The root mean square speed of molecules of a given mass of a gas at 27$^\circ$C and 1 atmosphere pressure is 200 ms^$-$1. The root mean square speed of molecules of the gas at 127$^\circ$C and 2 atmosphere pressure is ${{x \over {\sqrt 3 }}}$ ms^$-$1. The value of x will be _________.

Given, T₁ = 27$^\circ$C = 27 + 273 = 300K, p₁ = 1 atm, v₁ = 200 ms^$-$1, T₂ = 127$^\circ$C = 400 K, p₂ = 12 atm, v₂ = ?<br/><br/>As we know that,<br/><br/>Root mean square speed, ${v_{rms}} = \sqrt {{{3RT} \over m}}$<br/><br/>$\therefore$ $${{{v_1}} \over {{v_2}}} = \sqrt {{{{T_1}} \over {{T_2}}}} = \sqrt {{{300} \over {400}}} = \sqrt {{3 \over 4}} $$<br/><br/>$$ \Rightarrow {v_2} = \sqrt {{4 \over 3}} {v_1} = {2 \over {\sqrt 3 }} \times 200 = {{400} \over {\sqrt 3 }}$$ ms^$-$1<br/><br/>$\Rightarrow {x \over {\sqrt 3 }} = {{400} \over {\sqrt 3 }} \Rightarrow x = 400$