Question

Two flasks $X$ and $Y$ have capacity $1L$ and $2L$ respectively and each of them contains $1$ mole of a gas. The temperatures 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:

• 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 $Y$

According to Kinetic Theory of gases,

$PV=\frac{1}{3}mn{u}^2$

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
$P_X=\frac{1}{3}\frac{m{n}_X{u}_X^2}{V_X}$..........................(i)

$P_Y=\frac{1}{3}\frac{m{n}_Y{u}_Y^2}{V_Y}$..........................(ii)

According to question,
$u_X=2u_Y$ and $n_X=n_Y$ ($\because \text{1 mole of the gas is present in each flask}$)

$2V_X=V_Y$

Dividing eqn. (i) by (ii),

$\frac{P_X}{P_Y}=\frac{\left(1/3\right)n_X{u}_X^2{V}_Y}{\left(1/3\right)n_Y{u}_Y^2{V}_X}=4\times 2=8$

$P_X=8P_Y$

Option D is correct