First slide

Rms speed

Question

The molecules of a given mass of a gas have rms velocity of $200 {ms}^{- 1}$ at 27°C and $10 \times 10^{5} {Nm}^{- 2}$ pressure. When the temperature and pressure of the gas are respectively, 127°C and $0.05 \times 10^{5} {Nm}^{- 2}$ , the rms velocity of its molecules (in ${ms}^{- 1}$ ) is

Easy

A

$\frac{400}{\sqrt{3}}$
B

$\frac{100 \sqrt{2}}{3}$
C

$\frac{100}{3}$
D

$100 \sqrt{2}$

Solution

It is given that,
${(v_{rms})}_{1} = 200 {ms}^{- 1}, T_{1} = 300 K, p_{1} = 10^{5} {Nm}^{- 2}$
$T_{2} = 400 K, p_{2} = 0.05 \times 10^{5} {Nm}^{- 2}, {(v_{rms})}_{2} = ?$
As, rms velocity of gas molecules,
$v_{r m s} \propto \sqrt{T}$ $(∵ v_{rms} = \sqrt{\frac{3 R T}{M}})$
For two different cases,
$\Rightarrow$ $\frac{{(v_{rms})}_{1}}{{(v_{rms})}_{2}} = \sqrt{\frac{T_{1}}{T_{2}}} \Rightarrow \frac{200}{{(v_{rms})}_{2}} = \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}}$
$\Rightarrow$ ${(v_{rms})}_{2} = \frac{2}{\sqrt{3}} \times 200 = \frac{400}{\sqrt{3}} {ms}^{- 1}$

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