States of Matter - Result Question 13
13. For one mole of a van der waals’ gas when $b=0$ and $T=300 K$, the $p V \hspace {1mm} v s \hspace {1mm}1 / V$ plot is shown below. The value of the van der waals’ constant $a$ (atm $\left.L mol^{-2}\right)$
(2012)
(a) 1.0
(b) 4.5
(c) 1.5
(d) 3.0
Show Answer
Answer:
Correct Answer: 13. (c)
Solution:
The van der waals’ equation of state is
$ \left(p+\dfrac{n^{2} a}{V^{2}}\right)(V-n b)=n R T $
For one mole and when $b=0$, the above equation condenses to
$ \left(p+\dfrac{a}{V^{2}}\right) V=R T $
$ \Rightarrow \quad p V=R T-\dfrac{a}{V} $
Eq. (i) is a straight equation between $p V$ and $\dfrac{1}{V}$ whose slope is c $-a^{\prime}$. Equating with slope of the straight line given in the graph.
$-a =\dfrac{20.1-21.6}{3-2}=-1.5 $
$\Rightarrow a =1.5$
BETA
