In general, use the equation of state of an ideal gas,

PV = nkT

where

P = Pressure
V = Volume
n = the number of atoms or molecules
T = Temperature in degrees Kelvin
k = Boltzmann’s constant, a number selected to make the answer come out in whatever units you are using.

The Kelvin temperature scale is the same as the Celsius, except the zero point is at Absolute Zero, not at the freezing point of water.

One minor correction. The pint of gas created by the charge in a 9mm bullet at “very high” temperatures, instead of freezing. There is a much higher volume at high termperatures.

For example, I looked up the amount of gunpowder in a 9mm round. it is about a 1/3 of a gram. If we make the assumption that this is all carbon (that the oxidizer mass is negligible) we will wind up with 1/36 of a mole of CO2 which at STP is 22.4/36 = 0.62 liters of gas.

Of course, this pint of gas at freezing temperatures and 1 atmosphere is crammed into a tiny volume at very high pressure and temperature. A great deal of energy is available to move the bullet.

Hodgon Powder seems to be an open organization. I’d ask them and I’d bet they would figure it out.

C has a gram-molecular weight of 12, so 1 gram of C yields 1/12 of a mole of CO2. A molecular gas is mostly empty space, so the oxygen is irrelevant, it is only the number of molecules that matters, not how big they are. A mole of any gas occupies 22.4 liters at Standard Temperature and Pressure, so the gas produced is equivalent to 1.9 liters at STP.

So for just a very rough estimate, I would say a typical bullet charge produces a couple of quarts of gas at STP. Of course, in the barrel, that gas is extremely hot and highly compressed, so the pressure is enormous.

For example, an expanding gas tends to cool, which causes it to contract, (or not expand as fast). If you can keep the heat from escaping, all the energy can go into the expansion. This is called an isothermal process. But most expansions are at least partly adiabatic which means some energy is lost to cooling as the gas expands.

In the case of bullet ballistics, the situation is very complex, and it depends on a lot of things, such as how long the bullet is affected by friction in the barrel, how fast the powder burns, how fast the air in the barrel (which opposes the bullet) can get out of the way, and a host of other factors, such as the residual temperature and pressure of the gases after the bullet is on its way, and how much of that pressure is siphoned off to operate the mechanism (an M-14 is semi-automatic). Usually the manufacturers select the combustion rate of the powder so it is all burned up before the bullet exits the barrel. They usually tune the mix for the barrel length the round is most likely to encounter in the market, or in combat. The acceleration of the projectile is determined by how much powder has burned at every point along the way and how much room there is in the expanding cylindrical volume between bullet and chamber at every point along the way.

I remember one of the first problems I had in my Fortran class was to calculate bullet speed, location, and barrel pressure in the barrel of an M-14 rifle during the entire path of the bullet down the barrel. I was provided with the combustion properties of the powder. The speed reached a maximum at the muzzle, for obvious reasons, but the accleration was maximum in just the first tiny fraction of the journey (i.e., most of the work was done long before the powder was used up, and long before the bullet left the muzzle). I remember the curve for barrel pressure peaked when the bullet was about halfway out, but speed climbed quickly to near-max then leveled off long prior to exiting the muzzle.
Of course, the ammo was tuned to that piece, in another weapon, the performance might be very different. Ballistics weenies are always experimenting with different “loads” to optimize the performance of different rounds for different weapons and applications.

However, it is possible to calculate the volume of gas produced quite easily, IF you know the chemistry involved. Gunpowder is made up up carbon, which does the burning, and some compound like saltpeter (KNO3)which provides the oxygens needed to burn the carbon to CO2(all the other stuff is just traces of material used to help catalyze and moderate the reaction). If you can find out the chemical composition of the gunpowder, and how much carbon and saltpeter there are in each round, you can calculate the volume of CO2 produced. Of course, you will have to reduce everything to standard temperature and pressure (one Atmosphere, 0 deg C). Gases expand to fit any volume available for them.

If the manufacturers did their engineering right, most of the carbon will be converted to carbon dioxide and just enough oxygen will be left over to combine with the other substances in the propellant. I assume the other materials take up only a small proportion of the propellant, by mass. You will need the chemical formula for the combusion process to determine how much KNO3 was needed, and how the sulfur and other reactants interacted to produce the burn and to determine the oxides of the waste products.

Most commercial rounds are designed for the typical weapon they are likely to be used in, with a safety factor built in to guard against an old or damaged barrel from bursting. I would imagine a .22 or .38 would have as much of the chemical energy available in the propellant utilized with as little left over as possible, provided it was fired from a standard piece..

Hint: if you fire either of those rounds in a short-barreled weapon, note the muzzle flash. Pistol (.22) or snub-nose (.38) shooters can afford to go with a lighter load to reduce recoil, flash, smoke and report without affecting muzzle velocity. This is why putting a .357 Magnum round in a snub-nosed revolver is pointless. Most of the energy is wasted.