1. The Ideal Gas Equation
pV = nRT

2. The Ideal Gas Equation
Changing the temperature and pressure of a gas will change its volume.
If the volumes of gases are not at stp we need to use the ideal gas equation
What is an “ideal gas”?

3. An Ideal Gas Identical particles in rapid random motion
Particles = hard spheres of negligible size
Particles don’t react when they collide
Collisions between particles are elastic
Kinetic energy before = kinetic energy after
No intermolecular forces

4. The Effect of Pressure At constant temperature
Increasing pressure
Gas compressed into smaller volume
Volume decreases as pressure increases
V is indirectly proportional to p
V  1/p

5. The Effect of Temperature
At constant pressure
Gas increases in volume
Increasing temperature
Volume increases as temperature increases
V is directly proportional to T
V  T

6. The Effect of Number of moles
At constant temperature & pressure
“n” moles
2n moles
Volume increases as number of moles increases
V is directly proportional to n
V  n

7. If we combine these three equations
V T
V  1/p
V  n
V  nT p
R = gas constant
V = RnT p
pV = nRT

8. The Ideal Gas Equation pV = nRT p = pressure (Pa) V = volume (m3)
n = number of moles
R = the gas constant = 8.31JK-1mol-1
T = temperature (K)

9. Converting Units Temperature 0oC = 273K a OC → a + 273K Pressure
1kPa = 1000Pa
a kPa = a x 1000Pa

10. Converting Units Volume 1m = 10 dm = 100 cm 1m3 = 103 dm3 = 1003 cm3
= 1 x 10-3 m3 m3
1cm3 =
= 1 x 10-6 m3

11. What volume is occupied by 0.25 mol of a gas at 200kPa and 27oC?
1. Convert units
200kPa =
200 x 1000 Pa
= 2 x 105 Pa
27oC =
= 300K
2. Rearrange pV = nRT Equation
V = nRT p
V = 0.25 x 8.31 x x 105
V = 3.12 x 10-3 m3

12. At 571K a 0. 6g sample of He occupies a volume of 7
At 571K a 0.6g sample of He occupies a volume of 7.0 dm3, Calculate pressure.
1. Convert mass into moles n=m/Mr
n = 0.6 4
= 0.15
2. Convert units
= 7.0 x 10-3 m3
7.0 dm3 = 7
1000
3. Rearrange pV=nRT Equation
p = 0.15 x 8.31 x x 10-3
p = nRT V
p = 1.02 x 105 Pa

13. 2. Rearrange pV = nRT Equation
0.71g of a gas when contained in a vessel of 0.821dm3 exerted a pressure of 50.65kPa at 227oC. Use these data to calculate Mr of the gas
1. Convert units
0.821dm3 =
= 8.21 x 10-4 m3
0.821/1000 m3
227oC =
= 500K
5.065 x 104 Pa
50.65kPa =
50.65 x 1000 Pa =
2. Rearrange pV = nRT Equation
n = pV RT
n = x 104 x 8.21 x x 500
n = 0.01 mol

14. 3. Calculate Mr using n = m/Mr
0.71g of a gas when contained in a vessel of 0.821dm3 exerted a pressure of 50.65kPa at 227oC. Use these data to calculate Mr of the gas
3. Calculate Mr using n = m/Mr =
Mr = m n
= 70.94
4. Check final answer
Gases are small molecules
– they rarely have Mr values over 100