The Ideal Gas Equation

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# The Ideal Gas Equation - PowerPoint PPT Presentation

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

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### 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

V T

V  1/p

V  n

• If we combine these three equations

V nT p

R = gas constant

V = RnT p

pV = nRT

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

m3

• 1cm3 =1 1000 000

= 1 x 10-6 m3

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 = 0.6 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 571 7 x 10-3

p = nRT 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 =

= 8.21 x 10-4 m3

0.821/1000 m3

227oC =

227 + 273

= 500K

5.065 x 104 Pa

50.65kPa =

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

= 0.71 0.01

Mr = m n

= 70.94