Question

The van der Waals parameters for gases W, X, Y, Z are

Gases	${\mathbf{a}}\left( {{\mathbf{atm}}{{\mathbf{L}}^2}{\mathbf{mo}}{{\mathbf{l}}^{ - 2}}} \right)$	$b\left( {{\mathbf{Lmo}}{{\mathbf{l}}^{ - 1}}} \right)$
W	4	0.027
X	8	0.030
Y	6	0.032
Z	12	0.027

Which one of these gases has the highest critical temperature?
A.W
B.X
C.Y
D.Z

Answer 1

The Van der Waals equation is a state equation in chemistry and thermodynamics that generalises the ideal gas law based on possible explanations why particular gases do not behave ideally. Gas molecules are treated as point particles under the ideal gas law, which means they don't take up space or change kinetic energy through collisions (i.e. all collisions are perfectly elastic).

Complete answer: The van der Waals equation is a state equation that accounts for two properties of natural gases: gas particle excluded volume and gas molecule attractive powers.
The van der Waals equation is commonly stated as follows: $\left( {P + a\dfrac{1}{{V_m^2}}} \right)\left( {{V_m} - b} \right) = RT$
The constants a and b are unique to a given gas and represent the magnitude of intermolecular attraction and excluded length, respectively.
The Van der Waals equation replaces V in the ideal gas law with${V_m} - b$, where ${V_m}$ is the molar volume of the gas and b is the volume filled by one mole of the molecules, to account for the volume that a real gas molecule takes up.
The second change to the ideal gas law accounts for the fact that gas molecules do interact with one another (at low pressures, they normally feel attraction, and at high pressures, they typically experience repulsion), and that actual gases have different compressibility than ideal gases. Van der Waals added a parameter $\dfrac{a}{{{V_m}^2}}$ to the measured pressure P in the equation of state to account for intermolecular reaction, where a is a constant whose value varies depending on the gas.

Since the kinetic energies of the particles that make up the gas increase as the temperature rises, liquefying gases becomes more complicated. The critical temperature of a liquid is the temperature at which its vapour cannot be liquefied, regardless of how much pressure is applied.
${T_c} = \dfrac{{8a}}{{27Rb}}$
i.e, ${T_c}\propto \dfrac{a}{b}$

Gases	${\mathbf{a}}\left( {{\mathbf{atm}}{{\mathbf{L}}^2}{\mathbf{mo}}{{\mathbf{l}}^{ - 2}}} \right)$	$b\left( {{\mathbf{Lmo}}{{\mathbf{l}}^{ - 1}}} \right)$	$\dfrac{a}{b}$
W	4	0.027	148.148
X	8	0.030	266.67
Y	6	0.032	187.5
Z	12	0.027	444.444

Z has the highest $\dfrac{a}{b}$ value. Hence its critical temperature value is high.
Hence Option D is correct

Note:
A critical point is the end point of a process equilibrium curve in thermodynamics. The liquid–vapor critical point, which is the end point of the pressure–temperature curve and designates conditions under which a liquid and its vapour can coexist, is the most well-known example.