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:

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.