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Calculate the vapor pressure of an aerosol mixture containing HFA 134a and HFA 227, given their gauge pressures and mass percentages.

Chemistry

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Answer

58.7 Psig\text{58.7 Psig}

Step-by-step solution

6 steps

To calculate the vapor pressure of the aerosol mixture, we will use Raoult's Law and Dalton's Law of Partial Pressures. First, we need to convert the given gauge pressures to absolute pressures, then determine the mole fractions of each component, calculate their partial pressures, and finally sum them up.

  1. Convert the given gauge pressures (PsigP_{sig}) to absolute pressures (PabsP_{abs}).

    The atmospheric pressure is given as 14.7 Psia14.7 \text{ Psia}.

    Pabs=Psig+14.7 PsiaP_{abs} = P_{sig} + 14.7 \text{ Psia}

    For HFA 134a:

    P134a,abs=62.4 Psig+14.7 Psia=77.1 PsiaP_{134a, abs} = 62.4 \text{ Psig} + 14.7 \text{ Psia} = 77.1 \text{ Psia}

    For HFA 227:

    P227,abs=56.0 Psig+14.7 Psia=70.7 PsiaP_{227, abs} = 56.0 \text{ Psig} + 14.7 \text{ Psia} = 70.7 \text{ Psia}
  2. Calculate the moles of each component.

    Assume a total mass of 100 g100 \text{ g} for the mixture.
    Mass of HFA 134a = 30% of 100 g=30 g30\% \text{ of } 100 \text{ g} = 30 \text{ g}
    Mass of HFA 227 = 70% of 100 g=70 g70\% \text{ of } 100 \text{ g} = 70 \text{ g}

    n134a=30 g102 g/mol=0.2941 moln_{134a} = \frac{30 \text{ g}}{102 \text{ g/mol}} = 0.2941 \text{ mol} n227=70 g170 g/mol=0.4118 moln_{227} = \frac{70 \text{ g}}{170 \text{ g/mol}} = 0.4118 \text{ mol}
  3. Calculate the total moles and mole fraction (XX) of each component.

    ntotal=n134a+n227=0.2941 mol+0.4118 mol=0.7059 moln_{total} = n_{134a} + n_{227} = 0.2941 \text{ mol} + 0.4118 \text{ mol} = 0.7059 \text{ mol} X134a=n134antotal=0.2941 mol0.7059 mol=0.4166X_{134a} = \frac{n_{134a}}{n_{total}} = \frac{0.2941 \text{ mol}}{0.7059 \text{ mol}} = 0.4166 X227=n227ntotal=0.4118 mol0.7059 mol=0.5834X_{227} = \frac{n_{227}}{n_{total}} = \frac{0.4118 \text{ mol}}{0.7059 \text{ mol}} = 0.5834
  4. Calculate the partial vapor pressure (P′P') of each component using Raoult's Law (P′=X⋅PabsP' = X \cdot P_{abs}).

    P134a′=X134a×P134a,abs=0.4166×77.1 Psia=32.12 PsiaP'_{134a} = X_{134a} \times P_{134a, abs} = 0.4166 \times 77.1 \text{ Psia} = 32.12 \text{ Psia} P227′=X227×P227,abs=0.5834×70.7 Psia=41.25 PsiaP'_{227} = X_{227} \times P_{227, abs} = 0.5834 \times 70.7 \text{ Psia} = 41.25 \text{ Psia}
  5. Calculate the total absolute vapor pressure (Ptotal,absP_{total, abs}) of the mixture.

    Ptotal,abs=P134a′+P227′=32.12 Psia+41.25 Psia=73.37 PsiaP_{total, abs} = P'_{134a} + P'_{227} = 32.12 \text{ Psia} + 41.25 \text{ Psia} = 73.37 \text{ Psia}
  6. Convert the total absolute vapor pressure back to gauge pressure (Ptotal,sigP_{total, sig}).

    Ptotal,sig=Ptotal,abs−14.7 Psia=73.37 Psia−14.7 Psia=58.67 PsigP_{total, sig} = P_{total, abs} - 14.7 \text{ Psia} = 73.37 \text{ Psia} - 14.7 \text{ Psia} = 58.67 \text{ Psig}

    Rounding to one decimal place:

    Ptotal,sig=58.7 PsigP_{total, sig} = 58.7 \text{ Psig}

    The total vapor pressure of the aerosol mixture is 58.7 Psig\text{58.7 Psig}.

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