GCSE OCR Gateway Chemistry C5.1d-e Molar Volume of gases

GCSE OCR Gateway Chemistry C5.1d-e

C5.1d 

To be able to describe the relationship between the amount of a gas (in moles) and its volume (in litres) and vice versa

C5.1e 

To be able to calculate the volumes of gases involved in reactions using the molar gas volume at room temperature and pressure (assumed to be 24dm3  at r.t.p.)

These are higher tier learning objectives.

Molar Gas Volume

Let’s begin looking at the volume of different masses of several common gases.

Gas	Formula	Mass/g	Amount/mol	Volume/dm3	Molar Volume cm3/mol
Hydrogen	H2	1	0.5	12.0	24
Methane	CH4	8	0.5	12.0	24
Oxygen	O2	4	0.25	6.0	24
Nitrogen	N2	3.5	0.25	6.0	24
Carbon dioxide	CO2	5.5	0.125	3.0	24

Just look at the results of this simple analysis: whatever the gas, the volume of one mole of particles is always the same: 24dm³ or 24000cm³ at room temperature and pressure.

How to make a gas molar volume cube

Now here’s an interesting challenge if you want to get a handle on how big this volume is.  Apart from it being roughly half the volume of the average car's petrol tank, why not take the cube root of 24 or 24000cm³: ∛24000 = 28.4 cm and build a box to these dimensions.  Each side needs to be 28.4 cm long.  You’ll need to take down the net used to build a cube so here it is:

You can label each side with details about this volume: the molar volume of any gas at r.t.p. is 24,000cm^3.

This isn’t the only way of determining the volume of one mole of gas.  Here is an alternative, experimental method. 

How to find the molar volume of a gas by experiment

This experiment shows you how to determine the molar volume using a reaction that produces a gas like carbon dioxide or hydrogen.

For example, the reaction between sodium carbonate and hydrochloric acid produced carbon dioxide.

Na₂CO₃    +   2HCl    ⟶   2NaCl  +    H₂O    +  CO₂

So take a given mass of sodium carbonate in an ignition tube and locate it in a flask containing an excess of dilute hydrochloric acid.  Attach the flask via a delivery tube to a gas syringe.  See the diagram below:

Up end the flask to bring the sodium carbonate into contact with the acid and measure the volume of gas (Carbon dioxide) evolved. 

Calculation

Suppose there were 0.2000g of sodium carbonate in the ignition tube.  This reacts with excess hydrochloric acid according to the equation above. 

0.2000g of sodium carbonate is 0.2/106 moles =  0.001887 moles

According to the equation 0.001887 moles of sodium carbonate should produce the same number of moles of carbon dioxide 0.001887 mol. 

Suppose that 46cm³ of carbon dioxide collected in the gas syringe at room temperature and pressure. 

Then the molar volume of carbon dioxide can be calculated:

V_m   =  46/0.001887   =  24,380  cm³/mol

This value approximates well to the standard value of 24,000cm³ at r.t.p.

The molar volume of a gas from the ideal gas equation

You can also get at the molar volume of any gas if you use the ideal gas equation

pV = nRT

rearranging for V/n we obtain

V/n  =  RT/P

where R is the universal gas constant.

Some implications of the molar gas volume

One implication of the molar gas volume being constant at 24dm³ or 24,000cm³ is that equal volumes of all gases contain equal numbers of molecules.  In other words, there is the same number of moles of gas in 10cm³ of methane as there are in 10cm³ of hydrogen as there are in 10cm³ of ethene.

So if an experiment is carried out and it is found that 10 cm³ of nitrogen react completely with 30 cm³ of hydrogen and produce 20 cm³ of ammonia then we can say that 10 moles of nitrogen react with 30 moles of hydrogen producing 20 moles of ammonia and therefore the equation for the reaction has to be  N₂   +   3H₂  ⇌   2NH₃  and this would not work if nitrogen and hydrogen were not diatomic molecules since  N   +    3H   does not give 2NH₃ but just NH_3.

In other words, this argument (I think first proposed by Amadeo Avogadro) justifies the existence of nitrogen and hydrogen as diatomic molecules and not monoatomic molecules.