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?

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Van der Waals Equation concept. You can view video lessons to learn Van der Waals Equation. Or if you need more Van der Waals Equation practice, you can also practice Van der Waals Equation practice problems.