Question

Which of the following gases can be most easily liquified? Given, the value of $a$ in van der Waals equation for $N{H}_3=4.17,C{O}_2=3.59,S{O}_2=6.71\&C{l}_2=6.49$

• A. $N{H}_3$
• B. $C{l}_2$
• C. $S{O}_2$
• D. $C{O}_2$

Answer 1

The correct option is C $S{O}_2$

Van der Waals equation of state is given as:
$(P+\frac{n^{2}a}{V^2})(V-nb)=nRT......\left(1\right)$
Here, $a$ represents the attractive force between the gas molecules. Higher the value of $a$, more easily the gas can be liquified. In the given list of gases, $S{O}_2$ has the higher value of $a$ . Hence, it can be liquified more easily compared to the other gases given.
Thus, option (c) is the correct answer.