How to Calculate the Partial Pressure of Propane in a Mixture?

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SUMMARY

The calculation of the partial pressure of propane (C3H8) in a mixture with butane (C4H10) at 20 °C and a total pressure of 1918 mmHg is straightforward. Using the ideal gas law, the partial pressure can be determined by the formula P_i = n_iP/n, where n_i is the number of moles of the substance and P is the total pressure. Given equal moles of propane and butane, the mole fraction for each is 0.5, resulting in a partial pressure of 959 mmHg for both propane and butane.

PREREQUISITES
  • Understanding of the ideal gas law
  • Familiarity with the concept of partial pressure
  • Knowledge of mole fractions
  • Basic chemistry concepts related to gas mixtures
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  • Learn about Dalton's Law of Partial Pressures
  • Explore calculations involving gas mixtures
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Chemistry students, chemical engineers, and anyone involved in gas calculations or thermodynamics will benefit from this discussion.

Daizy5936
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I am having the hardest time figuring this one out:

"Calculate the partial pressure of propane in a mixture that contains equal numbers of moles of propane (C3H8) and butane (C4H10) at 20 °C and 1918 mmHg. Give your answer in mmHg"

Can anybody help me out? I'm so lost!
 
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In general the Total P=P1+P2+P3... where P1,2,3 etc are the pressures of each substance, which can be found using the ideal gas law
P_i=n_iRT/V. RT/V can be replaced by P/n (P is total pressure remember )giving you P_i=n_iP/n=yiP, where yi is the mole fraction of the substance i.
The problem gives you total pressure and says that you have equal amounts of moles of each substance. Therefore mole fraction for each substance is= to 1/2 because you have equal amounts. So P1=.5P=959 mmHg and P2=.5P=959mmHg.
 

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