Question

Consider the van der Waals constants, a and b, for the following gases.

Gas	${\text{Ar}}$	${\text{Ne}}$	${\text{Kr}}$	${\text{Xe}}$
a$\left( {{\text{atm d}}{{\text{m}}^6}{\text{ mo}}{{\text{l}}^{ - 2}}} \right)$	${\text{1}}{\text{.3}}$	${\text{0}}{\text{.2}}$	${\text{5}}{\text{.1}}$	${\text{4}}{\text{.1}}$
b$\left( {{\text{1}}{{\text{0}}^{ - 2}}{\text{ d}}{{\text{m}}^3}{\text{ mo}}{{\text{l}}^{ - 1}}} \right)$	${\text{3}}{\text{.2}}$	${\text{1}}{\text{.7}}$	${\text{1}}{\text{.0}}$	${\text{5}}{\text{.0}}$

Which gas is expected to have the highest critical temperature?
A. ${\text{Kr}}$
B. ${\text{Ne}}$
C. ${\text{Ar}}$
D. ${\text{Xe}}$

Answer 1

The gas cannot be liquefied above a certain temperature irrespective of the applied pressure. This temperature is known as the critical temperature. Above the critical temperature, the kinetic energy of the gas increases and below the critical temperature, when pressure is applied the gas molecules come closer and thus, gases liquefy.

Formula Used:
${{\text{T}}_{\text{c}}} = \dfrac{{8a}}{{27Rb}}$

Complete step by step solution:
We know that the relationship between van der Waals constant and critical temperature is given by the equation,
${{\text{T}}_{\text{c}}} = \dfrac{{8a}}{{27Rb}}$
Where ${{\text{T}}_{\text{c}}}$ is the critical temperature,
$a$ and $b$ are the van der Waals constants,
$R$ is the universal gas constant.
Thus, from the equation, we can conclude that,
${{\text{T}}_{\text{c}}} \propto \dfrac{a}{b}$
Thus, higher the ratio of the van der Waals constant higher is the critical temperature.
Calculate the ratio of the van der Waals constant higher for the argon gas as follows:
$\dfrac{a}{b} = \dfrac{{1.3}}{{3.2}} = 0.4062$
Thus, the ratio of the van der Waals constant higher for the argon gas is ${\text{0}}{\text{.4062}}$.
Calculate the ratio of the van der Waals constant higher for the neon gas as follows:
$\dfrac{a}{b} = \dfrac{{0.2}}{{1.7}} = {\text{0}}{\text{.1176}}$
Thus, the ratio of the van der Waals constant higher for the neon gas is ${\text{0}}{\text{.1176}}$.
Calculate the ratio of the van der Waals constant higher for the krypton gas as follows:
$\dfrac{a}{b} = \dfrac{{5.1}}{{1.0}} = {\text{5}}{\text{.1}}$
Thus, the ratio of the van der Waals constant higher for the krypton gas is ${\text{5}}{\text{.1}}$.
Calculate the ratio of the van der Waals constant higher for the xenon gas as follows:
$\dfrac{a}{b} = \dfrac{{4.1}}{{5.0}} = {\text{0}}{\text{.82}}$
Thus, the ratio of the van der Waals constant higher for the krypton gas is ${\text{0}}{\text{.82}}$.
From the ratios of van der Waals constants, we can conclude that the ratio for krypton is highest.
Thus, the gas that is expected to have the highest critical temperature is ${\text{Kr}}$.

**Thus, the correct option is (A) ${\text{Kr}}$.

Note: **
The van der Waals constant a represents the magnitude of intermolecular forces of attraction and b represents the effective size of the molecules.