Exam

JEE Advanced

Year

2012

Paper

Paper 1

Subject

Chemistry

Chapter

States of Matter

JEE Advanced 2012 · Paper 1 · Chemistry · States of Matter

For one mole of a van der Waal's gas when b=0 and T=300 ~K , the PV...

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

Step-by-step solution

$$\begin{array}{l} \left(P+\frac{\mathrm{a}}{V^{2}}\right)(V)=R T \\ P V+a / V=R T ; P V=R T-a(V) \\ y=R T-a(x) \end{array}$$ So, slope $=a=\frac{21.6-20.1}{3-2}=\frac{1.5}{1}=1.5$