Finding mass of propane in 2 sealed containers.

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SUMMARY

This discussion focuses on calculating the mass of propane in two sealed tanks using both the non-ideal gas law and the ideal gas law. The first tank (1 m3) contains propane at 100 kPa and 300 K, while the second tank (0.5 m3) has propane at 250 kPa and 400 K. The user initially calculated the mass in the first tank as 78.707 grams using the non-ideal gas law with a Z factor of 0.99, but found a discrepancy when comparing it to the ideal gas law, which yielded 40.09 moles. The confusion was resolved by correcting the use of propane's specific R constant, leading to consistent results.

PREREQUISITES
  • Understanding of non-ideal gas law (PV = ZnRT)
  • Familiarity with ideal gas law (PV = nRT)
  • Knowledge of propane's properties, including R constant and molecular weight
  • Ability to perform unit conversions in thermodynamic equations
NEXT STEPS
  • Study the differences between ideal and non-ideal gas behavior
  • Learn about the significance of the compressibility factor (Z) in real gases
  • Explore the calculation of mass using the non-ideal gas law with various substances
  • Investigate the impact of temperature and pressure on gas properties and calculations
USEFUL FOR

Students in thermodynamics, chemical engineering, or anyone involved in gas calculations and properties, particularly those working with propane or similar substances.

reed2100
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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.
 
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You're just messed up on units. Check your units more carefully.
 
Chestermiller said:
You're just messed up on units. Check your units more carefully.

Ah I think I see it - in my use of Propane's R constant I had kJ/mol-K when it was supposed to be kg. Taking the mass in tank A as 1.785 kg instead of moles it comes out to around the same mass as given by the ideal law. Thank you very much!
 

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