The molecules of a given mass of a gas have $$r.m.s.$$  velocity of $$200\,m{s^{ - 1}}$$  at $${27^ \circ }C$$  and $$1.0 \times {10^5}N{m^{ - 2}}$$    pressure. When the temperature and pressure of the gas are respectively, $${127^ \circ }C$$  and $$0.05 \times {10^5}N{m^{ - 2}},$$    the $$rms$$  velocity of its molecules in $$m{s^{ - 1}}$$ is

Solution :
It is given that
$$\eqalign{ & {v_{rms}} = 200\;m{s^{ - 1}},{T_1} = 300\;K,{P_1} = {10^5}\;N/{m^2} \cr & {T_2} = 400\;K,{P_2} = 0.05 \times {10^5}\;N/{m^2} \cr} $$
As, $$rms$$  velocity of gas molecules,
$$\because {V_{rms}} \propto \sqrt T \,\,\left( {\because {v_{rms}} = \sqrt {\frac{{3RT}}{m}} } \right)$$
For two different cases
$$\eqalign{ & \Rightarrow \quad \frac{{{{\left( {{v_{rms}}} \right)}_1}}}{{{{\left( {{v_{rms}}} \right)}_2}}} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} \Rightarrow \frac{{200}}{{{{\left( {{v_{rms}}} \right)}_2}}} = \sqrt {\frac{{300}}{{400}}} = \sqrt {\frac{3}{4}} \cr & \Rightarrow {\left( {{v_{rms}}} \right)_2} = \frac{2}{{\sqrt 3 }} \times 200 = \frac{{400}}{{\sqrt 3 }}\;m{s^{ - 1}} \cr} $$