Question

For one mole of a van der Waal's gas when b = 0 and T = 300 K, the PV vs $\frac{1}{V}$ plot is shown below. The value of the van der Waal's constant a $\left({\text{atm. liter}}^2{\text{mol}}^{-2}\right)$ is:

• A. 1
• B. 1.5
• C. 4.5
• D. 3

Answer 1

The correct option is B 1.5

As $b=0$, Van der waals equation for one mole of gas will be:
$(P+\frac{a}{V^2})\left(V\right)=RT$
$PV+\left(\frac{a}{V}\right)=RT$
$PV=RT-\left(\frac{a}{V}\right)...\left(1\right)$
Now ,in the graph , we have y-axis as PV and x-axis as $\left(\frac{1}{V}\right).$
Comparing Equation (1), with y = mx + c;
We obtain, slope $\left(m\right)=-a=\frac{[y_2-y_1]}{x_2-x_1}=\frac{[21.6-20.1]}{[2.0-3.0]}$
Slope (m) $=-a=-1.5$
Hence $a=1.5$