Gases and moles

Gas volumes

• It is easier to measure the volume of a gas than its mass.

• The volume of a gas depends on;

• The temperature.• The pressure.• The number of moles present.

Avogadro’s Law• Gases are not very dense, so the size

of gas particles is negligible compared to the distance between them.

• A gas consisting of single atoms, eg Argon, will behave the same as a gas composed of molecules of two atoms such as chlorine.

• Hence Avogadro’s Law;• Equal volumes of gases contain the

same number of particles under the same temperature and pressure.

The Kelvin Temperature Scale

• One degree of the Kelvin scale represents the same change in temperature as one degree on the Celsius scale.

• But it starts at -273 oC.• IE K = C + 273• Eg; A room is at 20 oC, what is its

temperature in K?• T = 20 + 273 = 293K.• NB units are K, not oK!

Standard conditions (STP)

• Standard conditions are defined as;

• A temperature of 273K (0oC)• A pressure of 101kPa (Kilo

Pascals) = 1 atmosphere. • One mole of any gas under

standard conditions will occupy a volume of 22.4 dm3.

A mole of gas!

Equation for stp

No moles =Volume in dm3

22.4

Number of moles

Volume (dm3)

n

V

22.4

Think of the equation as a triangle

At stp;

Number of moles =

Volume (dm3)

22.4

n

V

22.4

n = v/22.4

Rearranging;

Number of moles

Volume

n

V

22.4 V = n22.4

X 22.4

Rearranging;

Calculating the number of moles at stp.

• Number of moles = volume (dm3) / 22.4• Eg; How many moles are there in 150

cm3 of oxygen at stp? • First convert the volume to decimetres.• V = 150/1000 = 0.15 dm3

• Then divide by 22.4.• N = 0.15 / 22.4 = 0.0067 moles.

Calculating the volume of a gas at stp.

• Eg; What is the volume, at stp, of 2.5 moles of carbon dioxide?

• First rearrange the basic equation;• N = V /22.4• So V = N x 22.4• V = 2.5 x 22.4 = 56 dm3.

Room temperature and pressure.

• Room temperature is taken as being 298K (25oC).

• Room pressure is assumed to be 101 kPa (1 atm).

• Under these conditions one mole of any gas occupies a volume of 24 dm3.

Equation for rtp

No moles =Volume in dm3

24

Number of moles

Volume (dm3)

n

V

24

Think of the equation as a triangle

At rtp

Number of moles =

Volume (dm3)

24

n

V

24

n = v/24

Rearranging;

Number of moles

Volume

n

V

24 V = n24

X 24

Rearranging;

Calculating the number of moles at rtp.

• Number of moles = volume (in dm3) / 24• Eg; How many moles are there in 325

cm3 of hydrogen at rtp? • First convert the volume to decimetres.• V = 325 / 1000 = 0.325 dm3

• Then divide by 24• N = 0.325 / 24 = 0.0135 moles.

Calculating the volume of a gas at rtp.

• Eg; What is the volume, at rtp, of 0.75 moles of nitrogen dioxide?

• First rearrange the basic equation;• N = V / 24• So V = N x 24• V = 0.75 x 24 = 18 dm3.

Ideal gas equation

P V = n R T

P is pressure in Pascals (Pa) aka Newton/m2

V is volume in m3. NB 1m3 = 1,000 dm3.

n is the number of moles.

R is the gas constant = 8.31

T is the temperature in Kelvin.

Limitations of the ideal gas equation

• The ideal gas equation makes two assumptions;

• 1) Gas particles have mass, but zero volume.

• 2) There are no internal forces between particles.

• Gases conform to these assumptions unless the pressures are very high and/or the temperatures very low.

Using the ideal gas equation to calculate relative molecular mass.

•n = m/M• Where m = mass present

•M = molecular mass

•M = mRT/PV

Calculations on the ideal gas law

• 512cm3 of a gas has a mass of 1.236g at 30oC and 1atm pressure. What is its molecular mass?

• First write out what you know;• m = 1.236g V = 512cm3 P =1 atm T =

20oC• Then convert the units.• V = 512 x 1x 10-6 = 5.12x10-4 m3

• T = 20 + 273 = 293K.• P = 1 x 101375 = 101375 Nm-2

• Substitute into the ideal gas equation;

•M = mRT/PV• 1.236x8.31 x293 / 101325x 5.12x10-4

• = 58• NB Molecular mass is relative so

should have no units.