Question

When the temperature of a gas is raised from ${27}^{\circ }$ C to ${90}^{\circ }$ C, the percentage increase in the r.m.s. velocity of the molecules will be

• A. 10%
• B. 15%
• C. 20%
• D. 17.5%

Answer 1

The correct option is A. 10%.

The rms velocity of the gas molecules is given by,

$v_{rms}$ = $\sqrt{\frac{3RT}{M}}$

where R = Universal gas constant

T = Temperature of the gas

M = Molar mass of the gas

Thus, for a given gas $R$ and $M$ are constant.

$v_{rms}\propto \sqrt{T}$

$\Rightarrow$ $\frac{v_{2rms}}{v_{1rms}}$ = $\sqrt{\frac{T_2}{T_1}}$ = $\sqrt{\frac{(273+90)}{(273+27)}}==1.1$

% increase = ($\frac{v_2}{v_1}-1$) $\times 100$ = $(1.1-1)\times 100=0.1\times 100$= 10%.