The molecules of a given mass of a gas have root mean square speed of 100 ms^-1 at 27^oC and 1.00 atmospheric pressure. What will be the root mean square speed of the molecules of the gas at 127^oC and 2.0 atmospheric pressure?

We know that for a given mass of a gas

${\mathrm{v}}_{\mathrm{rms}}=\sqrt{\frac{3\mathrm{RT}}{\mathrm{M}}}$

where R is gas constant, T is temperature in kelvin and M is molar mass of the gas.
Clearly for a given gas, ${\mathrm{v}}_{\mathrm{rms}}\propto \sqrt{\mathrm{T}}$, as R, M are constants.

$\text{ Hence, }\frac{{\left({\mathrm{v}}_{\mathrm{rms}}\right)}_1}{{\left({\mathrm{v}}_{\mathrm{rms}}\right)}_2}=\sqrt{\frac{{\mathrm{T}}_1}{{\mathrm{T}}_2}}$ ……..(i)

$\text{ Given, }{\left({\mathrm{v}}_{\text{rms }}\right)}_1=100\mathrm{m}/\mathrm{s}$

    $\begin{array}{l}{\mathrm{T}}_1={27}^{\circ }\mathrm{C}=27+273=300\mathrm{K}\\ {\mathrm{T}}_2={127}^{\circ }\mathrm{C}=127+273=400\mathrm{K}\end{array}$

$\therefore \text{ From Eq. (i) }$

$\begin{array}{l} \frac{100}{{\left({\mathrm{v}}_{\mathrm{rms}}\right)}_2}=\sqrt{\frac{300}{400}}=\frac{\sqrt{3}}{2}\\ \Rightarrow {\left({\mathrm{v}}_{\mathrm{rms}}\right)}_2=\frac{2\times 100}{\sqrt{3}}=\frac{200}{\sqrt{3}}\mathrm{m}/\mathrm{s}\end{array}$