Question

Given Van Der Waals constant for $N{H_3},{H_2},{O_2}$ and $C{O_2}$ are respectively $4.17,0.244,1.36$ and $3.59$ , which one of the following gases is most easily liquefied?
A.${O_2}$
B.$N{H_3}$
C.$C{O_2}$
D.${H_2}$

Answer 1

Gases can convert into liquids when the temperature decreases. The process of changing from gas to liquid is known as condensation. Vander Waals constant is directly proportional to the critical temperature of the gases, Thus, the gas with high Vander Waal constant can be easily liquified.

Complete answer:
Given molecules are gases. Gases can be turned into liquids by the process of condensation. In the condensation process the gases reach the critical temperature and can be turned into liquids.
Given gases are ammonia with the molecular formula of $N{H_3}$ has the Vander Waal constant of $4.17$ , hydrogen gas with the molecular formula of ${H_2}$ has the Vander Waal constant of $0.244$ Carbon dioxide with the molecular formula of $C{O_2}$ has the Vander Waal constant of $3.59$ and oxygen gas with the molecular formula of ${O_2}$ has the Vander Waal constant of $1.36$ .
As Vander Waals constant is directly proportional to the critical temperature of the gases. The gas with high Vander Waal constant can be easily liquified. In the given gases ammonia has a high Vander Waal constant. Thus, ammonia can be easily liquified.
Hence, option (B) is the correct answer.

Note:
Critical temperature is a temperature at which a gas can be liquified by applying the pressure. Above the critical temperature a gas cannot be liquified by applying pressure. Higher is the Vander Waal constant, higher is the critical temperature and can be turned into liquid easily.