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NISA [10]
3 years ago
15

At 85°C, the vapor pressure of A is 566 torr and that of B is 250 torr. Calculate the composition of a mixture of A and B that b

oils at 85°C when the pressure is 0.60 atm. Also, calculate the composition of the vapor mixture. Assume ideal behavior.
Chemistry
1 answer:
Phantasy [73]3 years ago
7 0

Answer:

Composition of the mixture:

x_{A} =0.652=65.2 %

x_{B} =0.348=34.8 %

Composition of the vapor mixture:

y_{A} =0.809=80.9%

y_{B} =0.191=19.1%

Explanation:

If the ideal solution model is assumed, and the vapor phase is modeled as an ideal gas, the vapor pressure of a binary mixture with x_{A} and x_{B} molar fractions can be calculated as:

P_{vap}=x_{A}P_{A}+x_{B}P_{B}

Where P_{A} and P_{B} are the vapor pressures of the pure compounds. A substance boils when its vapor pressure is equal to the pressure under it is; so it boils when P_{vap}=P. When the pressure is 0.60 atm, the vapor pressure has to be the same if the mixture is boiling, so:

0.60*760=P_{vap}=x_{A}P_{A}+x_{B}P_{B}\\456=x_{A}P_{A}+(1-x_{A})P_{B}\\456=x_{A}*(P_{A}-P_{B})+P_{B}\\\frac{456-P_{B}}{P_{A}-P_{B}}=x_{A}\\\\\frac{456-250}{566-250}=x_{A}=0.652

With the same assumptions, the vapor mixture may obey to the equation:

x_{A}P_{A}=y_{A}P, where P is the total pressure and y is the fraction in the vapor phase, so:

y_{A} =\frac{x_{A}P_{A}}{P}=\frac{0.652*566}{456} =0.809=80.9 %

The fractions of B can be calculated according to the fact that the sum of the molar fractions is equal to 1.

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Explanation:

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Hence, molecular weight of the gas is 9.54 lb.

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  • Now, calculate the specific gravity of the gas as follows.

  Specific gravity relative to air = \frac{\text{density of methane}}{\text{density of air}}

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