If the number of moles present in a gas and its mass is known, then
the relative molecular mass of a gas can be calculated.
Syllabus reference S1.5.4
Structure 1.5.4 - The relationship between the pressure, volume, temperature and amount of an ideal gas is shown in the ideal gas equation PV = nRT and the combined gas law P1V1/T1= P2V2/T2.
- Solve problems relating to the ideal gas equation.
Guidance
- Units of volume and pressure should be SI only. The value of the gas constant R, the ideal gas equation, and the combined gas law, are given in the data booklet.
Tools and links
- Tool 1, Inquiry 2 - How can the ideal gas law be used to calculate the molar mass of a gas from experimental data?
Ideal gas equation
The equation of state refers to a fixed mass of gas. From Avogadro's law
we know that the same volume of all gases contain the same number of moles
and from this, it follows that the volume is proportional to the number of
moles.
Volume ∝ number of moles (n)
These two equations can be combined to obtain an expression involving all
the quantities:
After rearrangement, for 'n' moles of gas the proportionality constant is
called the Universal Gas Constant and is given the symbol 'R'
This gives the ideal gas equation:
Ideal Gas Equation: PV = nRT
where:
- P = pressure in Pa
- V = volume in m3
- n = number of moles of gas
- R = Universal Gas constant = 8.314 JK-1mol-1
- T = the absolute temperature in Kelvin
It is often more convenient to express the pressure in kPa and the volume
in litres (dm3). This leaves the value of R the same (see below).
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Example: Calculate
the number of moles of gas present in 2.6 dm3 at a pressure
of 1.01 x 105 Pa and 300 K.
PV = nRT
2.6 dm3 = 0.0026 m3
0.0026 x 1.01 x 105 = n x 8.314 x 300
n = 0.0026 x 1.01 x 105 / 8.314 x 300
n = 0.105 moles
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There are several units used for gas volume, gas pressure and temperature.
It is important to be consistent with the use of units when carrying out gas
law calculations. The Syllabus states that SI units will be used wherever
possible.
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Universal gas constant - R
Although called "Universal", its value depends on the units used
for P, V and T.
With the SI units of metres, kilograms, Kelvin and Joules, using P, V and
T values at STP gives:
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PV=nRT |
| therefore: |
R=PV/nT |
| for 1 mole of gas at STP (using accepted values of P = 1.00
x 105 Pa, V = 0.02271 m3, T = 273.15 K) |
| R = |
(1.00 x 105) x 0.02271)/273.15 |
| R = |
8.314 J K-1 mol-1 |
In chemistry, the units of volume used are the decimetre cubed (dm3)
and pressure in kiloPascals (kPa), so one unit is 100x greater and the other
100x smaller than the SI equivalent. Consequently the differences in the product,
PV, both cancel out (multiplying AND dividing by 1000), so that the final
value for R is the same as in SI units.
The Universal gas constant, R, calculated using atmospheres Pressure and
volume in litres, then:
| |
PV=nRT |
| |
R=PV/nT |
| at STP: |
P = 1 atm, V = 22.7 dm3, T = 273 |
| |
n = 1 |
| |
R = 0.0821 dm3 atm mol-1
K-1 |
There are, of course, several other values of R, as there are several ways of
measuring both the volume and the pressure of a gas.
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SI units and 'R'
The SI units of P, V and T give rise to the previously used value for the
universal gas constant, R = 8.314 J K-1 mol-1.
How does this happen when chemists do not use these SI units?
Remember:
1 litre = 1 dm3 = 1000 cm3
Consequently, if litres are used in the Ideal Gas equation then the pressure
units must also be divided by 1000 (as PV = constant). Pressure is measured
in Pa or Nm-1, and so the unit of the kPa correct for the difference
in volume units.
Atmospheric pressure in Pa = 1.00 x 105 Pa
Atmospheric pressure in kPa = 1.00 x 102 kPa
Provided that you are consistent with the application of units there will
be no problem. It is always a good idea when carrying out calculations to
look at the value of your answer and ask yourself, "does it seem reasonable?"
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Worked examples
Q154-01 What volume is needed
to store 50 moles of an ideal gas at 15 atmospheres and 25 ºC?
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PV = nRT
Convert pressure to kPa and temperature to Kelvin.
P = (15 x 100) = 1500 kPa, n = 50 mol, T = (273 + 25) = 298 K, R
= 8.314 J K-1 mol-1
V = nRT/P = (50 x 8.314 x 298)/1500
Therefore volume needed = 82.6 dm3
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Q154-02 What pressure will
be exerted by 200 moles of hydrogen gas in a 7.5L cylinder at 20ºC?
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PV = nRT
Convert temperature to Kelvin.
V = 7.5 dm3, n = 200 moles, T = (273 + 20) = 293 K,
P = nRT/V
Pressure exerted = (200 x 8.314 x 293)/7.5
Therefore pressure exerted = 64960 kPa
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Q154-03 The temperature in
Kelvin of 2 dm
3 of an ideal gas is doubled and its pressure is increased
by a factor of 4. Calculate the final volume of the gas.
