Question

For one mole of a van der Waals gas for which $b=0$ and $T=300\text{K}$, the $PV\text{vs}\frac{1}{V}$ plot is shown below. The value of the van der Waals constant $a$ (in ${\text{atm L mol}}^{-2}$) is:

• A. $1.0$
• B. $4.5$
• C. $1.5$
• D. $3.0$

Answer 1

The correct option is C $1.5$

Van der Waals equation of state is given as:
$(P+\frac{n^{2}a}{V^2})(V-nb)=nRT$
Given,
$n=1$ and $b=0$
$\Rightarrow (P+\frac{a}{V^2})V=RT$
$\Rightarrow PV=RT-\frac{a}{V}.......\left(1\right)$
Comparing this with the equation of straight line $Y=mX+c$, where $Y=PV$ & $X=\frac{1}{V}$, we can deduce that slope is equal to $-a$.
Comparing $\left(1\right)$ with the graph between $PV$ and $\frac{1}{V}$
Slope of the given graph is,
$-a=\frac{20.1-21.6}{3-2}=-1.5$
$\Rightarrow a=1.5$
Thus, option (c) is the correct answer.