Van der Waals Equation Video Lessons

Concept

# Problem: In Sample Exercise 10.16 in the textbook, we found that one mole of Cl2 confined to 22.41 L at 0 oC deviated slightly from ideal behavior. Calculate the pressure exerted by 1.00 mol  Cl2 confined to a smaller volume, 6.00 L , at 25 oC.Why is the difference between the result for an ideal gas and that calculated using van der Waals equation greater when the gas is confined to 6.00 L compared to 22.4 L?

###### FREE Expert Solution

Part A. We’re being asked to calculate the pressure exerted by a chlorine gas that deviates from ideal behavior. For this, we shall use the Van der Waal’s equation

The Van der Waals equation is shown below:

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

P = pressure, atm
V = volume, L
n = # of moles, mol
R = gas constant = 0.08206 (Latm)/(molK)
T = temperature, K
a = polarity coefficient
= size coefficient

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###### Problem Details

In Sample Exercise 10.16 in the textbook, we found that one mole of Cl2 confined to 22.41 L at 0 oC deviated slightly from ideal behavior. Calculate the pressure exerted by 1.00 mol  Cl2 confined to a smaller volume, 6.00 L , at 25 oC.

Why is the difference between the result for an ideal gas and that calculated using van der Waals equation greater when the gas is confined to 6.00 L compared to 22.4 L?