First slide

Rms speed

Question

The molecules of a given mass of a gas have root mean square speeds of 100 ${ms}^{- 1}$ at $27^{0} C$ and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at $127^{0} C$ and 2.0 atmospheric pressure?

Easy

A

$\frac{150}{\sqrt{3}} m / s$
B

$\frac{125}{\sqrt{3}} m / s$
C

$\frac{200}{\sqrt{3}} m / s$
D

$100 \sqrt{3} m / s$

Solution

We know that for a given mass of a gas $v_{rms} = \sqrt{\frac{3 RT}{M}}$
where R is gas constant, T is temperature in kelvin and M is molar mass of the gas.
Clearly for a given gas, $v_{rms} \propto \sqrt{T}, as R, M are constants .$
Hence, $\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} - - - - - - - - - (i)$
Given, ${(v_{rms})}_{1} = 100 m / s$
$T_{1} = 27^{0} C = 27 + 273 = 300 K$
$T_{2} = 127^{0} C = 127 + 273 = 400 K$
$∴ From Eq . (i) \frac{100}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{400}} = \frac{\sqrt{3}}{2}$
$\Rightarrow {(v_{rms})}_{2} = \frac{2 \times 100}{\sqrt{3}} = \frac{200}{\sqrt{3}} m / s$

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