Question

If the van der waals parameters of two gases are given as:
$a-d{m}^{6}barmo{l}^{-2}b-d{m}^{3}barmo{l}^{-1}$
$$\begin{array}{cc}GasA-6.5& 0.056\\ GasB-18.0& 0.011\end{array}$$
Then which of the above statements is / are correct?

• A. Critical presssure of A< critical pressure of B.
• B. Critical temperature of A< critical temperature of B.
• C. Both a and b
• D. None of the above

Answer 1

The correct option is C Both a and b

As we know,
$V_c=3b,T_c=\frac{8a}{27R{b}^{\prime }}{P}_c=\frac{a}{27b^2}$
For critical volume,
Since, $b_A>b_B\Rightarrow V_{c\left(A\right)}>V_{c\left(B\right)}$
Since ${\left(\frac{a}{b}\right)}_A<{\left(\frac{a}{b}\right)}_B\Rightarrow T_{c\left(A\right)}<T_{c\left(B\right)}$
For Critical temperature -
Since, ${\left(\frac{a}{b}\right)}_A<{\left(\frac{a}{b}\right)}_B\Rightarrow T_{c\left(A\right)}<T_{c\left(B\right)}$
Option B is correct
For critical pressure-
${\left(\frac{a}{b^2}\right)}_A<{\left(\frac{a}{b^2}\right)}_B\Rightarrow P_{c\left(A\right)}<P_{c\left(B\right)}$