For one mole of a van der Waals gas when b=0 and T=300K, the PV vs. 1/V plot is shown below. The value of the van der Waals constant a (atm.liter2mol−2) is
(A)1.0
(B)4.5
(C)1.5✓ Correct
(D)3.0
Answer: (C)
Solution
For a van der Waals gas with b=0, the equation becomes (P+V2a)V=RT, which rearranges to PV=−a×V1+RT. This is a linear equation in PV vs. 1/V with slope =−a and intercept =RT. From the graph, the slope is tan(π−θ)=−a, where θ is the angle the line makes with the positive x-axis. Reading from the graph: tanθ=3−221.6−20.1=1.5 or tanθ=3−024.6−20.1=1.5. Therefore, a=1.5 atm·L²·mol⁻².
JEE Advanced · 2012 · 08 Apr 2012 · Paper 1 · Single correct option