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Calculate the vapor pressure at 298K above an aerosol mixture consisting of 30% w/w of aerosol propellant 114 (mol.wt = 170.9) with vapor pressure of 1.90 imes 105 ext Nm⁻² and 70% w/w of propellant

Chemistry

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  • Calculate the vapor pressure at 298298K above an aerosol mixture consisting of 30%30\% w/w of aerosol propellant 114 (mol.wt = 170.9) with vapor pressure of 1.90×105 Nm−21.90 \times 10^5 \text{ Nm}^{-2} and 70%70\% w/w of propellant 12 (mol.wt = 120.9) with vapor pressure of 5.85×105 Nm−25.85 \times 10^5 \text{ Nm}^{-2}. Assume ideal behavior.
  1. Assume a total mass of the mixture and calculate the mass of each propellant.

    Let's assume a total mass of 100 g for the mixture.
    Mass of propellant 114 (m114m_{114}) = 30%30\% of 100 g = 30 g
    Mass of propellant 12 (m12m_{12}) = 70%70\% of 100 g = 70 g

  2. Calculate the moles of each propellant.

    Molar mass of propellant 114 (MW114MW_{114}) = 170.9 g/mol
    Molar mass of propellant 12 (MW12MW_{12}) = 120.9 g/mol
    Moles of propellant 114 (n114n_{114}):

    n114=30 g170.9 g/mol=0.17554 moln_{114} = \frac{30 \text{ g}}{170.9 \text{ g/mol}} = 0.17554 \text{ mol}

    Moles of propellant 12 (n12n_{12}):

    n12=70 g120.9 g/mol=0.57900 moln_{12} = \frac{70 \text{ g}}{120.9 \text{ g/mol}} = 0.57900 \text{ mol}
  3. Calculate the mole fractions of each propellant in the liquid mixture.

    Total moles (ntotaln_{total}):

    ntotal=n114+n12=0.17554 mol+0.57900 mol=0.75454 moln_{total} = n_{114} + n_{12} = 0.17554 \text{ mol} + 0.57900 \text{ mol} = 0.75454 \text{ mol}

    Mole fraction of propellant 114 (X114X_{114}):

    X114=n114ntotal=0.17554 mol0.75454 mol=0.23265X_{114} = \frac{n_{114}}{n_{total}} = \frac{0.17554 \text{ mol}}{0.75454 \text{ mol}} = 0.23265

    Mole fraction of propellant 12 (X12X_{12}):

    X12=n12ntotal=0.57900 mol0.75454 mol=0.76735X_{12} = \frac{n_{12}}{n_{total}} = \frac{0.57900 \text{ mol}}{0.75454 \text{ mol}} = 0.76735
  4. Calculate the partial vapor pressure of each propellant using Raoult's Law.

    Pure vapor pressure of propellant 114 (P114∘P_{114}^\circ) = 1.90×105 Nm−21.90 \times 10^5 \text{ Nm}^{-2}
    Pure vapor pressure of propellant 12 (P12∘P_{12}^\circ) = 5.85×105 Nm−25.85 \times 10^5 \text{ Nm}^{-2}
    Partial pressure of propellant 114 (P114P_{114}):

    P114=X114×P114∘=0.23265×(1.90×105 Nm−2)=4.420×104 Nm−2P_{114} = X_{114} \times P_{114}^\circ = 0.23265 \times (1.90 \times 10^5 \text{ Nm}^{-2}) = 4.420 \times 10^4 \text{ Nm}^{-2}

    Partial pressure of propellant 12 (P12P_{12}):

    P12=X12×P12∘=0.76735×(5.85×105 Nm−2)=4.489×105 Nm−2P_{12} = X_{12} \times P_{12}^\circ = 0.76735 \times (5.85 \times 10^5 \text{ Nm}^{-2}) = 4.489 \times 10^5 \text{ Nm}^{-2}
  5. Calculate the total vapor pressure.

    Total vapor pressure (PtotalP_{total}) = P114+P12P_{114} + P_{12}

    Ptotal=(4.420×104 Nm−2)+(4.489×105 Nm−2)P_{total} = (4.420 \times 10^4 \text{ Nm}^{-2}) + (4.489 \times 10^5 \text{ Nm}^{-2}) Ptotal=(0.4420×105 Nm−2)+(4.489×105 Nm−2)P_{total} = (0.4420 \times 10^5 \text{ Nm}^{-2}) + (4.489 \times 10^5 \text{ Nm}^{-2}) Ptotal=4.931×105 Nm−2P_{total} = 4.931 \times 10^5 \text{ Nm}^{-2}

    The total vapor pressure is 4.93×105 Nm−2\boxed{4.93 \times 10^5 \text{ Nm}^{-2}}.

