Ideal Gas Mixture Composition Calculation

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SUMMARY

The discussion focuses on calculating the composition of an ideal gas mixture containing methane (CH4) and ethane (C2H6) in a rigid cylinder of 1000 cm³ at 100°C and 500 kPa. The average molar mass of the mixture was determined to be 25.85 kg/kmol. Participants provided methods for calculating the mole fraction and mass fraction of each gas, emphasizing the use of the ideal gas equation (PV=nRT) and the importance of understanding the total number of moles in the mixture.

PREREQUISITES
  • Understanding of the ideal gas law (PV=nRT)
  • Knowledge of molar mass calculations for gases
  • Familiarity with mole fraction and mass fraction concepts
  • Basic principles of gas mixtures and Avogadro's hypothesis
NEXT STEPS
  • Learn how to apply the ideal gas law to calculate total moles in a gas mixture
  • Study the calculation of mole fractions and mass fractions in gas mixtures
  • Explore the implications of Avogadro's hypothesis in gas calculations
  • Investigate the properties of methane and ethane, including their molar masses and behaviors under varying conditions
USEFUL FOR

Chemistry students, chemical engineers, and anyone involved in gas mixture calculations or thermodynamics will benefit from this discussion.

Oblivion77
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Homework Statement



A rigid cylinder of 1000cm cubed contains 4.168grams of an ideal gas mixture. The gas mixture consists of methane and ethane at a temperature of 100C and a pressure of 500kPa

a) Estimate the average molar mass of the mixure
b) Calculate the composition of the mixture on a mole fraction basis
c) Express the mixture composition on a mass fraction basis

Homework Equations



PV=nRT

The Attempt at a Solution



I calculated part a) already and found that to be 25.85kg/kmol. I am stuck on part b and c.
Thanks for any help.
 
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Express mass of the sample in terms of numbers of moles of each gas.
 
Borek said:
Express mass of the sample in terms of numbers of moles of each gas.
Ok

I got mols of ethane to be 0.14 and mols of methane to be 0.26. Now what? I tried doing
0.14 / (0.14 + 0.26) to get the fraction of ethane but it is wrong.
 
Show how you calculated total number of moles of both gases.
 
Borek said:
Show how you calculated total number of moles of both gases.

For methane: CH4 molar mass is 16.04g/mol. We have 4.168g so M= m/n, n = m/M therefore
n = 4.168 / 16.04 = 0.259 ~ 0.26mols of methane

same thing for ethane.
 
No. You can't have 4.168g of methane plus 4.168g of ethane and 4.168g in total.

Use ideal gas equation to calculate total numebr of moles of gases.

Do you know Avogadro's hypothesis?
 
Borek said:
No. You can't have 4.168g of methane plus 4.168g of ethane and 4.168g in total.

Use ideal gas equation to calculate total numebr of moles of gases.

Do you know Avogadro's hypothesis?

Ok, so how would I calculate how much of each compound by just knowing the molar mass?

V/n = constant?
 
Mass is molar mass times number of moles.
 

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