Question

Given van der Waals constant for $NH _{3}, H _{2}, O _{2}$ and $CO _{2}$ are respectively $4.17, 0.244, 1.36$ and $3.59$, which one of the following gases is most easily liquefied?

• A. $CO_2 $
• B. $NH_3$
• C. $O_2 $
• D. $ H_2$

Answer 1

The Correct Option is B

The van der Waals equation provides a more accurate representation of the behavior of real gases compared to the ideal gas law by including constants that account for molecular size and intermolecular forces. These constants are denoted as a and b. Here, we focus on constant a, which is related to the attraction between particles.

In this question, we are given the van der Waals constant a for four gases: NH_3, H_2, O_2, and CO_2. The values are:

• NH_3: 4.17
• H_2: 0.244
• O_2: 1.36
• CO_2: 3.59

The van der Waals constant a reflects the strength of intermolecular forces. The larger the value of a, the stronger the intermolecular forces, which implies that the gas is more easily liquefied. This is because strong intermolecular attractions can pull the gas molecules closer together into a liquid state.

Among the gases listed, NH_3 has the highest value of a (4.17), indicating that it has the strongest intermolecular attractions. Consequently, NH_3 is the gas that is most easily liquefied among the options provided.

Let's conclude the answer with justification:

• NH_3 is most easily liquefied due to its highest value of the van der Waals constant a.
• Other gases have significantly lower values of a (e.g., H_2: 0.244, O_2: 1.36, and CO_2: 3.59), indicating weaker intermolecular forces.

Correct Answer: NH_3