JEE Challenger
More from States of Matter
ChemMedAdv 2012 • Paper 1

Calculate van der Waals Constant From PV Plot

Consider 1 mole1\text{ mole} of a van der Waals gas where b=0b = 0 at T=300 KT = 300\text{ K}. The diagram below shows the PVPV versus 1/V1/V relation. Determine the van der Waals parameter aa in atm⋅liter2 mol−2\text{atm}\cdot\text{liter}^2\text{ mol}^{-2}.

Question Diagram 1

Options

A

1.0

B

4.5

C

1.5

Correct
D

3.0

Topics & Concepts

States of MatterReal Gases

Step-by-Step Solution

The van der Waals equation of state for n=1 molen = 1\text{ mole} of a real gas is given by: (P+aV2)(V−b)=RT\left( P + \frac{a}{V^2} \right)(V - b) = RT

Given that b=0b = 0, the equation simplifies to: (P+aV2)V=RT\left( P + \frac{a}{V^2} \right) V = RT

Expanding the expression: PV+aV=RTPV + \frac{a}{V} = RT

Rearranging in terms of PVPV: PV=RT−a(1V)PV = RT - a \left( \frac{1}{V} \right)

This represents a linear equation of the form y=mx+cy = mx + c, where:

  • y=PVy = PV
  • x=1Vx = \frac{1}{V}
  • Slope, m=−am = -a
  • Y-intercept, c=RTc = RT

From the given PVPV versus 1V\frac{1}{V} plot, we can select two points lying on the line:

  1. (x1,y1)=(2.0,21.6)(x_1, y_1) = (2.0, 21.6)
  2. (x2,y2)=(3.0,20.1)(x_2, y_2) = (3.0, 20.1)

The slope mm of the line is calculated as: m=y2−y1x2−x1=20.1−21.63.0−2.0=−1.51.0=−1.5m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{20.1 - 21.6}{3.0 - 2.0} = \frac{-1.5}{1.0} = -1.5

Since m=−am = -a: −a=−1.5  ⟹  a=1.5 atm⋅liter2 mol−2-a = -1.5 \implies a = 1.5\text{ atm}\cdot\text{liter}^2\text{ mol}^{-2}

Thus, the correct option is C.

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