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Calculating Partial Pressure?

  1. Aug 6, 2004 #1
    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!
  2. jcsd
  3. Aug 6, 2004 #2
    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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