JEE Advanced · Chemistry · 5. States of Matter
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 :

- A \(1.0\)
- B \(4.5\)
- C \(1.5\)
- D \(3.0\)
Answer & Solution
Correct Answer
(C) \(1.5\)
Step-by-step Solution
Detailed explanation

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