Peter has more patience than I do, hats off.
For bonus points, you can actually calculate how much (by mass) of the CO2
will be solid and how much is still vapor. I would assume that the
entirety of the CO2 cartridge is saturated liquid.
On Sun, Aug 9, 2015 at 9:02 PM, Peter Fairbrother <zenadsl6186@xxxxxxxxx>
wrote:
On 08/08/15 23:55, John Dom wrote:
I would like to be able to see the equations which are involved.
Enthalpy balance and heat of evaporation, gas law etc..
hokay. will be long and complex ..
Enthalpy is a "thermodynamic potential", a quantity which can be measured
and/or calculated for a particular system, eg a bucket of water, a cubic
meter of air, or the CO2 gas in the cartridge. It is defined as the
internal energy of the system, plus PV, the product of pressure and volume.
It is a universal quantity, possessed by all substances - eg the unleashed
explosive power of TNT would be considered part of the internal energy and
thus part of the enthalpy of a block of TNT: the PV and heat of the hot gas
left immediately after a detonation would also be parts of the enthalpy.
In SI enthalpy is measured in Joules, or for substances in particular
conditions of pressure and volume it is often measured in Joules per gram
(or per mole) of substance.
An isoenthalpic (isenthalpic) process is a process in which the enthalpy
does not change during the process.
Generally, this means no mechanical work is done to the surroundings, and
no energy is transferred to or from the system in the form of heat.
The process of expansion in this case is isenthalpic, because no work is
done to the surroundings, only to the gas itself; and we assume it happens
quickly enough that no significant heat transfer takes place.
The process is isenthalpic, so you know the final enthalpy and
density. Look up P and T on the NIST webbook.
In cases like this, where a substance does not behave like an ideal or
even a non-ideal gas, and especially where it is close to its critical
and/or triple points, and the actual equations of state are verra
complicated Captain involving something like 18 constants; it is very much
easier to use NIST or similar data, and the results will be far more
accurate than using the more normal theoretical simplifications.
So. Final density is 8g/l (ignoring the cartridge volume, see below), or
0.1818 moles/litre (a mole of CO2 is 44.01 grams).
A constant-density graph for 0.1818 moles/litre is available at
http://webbook.nist.gov/cgi/fluid.cgi?D=0.181776869&TLow=216.592+K&THigh=300&TInc=1&Applet=on&Digits=5&ID=C124389&Action=Load&Type=IsoChor&TUnit=K&PUnit=MPa&DUnit=mol%2Fl&HUnit=kJ%2Fmol&WUnit=m%2Fs&VisUnit=uPa*s&STUnit=N%2Fm&RefState=DEF
But what is the initial (= final) enthalpy? In eg kJ/mol?
At 293.15K, with both liquid and vapour present in equilibrium, initial
pressure is 5.73 MPa. liquid density is 17.373 mol/l, vapour density is
4.413 mol/l. Enthalpy of the vapour is 17.950 kJ/mol, enthalpy of the
liquid is 11.261 kJ/mol.
All those figures can be found from the graphs on the NIST pages.
You have 0.1818 moles of CO2, but without knowing the volume of the
cartridge you cannot calculate the mass of liquid and/or vapour - you don't
know how much of the CO2 is in liquid form, how much is vapour - and so you
cannot calculate the initial enthalpy.
However, expanding to the final density given above, and with an overall
enthalpy somewhere in between those values, final conditions are off to the
left of the constant-density graph linked to above.
Try setting the X axis to enthalpy to see why.
The temperature will be below the triple point, and you _will_ get CO2
snow.
(that's why the constant-density graph won't go any lower - density
becomes a somewhat meaningless quantity. A volume could contain lots of
snow, or only a little, but still have the same pressure and temperature)
-- Peter Fairbrother