Question

The value of van der Waals' constant $a$ for various gases are given
$\begin{array}{cc}\text{Gas}& a\left(L^{2}atmmo{l}^{-2}\right)\\ N_2& 1.390\\ O_2& 1.360\\ N{H}_3& 4.170\\ C{H}_4& 2.253\end{array}$
The order of liquefaction of these gases is

• A. $N_2>O_2<N{H}_3>C{H}_4$
• B. $N{H}_3<C{H}_4<N_2>O_2$
• C. $N{H}_3>C{H}_4>N_2>O_2$
• D. $N_2<O_2<N{H}_3<C{H}_4$

Answer 1

The correct option is C $N{H}_3>C{H}_4>N_2>O_2$

Higher the value of van der Waals' constant ${}^{\prime }{a}^{\prime }$, higher the force of attraction between the molecules, therefore, liquefaction will be easier. $N{H}_3$ has highest value of 'a' due to H-bonding leading to strong intermolecular forces.
Hence, the order of liquefaction of the gases is $N{H}_3>C{H}_4>N_2>O_2$.