How do you calculate the pressure in a vessel (dubbed pv from here on) based on the “normal” volume of a gas to the volume of gas in the pv?
Say I’ve got a 10cc pellet of dry ice (this particular pellet of dry ice expands to 554x that as a vapor), and I put that pellet in a 1 liter container. How do I calculate how much pressure that container will be subject to?
Also, in converting that number to psi, would I square, cube, or simply multiply the conversion factor?
Thanks for posting this. I have been trying to figure out the exact same thing but didn’t know how to phrase the question.
I am going to take an uneducated stab just for fun.
1 liter=1000 cc
pellet =10ccX554=5540 cc’s
6540 total cc’s
1000cc=14.7
6540=X 6540X14.7/1000=96.138 Not sure how air temp would affect this.
A lot of people would use the universal gas law, PV=nRT. The volume is 1 liter, the number of moles is approximately 0.34, assuming a 1.5 g/cc density for dry ice, R is the universal gas constant, which is this case would be 0.08206 L atm /mol K, and T is temperature in degrees Kelvin. That accounts for the pressure from the dry ice. If the container was at atmospheric pressure when you started that would have to be added in, but the total pressure will be the sum of the partial pressures.
Or to put it more simply, if you have the same amount of gas at the same temperature in two containers, the container with half the volume has twice the pressure.
The good part about the universal gas law is that the result is in atmospheres. You can multiply by your preferred units. At zero C I come up with 7.6 atmospheres, or around 112 psig. Again, that’s just the CO2. Add another 14.696 psig for the nitrogen and oxygen and such.
thanks for the answers, all. Looks like it’s nowhere near as simple as I’d hoped.
So, how do you find out the number of moles? Is that a constant for each element or compound (e.g. graphite and diamond would have the same #, but cesium and rubidium would have different #'s)?
Moles are Avogadro’s number of molecules and can be calculated by dividing grams by the molecular weight.
The result is in whatever units correspond to the constant you use. It’s nothing inherent to the formula itself.
And while we’re at it, the formula is also much simpler if you forget about moles, and just do it in terms of number of particles: PV = NkT, where N is the number of particles, and k is Boltzmann’s constant. I don’t know why chemists insist on using moles; scientific notation really isn’t that scary, and it saves a layer of complication.
If you like solving for things like this, I highly encourage you to take up Chemical engineering. You’ll find good jobs too since there is a natural gas boom.
Back to the question.
Here’s what the symbols mean : V = Volume, P = Pressure, M = mass, RHO = density, N = Number of moles, MW = Molecular Weight, R = Universal Gas Constant, T = Absolute temperature
Here’s what the subscripts mean: CO2 = dry ice (assumed pure), AIR = air that fills up the balance volume of the vessel, V = Vessel, amb = Ambient temperature (assumed to be same initially and when the system reaches equilibrium), ATM = atmospheric pressure (assumed initially), FINAL = final stage after equilibrium i.e. the CO2 evaporates and the vessel temperature comes to equilibrium with ambient
So with that
1> Initially (Calculate the number of moles in the system - assume the total volume consists of the volume of solid co2 and the rest is air. Assume the air temperature is ambient and pressure is atmospheric)
V[sub]V[/sub] = V[sub]CO2[/sub] + V[sub]AIR[/sub]
=> V[sub]AIR[/sub] = V[sub]V[/sub] - V[sub]CO2[/sub] …(1)
N[sub]AIR[/sub] = (P[sub]ATM[/sub] x V[sub]AIR[/sub])/ (R x T[sub]AMB[/sub]) …(2)
Using (1), (2) reduces to
N[sub]AIR[/sub] = (P[sub]ATM[/sub]/(R x T[sub]AMB[/sub])) x ( V[sub]V[/sub] - V[sub]CO2[/sub]) …(3)
Now:
Mass[sub]CO2[/sub] = RHO[sub]CO2[/sub] x V[sub]CO2[/sub] …(4)
So moles of CO2, N[sub]CO2[/sub] = Mass[sub]CO2[/sub] / MW[sub]CO2[/sub] …(5)
Using (4), (5) reduces to
N[sub]CO2[/sub] = RHO[sub]CO2[/sub] x V[sub]CO2[/sub] / MW[sub]CO2[/sub] …(6)
Equation (3) gives the moles of air and equation (6) gives the moles of CO2
2> Now look at the final stage when it has equilibriated (assume ideal gas law although CO2 is non ideal)
P[sub]FINAL[/sub] x V[sub]V[/sub] = (N[sub]CO2[/sub] + N[sub]AIR[/sub]) x R x T[sub]AMB[/sub] … (7)
Substituting from Equation N[sub]AIR[/sub] from (3) and N[sub]CO2[/sub] from (6)
P[sub]FINAL[/sub] x V[sub]V[/sub] = (RHO[sub]CO2[/sub] x V[sub]CO2[/sub] / MW[sub]CO2[/sub] + (P[sub]ATM[/sub]/(R x T[sub]AMB[/sub])) x ( V[sub]V[/sub] - V[sub]CO2[/sub])) x R x T …(8)
Dividing both sides of equation (8) by V[sub]V[/sub] and rearranging, we have
P[sub]FINAL[/sub] = (RHO[sub]CO2[/sub]/MW[sub]CO2[/sub])x(V[sub]CO2[/sub]/V[sub]V[/sub]) x R x T[sub]AMB[/sub] + P[sub]ATM[/sub](1 - V[sub]CO2[/sub]/V[sub]V[/sub])
Now - put in the values (Assume T[sub]AMB[/sub] = 300 K or 80 F)
RHO[sub]CO2[/sub] = 1.6 g/cm3 (it can vary between 1.4 and 1.6 per wiki)
MW[sub]CO2[/sub] = 44 44.01 g/mol
V[sub]CO2[/sub] = 10 cm3, V[sub]V[/sub] = 1000 cm3
R = 82 (cm3 atm)/( K mol)
P[sub]ATM[/sub] = 1 atm
T[sub]AMB[/sub] = 300 K
Then
P[sub]FINAL[/sub] = 9.94 atm = 146 PSIA = 1007 kPa
I hope I did not make a mistake in the calcs since it is frustrating to type in code
I do not know about Chemists, but chemical engineers use it for the following reasons:
1> Vapors are ideal only for a small window of (T P) - Non ideal gases are traditionally represented by Peng Robinson or Soave Redlich Kwong or other equations of state in which traditional units are molar basis
2> Molar fractions equal volume fractions in gases
3> Most thermodynamic properties : Molar Volume, Molar Enthalpy, Gibbs free energy … databases are in molar units
We calculate the production of a refinery (or other chemical processes) in moles or kilo-moles or lb-moles and it is easy to visualize how the number of moles of carbon remain unchanged and how they change forms.
On similar lines, it can be said that since the velocity of light is the fundamental unit - physicists should stop using the meter (or foot) and start representing length as c.t (We know that’s not happening)
Point 3 is basically just the same as Point 1: It’s traditional. And Point 2 is equally true for numbers of particles as it is for numbers of moles.
In my opinion, the molar method is more intuitive and easy to use.
I came up with the answer using molar methods. You didn’t. - QED