Finding Mass Through Percent by Volume

Click For Summary

Discussion Overview

The discussion revolves around calculating the pressure of a gaseous mixture of methane and ethylene in a cylinder, given their volume percentages and total mass. Participants explore methods based on the ideal gas law and the compressibility factor, while addressing the implications of using volume percentages versus mass fractions in their calculations.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant initially calculates the mass of methane and ethylene by directly applying the volume percentages to the total mass, but questions the validity of this approach.
  • Another participant suggests that since the mixture behaves ideally, the volume ratio corresponds to the mole ratio, leading to a relationship between the number of moles of each gas.
  • A third participant expresses confusion about how to apply the molar masses of the gases to the volume percentages, despite obtaining similar results using both methods.
  • A participant mentions a method involving the weighted average molar mass of the gas mixture to find the number of moles and subsequently the pressure using the ideal gas law.
  • It is noted that using a weighted average molar mass should yield consistent results, similar to calculating the number of moles of air using its average molar mass.

Areas of Agreement / Disagreement

Participants exhibit uncertainty regarding the proper method to apply volume percentages in the context of mass and molar calculations. There is no consensus on which approach is more intuitive or correct, as differing methods yield similar results.

Contextual Notes

The discussion highlights the potential confusion arising from the distinction between volume percentages and mass fractions, as well as the assumptions made in applying the ideal gas law and the compressibility factor method.

JeweliaHeart
Messages
67
Reaction score
0
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?
 
Physics news on Phys.org
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.
 
Last edited:
Using weighted average molar mass should yield the same answer (it is like calculating number of moles of air using molar mass of 29).
 

Similar threads

Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
9K
Replies
1
Views
2K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 10 ·
Replies
10
Views
15K
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
4K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
4K