**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.