Two flasks X and Y have capacity 1L and 2L respectively and each of them contains 1mole of a gas. The temperature of the flasks are so adjusted that average speed of molecules in X is twice as those in Y. The pressure in flask X would be

According to Kinetic theory of gases,

$\mathrm{PV}=\frac{1}{3}{\mathrm{mnu}}^2$

where, P is pressure of the gas, m is mass of the gas, n is no. of molecules and u is the average speed of the molecules of the gas.

So, for gas present in flask X and Y will have the pressure as follows

${\mathrm{P}}_{\mathrm{X}}=\frac{1{\mathrm{mn}}_{\mathrm{X}}{\mathrm{u}}_{\mathrm{X}}^2}{3{\mathrm{V}}_{\mathrm{X}}}-----\left(\mathrm{i}\right)$

${\mathrm{P}}_{\mathrm{X}}=\frac{1{\mathrm{mn}}_{\mathrm{Y}}{\mathrm{u}}_{\mathrm{Y}}^2}{3{\mathrm{V}}_{\mathrm{Y}}}-----\left(\mathrm{ii}\right)$

According to question,
${\mathrm{u}}_{\mathrm{X}}=2{\mathrm{u}}_{\mathrm{Y}} \mathrm{and} {\mathrm{n}}_{\mathrm{X}}={\mathrm{n}}_{\mathrm{Y}}$$\left(\because 1 \mathrm{mole} \mathrm{of} \mathrm{the} \mathrm{gas} \mathrm{is} \mathrm{present} \mathrm{in} \mathrm{each} \mathrm{flask}\right)$

$2{\mathrm{V}}_{\mathrm{X}}={\mathrm{V}}_{\mathrm{Y}}$

Dividing eqn. (i) by (ii),

$\frac{{\mathrm{P}}_{\mathrm{X}}}{{\mathrm{P}}_{\mathrm{Y}}}=\frac{\left(1/3\right){\mathrm{n}}_{\mathrm{X}}{\mathrm{u}}_{\mathrm{X}}^2{\mathrm{V}}_{\mathrm{Y}}}{\left(1/3\right){\mathrm{n}}_{\mathrm{Y}}{\mathrm{u}}_{\mathrm{Y}}^2{\mathrm{V}}_{\mathrm{X}} }=4\times 2=8$

${\mathrm{P}}_{\mathrm{X}}=8{\mathrm{P}}_{\mathrm{Y}}$