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Homework Statement
A 1 m3 rigid tank has propane at 100 kPa and 300 K. The tank is connected to another 0.5 m3rigid tank which has propane at 250 kPa and 400 K by a ball valve. The valve is opened and both tanks come to a uniform state at 325 K.
Calculate the mass of propane before and after opening the valve in each tank.
Homework Equations
Reduced Pressure and Temperature = current state / critical state
Non-ideal Gas law - PV = ZnRT , or PV = ZmR(specific to substance)T
can rearrange the above for n or m
The Attempt at a Solution
So using the above given pressures and temperatures I have tank A with a z factor of .99 from charts in my book, and tank B with a z factor of .98. The book also says that at 25 C and 100 kPa, propane has an R constant of .1886 kJ/kg-K. The professor also just said that a substance's specific R constant is just the universal R constant divided by it's molecular weight. And a kPa-m^3 is equivalent to a kJ, so I can leave the given units as is or rewrite as kJ on the PV side of the gas law.
For tank A, rearranging the non ideal gas law - (100 kPa-m^3) / (.99 * .1886 kJ/mol-K * 300K) = m[A] = 1.785 mol = 78.707 grams of propane.
Now my confusion and reason for posting this is that when I use the ideal gas law as a reference point, just to see if I'm really far off, I get a pretty different answer. With the ideal gas law , I calculate
(100,000 Pa-m^3) / ( 8.314 (Pa-m^3 / mol-K) * 300 K) = 40.09 mol of propane, which obviously equals a ton more propane than 78.07 grams.
I just want to verify if my approach is the correct idea. You don't need to tell me the answer, but I'd like to know if I'm doing it incorrectly and if so, where. Am I just not using the correct version of the non-ideal gas law or something, or does it look like a confusion with units? Any and all help is appreciated, thank you.