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.

### Van der Waals Equation 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:

a_{He} = 0.0341 atm·L^{2}/mol^{2}

b_{He} = 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

P_{ideal} = 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·L^{2}/mol^{2} x [0.3000 mol/0.2000 L]^{2}

Y = 0.0341 atm·L^{2}/mol^{2} x (1.5 mol/L)^{2}

Y = 0.0341 atm·L^{2}/mol^{2} x 2.25 mol^{2}/L^{2}

Y = 0.077 atm

Recombine to find pressure

P = X - Y

P = 32.152 atm - 0.077 atm

P_{non-ideal} = 32.075 atm**Part 3** - Find the difference between ideal and non-ideal conditions

P_{non-ideal} - P_{ideal} = 32.152 atm - 30.55 atm

P_{non-ideal} - P_{ideal} = 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.

### Ideal vs Non-Ideal Gases

An ideal gas is one in which the molecules don't interact with each other and don't take up any space. In an ideal world, collisions between gas molecules are completely elastic. All gases in the real world have molecules with diameters and which interact with each other, so there's always a bit of error involved in using any form of the Ideal Gas Law and van der Waal's equation.

However, noble gases act much like ideal gases because they don't participate in chemical reactions with other gases. Helium, in particular, acts like an ideal gas because each atom is so tiny.

Other gases behave much like ideal gases when they are at low pressures and temperatures. Low pressure means few interactions between gas molecules occur. Low temperature means the gas molecules have less kinetic energy, so they don't move around as much to interact with each other or their container.