I have a question regarding ideal gasses and pressures. The end goal is to determine the amount of gas (volume) evolved when a specimen within the vessel has been introducted to heat.

The specimen is kept in a fixed container, so the volume is fixed. As the specimen heats up in a closed system, it will combust and increase gas concentration, and thus increase the pressure. If I am able to determine the pressure of the gas, and I can determine the volume of the vessel, is there a means I could correlate this to the volume of gas evolved?

Use of the ideal gas law, and/or Boyle's law, would reference an external force acting on the gas to change it's volume.pressure etc. I only know the pressure of the gas that exerts on the walls at a fixed volume.


It's not as simple as you may think. There are two main obstacles.

Firstly, look at the combustion of carbon (coal) as an example:

$$C(s)+O_2(g)\to CO_2(g)$$

The stoichiometry tells us that a mere $12\ \mathrm{g}$ of coal requires about $22.4\ \mathrm{L}$ of oxygen (at Standard conditions), that is over $100\ \mathrm{L}$ of air!

Secondly, the reaction equation also tell us that the same volume of gas (oxygen) is being consumed as is being generated as carbon dioxide. So there no net increase in gas, thus no net increase in volume. Nothing to measure here!

Or take the example of a generic, solid hydrocarbon $C_nH_{n+2}$ and assume the water vapour is allowed to condense to water ($(l)$), by cooling the device back to room temperature:

$$C_nH_{n+2}(s)+\frac{3n+1}{2}O_2(g)\to nCO_2(g)+\frac{n+2}{2}H_2O(l)$$

Neglecting the volumes of the solid hydrocarbon and the water, here the gas consumed in the combustion is actually more voluminous than the carbon dioxide generated, because always:


Of course there are substances that generate gas without consuming any, when heated. The burning of limestone springs to mind:

$$CaCO_3(s) \to CaO(s)+CO_2(g)$$

This requires temperatures in excess of $1000\ \mathrm{K}$ though.

Suppose we have such a device, of volume $V$, at atmospheric pressure $P_0$, at room temperature $T_0$ and loaded with a known quantity of combustible material and enough air.

Using $P_0V=n_0RT_0$ we know the initial number of moles $n_0$ of gas inside the apparatus.

Now we carry out the "burn" and when complete allow cooling back to $T_0$. The pressure will now be different, say $P$.

Using $PV=n_1RT_0$ we can determine $n_1$.

The difference $\Delta n=n_1-n_0$ is the number of moles of gas that was created or disappeared during the combustion of the known quantity of combustible material. How useful that information is will depend to some extent on one's knowledge of stoichiometry.

  • $\begingroup$ Say I were to neglect stoichiometry entirely, (and maybe I'm missing one of your points here) and focus only on absolute amount of pressure that's created. If a 1L vessel starts at 1 atm, and upon heating of sample, ends at 2 atm. Given the known volume of the container, the increase in pressure, and (if beneficial the mass of sample combusted), is there any equational relationship between these parameters that could get to me gaseous volume? $\endgroup$ – Kevin K Nov 28 '17 at 19:27
  • $\begingroup$ Assuming all gases behave more or less ideally, you can simply use the Ideal Gas Law: $PV=nRT$. This should allow you to evaluate the increase (or even decrease) of the number of moles of gas, in the process studied. But you'd still need stoichiometry in any case. $\endgroup$ – Gert Nov 28 '17 at 19:30
  • $\begingroup$ I'll put in an addendum to my answer. $\endgroup$ – Gert Nov 28 '17 at 19:41
  • $\begingroup$ Yeah, any way I'm working it I'm running into a problem of limited information. The sample that is heated is of unknown composition, so stoichiometry is ruled out. The best I am able to do is determine the mass of the gas produced and see if I can come up with some pressure/mass relationship. $\endgroup$ – Kevin K Nov 28 '17 at 19:48
  • $\begingroup$ Yes but this kind of experiment can help understanding the unknown stoichiometry of a sample of known mass. That's the whole purpose of it, IMO. $\endgroup$ – Gert Nov 28 '17 at 19:51

Your Answer

