States of Matter - Result Question 13

####13. For one mole of a van der Waals’ gas when $b=0$ and $T=300 K$, the $p V v s 1 / V$ 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

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Solution:

  1. The van der Waals’ equation of state is

$$ \left(p+\frac{n^{2} a}{V^{2}}\right)(V-n b)=n R T $$

For one mole and when $b=0$, the above equation condenses to

$$ \left(p+\frac{a}{V^{2}}\right) V=R T $$

$$ \Rightarrow \quad p V=R T-\frac{a}{V} $$

Eq. (i) is a straight equation between $p V$ and $\frac{1}{V}$ whose slope is c $-a^{\prime}$. Equating with slope of the straight line given in the graph.

$$ \begin{array}{rlrl} & & -a & =\frac{20.1-21.6}{3-2}=-1.5 \ \Rightarrow & a & =1.5 \end{array} $$



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