Interpretation:
The partial pressures of methylbenzene (A) and butanone have to be calculated. The calculated partial pressures have to be plotted against the mole fraction of the liquid mixture. Henry’s law constant for both the components has to be calculated.
Concept introduction:
Henry’s law states that the partial vapour pressure is directly proportional to the amount of substance dissolved. It involves an empirical constant KH called Henry’s law constant.
Answer to Problem 5A.4P
The partial pressures of methylbenzene (A) are calculated in the table given below.
yA | xA | p/kPa | pA/kPa |
0 | 0 | 36.066 | 0 |
0.041 | 0.0898 | 34.121 | 1.398961 |
0.1154 | 0.2476 | 30.9 | 3.56586 |
0.1762 | 0.3577 | 28.626 | 5.043901 |
0.2772 | 0.5194 | 25.239 | 6.996251 |
0.3393 | 0.6036 | 23.402 | 7.940299 |
0.445 | 0.7188 | 20.6984 | 9.210788 |
0.5435 | 0.8019 | 18.592 | 10.10475 |
0.7284 | 0.9105 | 15.496 | 11.28729 |
1 | 1 | 12.295 | 12.295 |
The partial pressures of butanone are calculated in the table given below.
yB | xB | p/kPa | pB/kPa |
1 | 1 | 36.066 | 36.066 |
0.959 | 0.9102 | 34.121 | 32.72204 |
0.8846 | 0.7524 | 30.9 | 27.33414 |
0.8238 | 0.6423 | 28.626 | 23.5821 |
0.7228 | 0.4806 | 25.239 | 18.24275 |
0.6607 | 0.3964 | 23.402 | 15.4617 |
0.555 | 0.2812 | 20.6984 | 11.48761 |
0.4565 | 0.1981 | 18.592 | 8.487248 |
0.2716 | 0.0895 | 15.496 | 4.208714 |
0 | 0 | 12.295 | 0 |
The partial pressure of methylbenzene (A) is plotted against the mole fraction of the liquid mixture in the graph shown below.
Figure 1
The partial pressure of butanone is plotted against the mole fraction of the liquid mixture in the graph shown below.
Figure 2
The value of Henry’s law constant for methylbenzene (A) is 15.578 kPa_. The value of Henry’s law constant for butanone is 47.025 kPa_.
Explanation of Solution
The data for the mole fraction of methylbenzene (A) in the liquid phase, xA, vapour phase, yA and the total pressure, p is given below.
xA | yA | p/kPa |
0 | 0 | 36.066 |
0.0898 | 0.041 | 34.121 |
0.2476 | 0.1154 | 30.9 |
0.3577 | 0.1762 | 28.626 |
0.5194 | 0.2772 | 25.239 |
0.6036 | 0.3393 | 23.402 |
0.7188 | 0.445 | 20.6984 |
0.8019 | 0.5435 | 18.592 |
0.9105 | 0.7284 | 15.496 |
1 | 1 | 12.295 |
According to Dalton’s law of partial pressures, the partial pressure of methylbenzene (A) (pA) is calculated by the formula given below.
pA=yAp (1)
Where,
- yA is the mole fraction of methylbenzene (A) in the vapour phase.
- p is the total vapour pressure.
The total pressure (p) at yA=0.0410 is 34.121 kPa.
Substitute the value of p and yA in equation (1).
pA=0.0410×34.121 kPa=1.398961 kPa_
Therefore, the partial pressure of methylbenzene (A) (pA) is calculated for all vapour phase composition (yA) in the table below.
yA | xA | p/kPa | pA/kPa |
0 | 0 | 36.066 | 0 |
0.041 | 0.0898 | 34.121 | 1.398961 |
0.1154 | 0.2476 | 30.9 | 3.56586 |
0.1762 | 0.3577 | 28.626 | 5.043901 |
0.2772 | 0.5194 | 25.239 | 6.996251 |
0.3393 | 0.6036 | 23.402 | 7.940299 |
0.445 | 0.7188 | 20.6984 | 9.210788 |
0.5435 | 0.8019 | 18.592 | 10.10475 |
0.7284 | 0.9105 | 15.496 | 11.28729 |
1 | 1 | 12.295 | 12.295 |
The graph between the mole fraction in the liquid phase, xA and the partial pressure, pA is given below.
Figure 1
In the above graph, when xA is 0.0898, pA is 1.398961 kPa.
According to Henry’s law, the partial pressure of methylbenzene (A) is given as,
pA=xAKH (2)
Where,
- pA is the partial pressure of methylbenzene (A).
- KH is Henry’s law constant.
- xA is the mole fraction of methylbenzene (A) in the liquid phase.
Substitute the value of xA and pA in equation (2).
1.398961 kPa=0.0898×KHKH=1.398961 kPa0.0898=15.578 kPa_
Therefore, the value of Henry’s law constant for methylbenzene (A) is 15.578 kPa_.
The total mole fraction in the vapour phase equal to 1.
yA+yB=1yB=1−yA (3)
Where,
- yB is the mole fraction of butanone in the vapour phase.
The mole fraction of methylbenzene (A) in the vapour phase (yA) is 0.0410.
Substitute the value of yA in equation (3).
yB=1−0.0410=0.959
According to Dalton’s law of partial pressures, the partial pressure of butanone (B) (pB) is calculated by the formula given below.
pB=yBp (4)
Where,
- yB is the mole fraction of butanone (B) in the vapour phase.
- p is the total vapour pressure.
The total pressure (p) at yB=0.959 is 34.121 kPa.
Substitute the value of p and yB in equation (4).
pB=0.959×34.121 kPa=32.722 kPa_
Therefore, the partial pressure of butanone (B) (pB) is calculated for all vapour phase composition (yB) in the table below.
yB | xB | p/kPa | pB/kPa |
1 | 1 | 36.066 | 36.066 |
0.959 | 0.9102 | 34.121 | 32.72204 |
0.8846 | 0.7524 | 30.9 | 27.33414 |
0.8238 | 0.6423 | 28.626 | 23.5821 |
0.7228 | 0.4806 | 25.239 | 18.24275 |
0.6607 | 0.3964 | 23.402 | 15.4617 |
0.555 | 0.2812 | 20.6984 | 11.48761 |
0.4565 | 0.1981 | 18.592 | 8.487248 |
0.2716 | 0.0895 | 15.496 | 4.208714 |
0 | 0 | 12.295 | 0 |
The graph between the mole fraction in the liquid phase, xB and the partial pressure, pB is given below.
Figure 2
In the above graph, when xB is 0.0895, pB is 4.208714 kPa.
According to Henry’s law, the partial pressure of butanone (B) is given as,
pB=xBKH (5)
Where,
- pB is the partial pressure of butanone (B).
- KH is Henry’s law constant.
- xB is the mole fraction of butanone (B) in the liquid phase.
Substitute the value of xB and pB in equation (5).
4.208714 kPa kPa=0.0895×KHKH=4.208714 kPa0.0895=47.025 kPa_
Therefore, the value of Henry’s law constant for butanone is 47.025 kPa_.
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