
or $t+273=1200$$t=1200-273$$=927^{\circ}C$
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