Lupin said:
I didn't realize this was going to be so complicated. I'm learning a lot. I figured there would be an easy calculation that was based upon knowing the mass of the CO2: 12g in liquid state; the temperature: assumed to be constant at room temp 24C; and atmospheric pressure: 1 bar. Then there would be some calculation that either started with 12 grams of CO2 at STP, 6.1 liters, and slowly compressed it to liquid state at constant temperature calculating the required energy to do it. Or, starting with liquid CO2 and letting it slowly expand and push a piston until the pressure under the piston is at or close to 1 atmosphere.
Or counting shots from a Daisy BB gun. 200 joules.
For the "easy" calculation I'd use the approximation from
@jbriggs444 in post #4. An isothermal expansion of an ideal gas.
Regarding my contribution in post #5, I think it's better to just ignore it. I don't think it's wrong but I don't see how it could have some physical meaning here.
Lastly, for a more accurate but complex approach,
@Chestermiller provided a method in post #8. Again it's an isothermal expansion but includes real gas effects. Not all the gas will be expanding like that as he said but most of the work will come from that expansion so I'll treat it as if all the gas followed that process. It's a new method for me so I'm interested in seeing the differences between the 2 methods.
Properties of the gas.
##R_{CO_2}=0.1889 \frac{kJ}{kgK}##; ##T_{cr}=304.2K##; ##P_{cr}=7.39MPa##.
Known information from the problem definition.
##P_i=850PSI + 1atm = 5.961MPa##; ##P_f=1atm=0.101MPa##; ##T_i=T_f=T=24^{\circ}C=297K##; ##m=12g=12*10^{-3}kg##; ##V_i=\frac{mRT}{P_i}=0.00011276 m^3##; ##V_f=\frac{mRT}{P_f}=0.00663381 m^3##.
ISOTHERMAL EXPANSION OF AN IDEAL GAS
$$W=\int_{i}^{f}PdV=\int_{i}^{f}\frac{mRT}{V}dV=mRT\ln\frac{V_f}{V_i}=mRT\ln\frac{P_i}{P_f}=2738.9J$$
ISOTHERMAL EXPANSION INCLUDING REAL EFFECTS.
First I need to define the ##z## factor for this problem.
Temperature is constant so using ##T_R=\frac{T}{T_{cr}}=0.976 \approx 1## I can choose the lowest curve which is the closest to that value.
Second, I find in which interval of ##P_R## the expansion happens to linearize ##z## with either one or two straight lines. ##P_{R_i}=\frac{P_i}{P_{cr}}=0.806## and ##P_{R_f}=\frac{P_f}{P_{cr}}=0.0137##. It's then possible to conclude we're on the left side of that curve and only one straight line will be enough to approximate the problem.
I'm eyeballing the values from the graph which is not ideal but I see this mostly as an exercise to learn. Feel free to try more accurate measurements if necessary and do the proper integral without the linearization if necessary.
Now that ##z## is known, we can tackle the expression for work.
$$W=mRT\ln{(P_i/P_f)}+W^R$$
$$W^R=mRT(z_f-z_i)+mRT\int_{P_i}^{P_f}{(z-1)\frac{dP}{P}}$$
For clarity, I'll solve it in chunks. First, this part is the same as the ideal isothermal expansion
$$mRT\ln{(P_i/P_f)}=2738.9J$$
Then, the first component of ##W_R##.
$$mRT(z_f-z_i)=262.14J$$
Finally, the last component of ##W_R## which contains the integral. First, I'll express ##z## in terms of ##P## to solve the integral.
$$z=-0.491P_R+0.996= -0.491\frac{P}{P_{cr}}+0.996$$
$$mRT\int_{P_i}^{P_f}{(z-1)\frac{dP}{P}}=mRT\int_{P_i}^{P_f}{(-0.491\frac{P}{P_{cr}}+0.996-1)\frac{dP}{P}}=272.5J$$
That's easily solvable analytically but I cheated to be a little faster and solved it numerically online to avoid writing the steps.
Anyway, assuming the previous calculations are OK, now we have all the components of ##W##.
$$W=2738.9+262.14+272.5=3273.54J$$
So it seems it's possible to extract about 20% more energy from the "real expansion" than from the "ideal expansion". Whether that's right or not I can't judge due to my lack of experience in these problems. Maybe I committed an error while inputting the numbers or in the derivation but I couldn't find it while checking the numbers myself.