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Ideal Gas vs Non-ideal Gas Example Problem

Van Der Waal's Equation Example Problem

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At low temperatures, real gases behave as ideal gases.

At low temperatures, real gases behave as ideal gases.

Tetra Images - Jessica Peterson, Getty Images

This example problem demonstrates how to calculate the pressure of a gas system using the ideal gas law and the van der Waal's equation. It also demonstrates the difference between an ideal gas and a non-ideal gas.

Problem:

Calculate the pressure exerted by 0.3000 mol of helium in a 0.2000 L container at -25 °C using

a. ideal gas law
b. van der Waal's equation

What is the difference between the non-ideal and ideal gases?

Given:

aHe = 0.0341 atm·L2/mol2
bHe = 0.0237 L·mol

Solution

Part 1: Ideal Gas Law

The ideal gas law is expressed by the formula:

PV = nRT

where
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant = 0.08206 L·atm/mol·K
T = absolute temperature

Find absolute temperature

T = °C + 273.15
T = -25 + 273.15
T = 248.15 K

Find the pressure

PV = nRT
P = nRT/V
P = (0.3000 mol)(0.08206 L·atm/mol·K)(248.15)/0.2000 L
Pideal = 30.55 atm

Part 2: Van der Waal's Equation

Van der Waal's equation is expressed by the formula

P + a(n/V)2 = nRT/(V-nb)

where
P = pressure
V = volume
n = number of moles of gas
a = attraction between individual gas particles
b = average volume of individual gas particles
R = ideal gas constant = 0.08206 L·atm/mol·K
T = absolute temperature

Solve for pressure

P = nRT/(V-nb) - a(n/V)2

To make the math easier to follow, the equation will be broken into two parts where

P = X - Y

where
X = nRT/(V-nb)
Y = a(n/V)2

X = P = nRT/(V-nb)
X = (0.3000 mol)(0.08206 L·atm/mol·K)(248.15)/[0.2000 L - (0.3000 mol)(0.0237 L/mol)]
X = 6.109 L·atm/(0.2000 L - .007 L)
X = 6.109 L·atm/0.19 L
X = 32.152 atm

Y = a(n/V)2
Y = 0.0341 atm·L2/mol2 x [0.3000 mol/0.2000 L]2
Y = 0.0341 atm·L2/mol2 x (1.5 mol/L)2
Y = 0.0341 atm·L2/mol2 x 2.25 mol2/L2
Y = 0.077 atm

Recombine to find pressure

P = X - Y
P = 32.152 atm - 0.077 atm
Pnon-ideal = 32.075 atm

Part 3 - Find the difference between ideal and non-ideal conditions

Pnon-ideal - Pideal = 32.152 atm - 30.55 atm
Pnon-ideal - Pideal = 1.602 atm

Answer:

The pressure for the ideal gas is 30.55 atm and the pressure for van der Waal's equation of the non-ideal gas was 32.152 atm. The non-ideal gas had a greater pressure by 1.602 atm.

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