The Ideal Gas Equation

pV = nRT

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”?

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

The Effect of Pressure • At constant temperature

Volume decreases as pressure increases

Gas compressed into smaller volume

Increasing pressure

V 1/p

V is indirectly proportional to p

The Effect of Temperature• At constant pressure

Volume increases as temperature increases

Gas increases in volumeIncreasing temperature

V is directly proportional to T

V T

The Effect of Number of moles• At constant temperature & pressure

Volume increases as number of moles increases

2n moles“n” moles

V is directly proportional to n

V n

• If we combine these three equations

V 1/pV TV n

V nT p

V = RnT p

pV = nRT

R = gas constant

The Ideal Gas Equation

• p = pressure (Pa)

• V = volume (m3)

• n = number of moles

• R = the gas constant = 8.31JK-1mol-1

• T = temperature (K)

pV = nRT

Converting Units

• Temperature

• 0oC = 273K

• a OC → a + 273K

• Pressure

• 1kPa = 1000Pa

• a kPa = a x 1000Pa

Converting Units• Volume

• 1m = 10 dm = 100 cm

• 1m3 = 103 dm3 = 1003 cm3

• 1m3 = 1000 dm3 = 1 000 000 cm3

• 1dm3 = 1 1000

m3 = 1 x 10-3 m3

• 1cm3 = 1 1000 000

m3

= 1 x 10-6 m3

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 = 27 + 273 = 300K

2. Rearrange pV = nRT Equation

V = nRT p

V = 0.25 x 8.31 x 300 2 x 105

V = 3.12 x 10-3 m3

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 = 0.64 = 0.15

7.0 dm3 = 7 1000

= 7.0 x 10-3 m3

3. Rearrange pV=nRT Equation

p = nRT V

p = 0.15 x 8.31 x 571 7 x 10-3

p = 1.02 x 105 Pa

2. Convert units

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 units0.821dm3 = 0.821/1000 m3 = 8.21 x 10-4 m3

227oC = 227 + 273 = 500K

2. Rearrange pV = nRT Equation

n = pV RT

n = 5.065 x 104 x 8.21 x 10-4 8.31 x 500

n = 0.01 mol

50.65kPa = 50.65 x 1000 Pa = 5.065 x 104 Pa

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 = mn

= 0.710.01

= 70.94

4. Check final answer

Gases are small molecules

– they rarely have Mr values over 100