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The ideal gas equation: PV = nRT
Shows that when the temperature is doubled so the value of the volume
must double also, ie a twofold increase (initial volume x 2).
However, when nRT is constant, a fourfold increase in pressure results
in a fourfold (x4) decrease in volume (= initial volume/4)
Combining the two effects (that of the temperature and the pressure
change) the volume increases by a factor of 2 and decreases by a factor
of 4 (volume = initial volume x 2/4)
Therefore the overall change in gas volume is a twofold decrease
(volume = initial volume/2)
Final volume is therefore 2/2 = 1 dm3
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Q154-04 Calculate the volume
occupied by 0.01 moles of hydrogen gas at 20ºC and atmospheric pressure.
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Gas molar volume at STP = 22.7 dm3
0.01 moles of gas occupies 0.01 x 22.7 = 0.227 dm3
20ºC is equivalent to 20 + 273 K = 293 K
Using V1/T1 = V2/T2 (where
V1 and T1 are under STP conditions)
0.227/273 = V2/293
Therefore V2 = (0.227 x 293)/273 = 0.2436 dm3
= 243.6 cm3
Therefore 0.01 moles of hydrogen occupies 243.6
cm3 at 20ºC
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Q154-05 How many moles of
gas are present in a gas volume of 24 dm
3 at 100ºC and atmospheric
pressure?
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Using the ideal gas equation PV=nRT
100ºC = 373K, R = 8.314, atmospheric pressure = 100 kPa
n = PV/RT= (100 x 24)/(8.314 x 373)
n = 0.774 moles
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Q154-06 For which set
of conditions does a fixed mass of an ideal gas have the greatest volume?
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Temperature
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Pressure
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A
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low
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low
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B
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low
|
high
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C
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high
|
high
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D
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high
|
low
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From the ideal gas equation, PV = nRT
When T is high, V must also be high
but as nRT is equal to a constant value, any increase in P must cause
a decrease in V
Therefore V is high when P is low.
Correct response = D
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Q154-07 A 0.450g sample of
gaseous aluminium chloride occupies a volume of 51.2 cm
3 at 100ºC
and 102 kPa. Calculate its relative molecular mass.
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100ºC is equivalent to 373K, 51.2 cm3 = 0.0512 dm3
Using PV=nRT, n = PV/RT
n = (102 x 0.0512)/(8.314 x 373)
n = 1.684 x 10-3
The mass of 1.684 x 10-3 moles = 0.450 g
Therefore the mass of 1 mole = 0.450/1.684 x 10-3 = 267
The relative molecular mass of aluminium chloride = 267
Note: The formula of aluminium chloride
is AlCl3, however this gives a relative formula mass of
27 +(3 x 35.5) = 133.5 exactly half the calculated value. The Mr value
produced in the above calculation suggests that the correct formula
of aluminium chloride is 2 x AlCl3 or Al2Cl6
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Q154-08 Calculate the number
of moles of helium in a weather balloon measuring 200 m3 at 5ºC and 80
kPa pressure.
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Using PV=nRT
n=PV/RT
Volume = 200 m3 = 2 x 105 dm3
Number of moles of helium = (80 x 2 x 105 )/(8.314 x 278)
Therefore the number of moles of helium = 6923
moles (to 4 significant figures)
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Q154-09 The molar mass of
an unknown gas is to be determined by weighing a sample. As well as its mass
which of the following must be known.
- I. Pressure
- II. Temperature
- III. Volume
- I only
- II only
- I and II only
- I, II, and III
Answer
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To calculate the molar mass of a gas the number of moles must be
known as well as the mass of gas to use the relationship:
Molar mass = Mass/moles
The number of moles is obtained from PV=nRT
Therefore the Pressure, Volume and Temperature must be known.
Correct response = D
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Q154-10 In an experiment
to determine the relative molecular mass of an unknown hydrocarbon, 0.15 g of
the liquid hydrocarbon was injected through a rubber
septum into a gas syringe. The gas syringe was placed
in an oven at 120ºC and left to reach equilibrium. The final volume occupied
by the vapour in the syringe was measured and found to be 67.4 cm
3.
Calculate the relative molecular mass of the hydrocarbon, if the atmospheric
pressure was 101 kPa on that day.
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Temperature 120ºC is equal to 120 + 273 = 393K
Volume 67.4 cm3 = 0.0674 dm3
Using PV = nRT
n = PV/RT
n = (101 x 0.0674)/(8.314 x 393) = 2.083 x 10-3
This number of moles has a mass = 0.15 g
Therefore mass of 1 mole = 0.15/ (2.083 x 10-3) = 72
Therefore the relative molecular mass of the hydrocarbon =
72
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Now test yourself
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