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
Increasing pressure
Gas compressed into
smaller volume
Volume decreases as pressure increases
V is indirectly proportional to p
V  1/p
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
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
• If we combine these three equations
Vn
V T
V  1/p
V  nT
p
V = RnT
p
R = gas constant
pV = nRT
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)
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 =
•
1cm3
1 m3
1000
= 1 x 10-3 m3
3
m
=
1
= 1 x 10-6 m3
1000 000
What volume is occupied by 0.25 mol of a
gas at 200kPa and 27oC?
1. Convert units
200kPa = 200 x 1000 Pa
27oC =
27 + 273
= 2 x 105 Pa
= 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 = 0.6
= 0.15
4
2. Convert units
-3 m3
3
=
7.0
x
10
7.0 dm = 7
1000
3. Rearrange pV=nRT Equation
p = 0.15 x 8.31 x 571
p = nRT
-3
7
x
10
V
p = 1.02 x 105 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
1. Convert units
0.821dm3 = 0.821/1000 m3
227oC = 227 + 273
50.65kPa =
= 8.21 x 10-4 m3
= 500K
4 Pa
5.065
x
10
50.65 x 1000 Pa =
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
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
=
0.71
0.01
= 70.94
4. Check final answer
Gases are small molecules
– they rarely have Mr values over 100