Question

Select the incorrect statements :

• A. the fraction of molecules having speed in the range of u to (u+du) of a gas of molar mass m at temperature T is the same as that of the gas of molar mass 2m at temperature T/2
• B. it is possible to liquefy an ideal gas
• C. vapour phase of a liquid exist above its critical temperature
• D. the excluded volume b is approximately four times the actual volume occupied by the molecules

Answer 1

Answer: (A), (B), (C)

The correct options are
A the fraction of molecules having speed in the range of u to (u+du) of a gas of molar mass m at temperature T is the same as that of the gas of molar mass 2m at temperature T/2
B it is possible to liquefy an ideal gas
C vapour phase of a liquid exist above its critical temperature

A.The fraction of gaseous molecules having their speed within the range u to (u+du) is given by the expression-

$\frac{d{N}_u}{dt}=4\pi \left(\frac{M}{2\pi RT}{)}^{3/2}exp\right(-M{u}^2/2RT)u^{2}du$.

As long as the ratio $\frac{M}{T}$ is constant, the fraction of molecules having speed in the range u and (u+du) of a gas will be same. Here, the ratio of $\frac{M}{T}=\frac{2M}{T/2}=\frac{4M}{T}$ is not constant. Therefore the speed won't be the same. Option A is an incorrect statement.

B. An ideal gas can not be liquefied as there is no attraction forces exists in ideal gas.

C. . Critical temperature is the highest temperature above which liquid phase doesn't exist by pressure alone. Vapor phase of a liquid exists above its critical temperature.

D. The excluded volume of molecule in motion is $4$ times the actual volume of a molecule at rest.

Explanation:

Consider $1$ mole of a gas is satisfying ideal gas law,

$P=\frac{RT}{V_m}=\frac{RT}{V}$

Assuming all particles are hard spheres of the same finite radius $r$. Replacing $V$ with $V_m-b$ where $b$ is the excluded volume,

$P=\frac{RT}{V_m-b}$

We must realise that a particle is surrounded by a sphere of radius $2r$ that is forbidden for the centres of other particles.

The excluded volume for $2$ particles,

$b_2^1=4\pi \frac{d^3}{3}=8\left(\frac{4\pi r^3}{3}\right)$

which divided by $2$ gives the excluded volume per particle.

$b^1=\frac{b_2^1}{2}\Rightarrow b^1=4\left(\frac{4\pi r^3}{3}\right)$
Option A and B are incorrect statements.
Hence Option A and B are the correct answer.