Finding Mass Through Percent by Volume

AI Thread Summary
The discussion focuses on calculating the pressure of a gas mixture containing 60% methane and 40% ethylene by volume, with a total mass of 12.3 kg stored in a cylinder. The initial approach of using volume percentages to find mass was challenged, as the correct method involves using molar masses to determine the average molecular weight of the mixture. Both methods, including the ideal gas law and the compressibility factor, yield similar results, although the weighted average molar mass approach is preferred for its clarity. The relationship between volume ratio and molar ratio is emphasized, indicating they are equivalent under ideal gas conditions. Ultimately, understanding how to apply these concepts is crucial for accurate pressure calculations in gas mixtures.
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A gas analysis on a gaseous mixture gives 60% methane and 40% ethylene by volume. You need to store 12.3 kg of the mixture in a cylinder of volume 0.0514 m^3 at a maximum temperature of 45°C. Determine the pressure (kPa) inside the cylinder by:

a. assuming that the mixture obeys ideal gas law

b. using the compressibility factor method


I approached part a of this problem by taking both 60 and 40 percent of 12.3 kg to find the mass of the methane and ethylene, respectively in the ideal gas mixture. A friend has informed me that this is not the proper approach because 60% and 40% are percentages by volume, not mass fractions, and therefore he recommended that I should instead take 60% and 40% of the molar masses of methane and ethylene instead.


I don't understand what percent volume has to do with molar mass. Why would I multiply the percentage by volume times the respective molar masses if I want to find the number of moles methane and ethylene in the compound?
 
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I am not sure I follow your friend advice (I am not saying it is wrong).

Assuming ideal gas behavior volume ratio and molar ratio are identical. That means you have

\frac{n_{methane}}{n_{ethane}}=\frac{6}{4}

M_{methane}\times{n_{methane}}+M_{ethane}\times{n_{ethane}}=12.3~kg

Can you go from there?
 
Yes. That makes sense above. I am just trying to figure out how what you are saying translates over to multiplying the molar masses of methane and ethylene by the volume percentages, which is what my friend was doing.

I tried solving part a using what you did above and what my friend did and got nearly identical answers using both methods. However, his method is still not as intuitive to me.
 
My friend advises I do it his way:0.6(16.04 kg/kmol of Methane) +0.4(28.04 kg/kmol of Ethylene)

to get the average molecular weight of the combined gas mixture.

After that, he uses 12.3 kg to find the number of moles of gas in the mixture and uses pv=nrt to find pressure.
 
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Using weighted average molar mass should yield the same answer (it is like calculating number of moles of air using molar mass of 29).
 
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