    • Calculate the vapor pressure (in Pa) at 20∘C20^\circ\text{C} above an aerosol mixture consisting of 30%30\% w/w of HFA 134a (tetrafluoroethane, molecular weight 102) with a vapor pressure of 68.4 psig and 70%70\% w/w of HFA 227 (heptafluoropropane, molecular weight 170) with a vapor pressure of 56.0 psig. Assume ideal behaviour. NOTE: Psia = Psig + 14.7, 1 Psia = 6894.74 Pa
  6. Convert the pure vapor pressures from psig to psia, then to Pa.

    For HFA 134a:
    P134a∘(psia)=68.4 psig+14.7=83.1 psiaP_{134a}^\circ (\text{psia}) = 68.4 \text{ psig} + 14.7 = 83.1 \text{ psia}
    P134a∘(Pa)=83.1 psia×6894.74 Pa/psia=572953.554 PaP_{134a}^\circ (\text{Pa}) = 83.1 \text{ psia} \times 6894.74 \text{ Pa/psia} = 572953.554 \text{ Pa}
    For HFA 227:
    P227∘(psia)=56.0 psig+14.7=70.7 psiaP_{227}^\circ (\text{psia}) = 56.0 \text{ psig} + 14.7 = 70.7 \text{ psia}
    P227∘(Pa)=70.7 psia×6894.74 Pa/psia=487430.798 PaP_{227}^\circ (\text{Pa}) = 70.7 \text{ psia} \times 6894.74 \text{ Pa/psia} = 487430.798 \text{ Pa}

  7. Assume a total mass of the mixture and calculate the mass of each HFA.

    Let's assume a total mass of 100 g for the mixture.
    Mass of HFA 134a (m134am_{134a}) = 30%30\% of 100 g = 30 g
    Mass of HFA 227 (m227m_{227}) = 70%70\% of 100 g = 70 g

  8. Calculate the moles of each HFA.

    Molar mass of HFA 134a (MW134aMW_{134a}) = 102 g/mol
    Molar mass of HFA 227 (MW227MW_{227}) = 170 g/mol
    Moles of HFA 134a (n134an_{134a}):

    n134a=30 g102 g/mol=0.29412 moln_{134a} = \frac{30 \text{ g}}{102 \text{ g/mol}} = 0.29412 \text{ mol}

    Moles of HFA 227 (n227n_{227}):

    n227=70 g170 g/mol=0.41176 moln_{227} = \frac{70 \text{ g}}{170 \text{ g/mol}} = 0.41176 \text{ mol}
  9. Calculate the mole fractions of each HFA in the liquid mixture.

    Total moles (ntotaln_{total}):

    ntotal=n134a+n227=0.29412 mol+0.41176 mol=0.70588 moln_{total} = n_{134a} + n_{227} = 0.29412 \text{ mol} + 0.41176 \text{ mol} = 0.70588 \text{ mol}

    Mole fraction of HFA 134a (X134aX_{134a}):

    X134a=n134antotal=0.29412 mol0.70588 mol=0.41666X_{134a} = \frac{n_{134a}}{n_{total}} = \frac{0.29412 \text{ mol}}{0.70588 \text{ mol}} = 0.41666

    Mole fraction of HFA 227 (X227X_{227}):

    X227=n227ntotal=0.41176 mol0.70588 mol=0.58334X_{227} = \frac{n_{227}}{n_{total}} = \frac{0.41176 \text{ mol}}{0.70588 \text{ mol}} = 0.58334
  10. Calculate the partial vapor pressure of each HFA using Raoult's Law.

    Partial pressure of HFA 134a (P134aP_{134a}):

    P134a=X134a×P134a∘=0.41666×572953.554 Pa=238728.9 PaP_{134a} = X_{134a} \times P_{134a}^\circ = 0.41666 \times 572953.554 \text{ Pa} = 238728.9 \text{ Pa}

    Partial pressure of HFA 227 (P227P_{227}):

    P227=X227×P227∘=0.58334×487430.798 Pa=284399.7 PaP_{227} = X_{227} \times P_{227}^\circ = 0.58334 \times 487430.798 \text{ Pa} = 284399.7 \text{ Pa}
  11. Calculate the total vapor pressure.

    Total vapor pressure (PtotalP_{total}) = P134a+P227P_{134a} + P_{227}

    Ptotal=238728.9 Pa+284399.7 Pa=523128.6 PaP_{total} = 238728.9 \text{ Pa} + 284399.7 \text{ Pa} = 523128.6 \text{ Pa}

    Rounding to three significant figures:

    Ptotal=5.23×105 PaP_{total} = 5.23 \times 10^5 \text{ Pa}

    The total vapor pressure is 5.23×105 Pa\boxed{5.23 \times 10^5 \text{ Pa}}.

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