Question

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

• A. Same as that in $Y$
• B. Half of that in $Y$
• C. Twice of that in $Y$
• D. $8$ times of that in $Y$

Answer 1

The correct option is D $8$ times of that in $Y$

Ideal gas law: $PV=nRT$

For same no. of moles

$\frac{PV}{T}=constant$

$Averagespeed\left(v\right)\propto \sqrt{T}$

Gas X
$v_x=2v_y\propto \sqrt{T_x}................\left(i\right)$
Gas Y
$v_y\propto \sqrt{T_y}...............\left(ii\right)$

$\frac{\left(i\right)}{\left(ii\right)}$

$\Rightarrow 2=\sqrt{\frac{T_x}{T_y}}$

$\Rightarrow \frac{T_x}{T_y}=4\Rightarrow T_x=4T_y$

Now $\frac{P_x{V}_x}{T_x}=\frac{P_y{V}_y}{T_y}$

$\Rightarrow \frac{P_x\left(1\right)}{4T_y}=\frac{P_y\left(2\right)}{T_y}$

$\Rightarrow \frac{P_x}{P_y}=8$

Thus (D) is the correct answer.