# How much gas is inside a pressurised gas cylinder?

Given a 19kg Liquid Petroleum Gas (LPG) cylinder, with this specification:

• LPG inside the cylinder weighs 19kg.
• The internal volume of the cylinder is 44.5 litres.
• The LPG is a mixture of propane gas (60%) and butane (40%).

I would like to calculate the volume of LPG contained in the cylinder. This is the volume the pressurised gas will uncompress to, if it escapes the cylinder.

I have looked at the Ideal Gas Law:

Pressure x Volume = Number of Moles x Gas constant x Temperature

But I cannot use it as I do not know about the pressure inside the cylinder and volume is the unknown I am interested in.

How do I proceed with this problem?

The cylinder contains $\pu{19kg}$ of LPG, $\pu{60\%}$ of which is propane (presumably by mass) so we have $\pu{11.4kg}$ propane and the rest is butane.

The molar mass of propane $\ce{C3H8}$ is $\pu{3*12+8*1 = 44 g/mol}$, so we can work out how many moles of propane there are. Then do the same for butane.

Now we use the ideal gas law:

But I cannot use it as I do not know about the pressure inside the cylinder and volume is the unknown I am interested in.

You are not interested in the pressure inside the cylinder. You are interested in what the volume is when the LPG uncompresses. Therefore the pressure you need to use is atmospheric pressure.

If you apply the formula for the ideal gas law $PV=nRT$ you will find that $1$ mole of gas occupies about $\pu{24 L}$ at $\pu{25^oC}$ ($\pu{298K}$) and atmospheric pressure. That is a good shortcut to remember for cases like this.

Note that the ideal gas law only works when the conditions are in the range where the gas behaves like an ideal gas. You know the volume of the cylinder, but you cannot use the ideal gas law to find the pressure inside the cylinder, because the gas behaves non-ideally at high pressure (It liquifies!)

LPG inside the cylinder weighs $$\mathrm{19\,kg}$$.

The LPG is a mixture of propane gas (60%) and butane (40%).

With other words, you know:

• the total mass of the gas mixture
• the mass fractions of propane and butane
• the molar masses of both gases
• the molar volume of a gas under standard conditions
• so, how do I calculate the volume then? Do I use the Ideal Gas Law? – Chirayu Shishodiya Mar 10 '15 at 18:45
• @ChirayuShishodiya We have a policy here that says: "Homework questions should be hinted at, and are not to be answered directly." You can learn more here: meta.chemistry.stackexchange.com/questions/141/…‎on-chemistry-stack-exchange – M.A.R. Mar 10 '15 at 19:31
• Sorry that wasn't enough of a hint for me, hence I was trying to clarify. Besides, this is not a homework question. – Chirayu Shishodiya Mar 10 '15 at 21:46

Ideal Gas Law only works when a gas behaves like an ideal gas. It cannot be used in this case for finding pressure inside the cylinder as LPG behaves non-ideally at high pressure (it liquefies). Thanks to @steveverrill for pointing this out.

So, we will use Ideal Gas Law to find out what the volume of LPG will be at normal atmospheric pressure (1 atm) and a temperature of 20 degrees Celsius, when it is still a gas:

A 19kg cylinder contains 19kg of LPG.
Mass of propane                    = 60% of 19kg = 11.4kg = 11400g
Mass of butane                     = 40% of 19kg = 7.6kg  = 7600g
Molar mass of propane              = 44.1g/mol
Number of moles of propane         = 11400g / 44.1g/mol = 258.5mol
Molar mass of butane               = 58.12g/mol
Number of moles of butane          = 7600g / 58.12g/mol = 130.8mol
Total number of moles in 19kg LPG  = 258.5 + 130.8 = 389.3mol

Pressure x Volume = Number of moles x Gas constant x Temperature
1 atm x V = 389.3mol x 0.00008205736 m3 atm K-1 mol-1 x 293.15 K
V = 9.365 m3


So, there is 9.365 cubic metres of LPG at normal atmospheric pressure and temperature of 20 degrees Celsius.

Let me give you a simple method of calculation:
Total weight of LPG in the cylinder: 19 kg
Density of LPG in vapour state at atmospheric conditions: 2 kg/m3 approx.
Hence, the volume of LPG in gaseous state is: 19/2 = 9.5 m3