Question

The molecules of a given mass of a gas have root mean square speeds of 100 ms⁻¹ at 27°C and 1.00 atmospheric pressure. What will be the root mean square speeds of the molecules of the gas at 127°C and 2.0 atmospheric pressure?

Answer 1

We know that for a given mass of a gas

${\displaystyle V_{rms}=\sqrt{\frac{3RT}{M}}}$

Where R is gas constant

T is the temperature in Kelvin

M is the molar mass of the gas

Clearly, ${\displaystyle V_{rms}\propto \sqrt{T}}$

As R and M are constants,

${\displaystyle \frac{{\left(V_{rms}\right)}_1}{{\left(V_{rms}\right)}_2}=\sqrt{\frac{T_1}{T_2}}}$

Given, ${\displaystyle {\left(V_{rms}\right)}_1}$ = 100 m/s

T₁ = 27°C = 27 + 273 = 300 K

T₂ = 127°C = 127 + 273 = 400 K

∴ From equation (i)

${\displaystyle \frac{100}{{\left(V_{rms}\right)}_2}=\sqrt{\frac{300}{400}}=\frac{\sqrt{3}}{2}}$

⇒ ${\displaystyle {\left(V_{rms}\right)}_2=\frac{2\times 100}{\sqrt{3}}=\frac{200}{\sqrt{3}}}$ m/s