Calculating mass of vapor from vapor pressure

[Bacat]

Homework Statement

An open vessel containing water, benzene, and mercury stands in a laboratory measuring 5.0 m by 5.0 m by 3.0 m at 25 degrees C. The vapor pressures are 3.2 kPa, 13.1 kPa, and 0.23 Pa, respectively. What mass of each substance will be found in the air if there is no ventilation?

Homework Equations

Raoult's Law
P_A = X_A * P^*_A

Where the P_A is the partial pressure of the diluted liquid, X_A is the mole fraction of the substance, and P^*_A is the partial pressure of the pure liquid.

P_{TOT} = P_{Air} + P_{Water} + P_{Benzene} + P_{Mercury}

Where P denotes the partial pressure of the specific substance such that all partial pressures add to the total pressure, which we take to be the pressure of the atmosphere at sea level: 101325 Pa.

The Attempt at a Solution

1) I tried calculating relative humidity for water by:

x = \frac{0.62198 * P_{Water}}{101325 Pa - P_{Water}}

And then calculated the mass of the water as: m_{water} = x*mass_{air}

This failed.

2) I calculated P_{TOT} by adding the partial pressures to 101325 Pa and then divided:

\frac{P_{water}}{P_{TOT}}

I then multiplied this result by the total number of moles of air, calculated as...

75 m^3 * \frac{1.2041 kg}{m^3} * \frac{1 mole}{28.97 kg} = 3.12 moles of air

And then converted the moles of water to kilograms. This also failed.

QUESTIONS

1) Do I need to consider the three partial pressures at the same time? Or all separately? In other words, does the partial pressure of the benzene interact with the air and the mercury? Or can I treat them all separately?

2) Which equations should I be using to find the number of moles of a substance that will vaporize at a particular temperature? None of the techniques I have applied above seem to include temperature...but I believe that a higher temperature should increase the amount of vapor in the air.

3) If I am supposed to use Raoult's Law, how can I calculate the mole fraction without knowing how many of each substance there is?

 

[Borek]

This is simple application of PV=nRT.

 

[rbsaway]

My reply to questions:

1) For this problem, it is safe to assume that all three gases are ideal (that is, they do not interact), and they occupy the exact same volume. So you can simply consider the three partial pressures separately.

2) All you need is the good old PV = nRT, where
R = 8.314472 \frac{Pa*m^3}{mol * K},
T = 25^\circ C = 298K,
V = 5.0m * 5.0m * 3.0m = 75m^3, and
P =each individual vapor pressure, in order to calculate n for each gas and their respective masses:

n_{water} = \frac{P_{water}V}{RT}
n_{benzene} = \frac{P_{benzene}V}{RT}
n_{mercury} = \frac{P_{mercury}V}{RT}

m_{water} = n_{water}*M_{water}
m_{benzene} = n_{benzene}*M_{benzene}
m_{mercury} = n_{mercury}*M_{mercury}


3) This problem can easily be solved without using Raoult's Law, although its concept is implied. You can check that
x_{water} = \frac{P_{water}}{P_{tot}} = \frac{n_{water}}{n_{tot}}
x_{benzene} = \frac{P_{benzene}}{P_{tot}} = \frac{n_{benzene}}{n_{tot}}
x_{mercury} = \frac{P_{mercury}}{P_{tot}} = \frac{n_{mercury}}{n_{tot}}

Where
P_{tot} = P_{water} + P_{benzene} + P_{mercury}
n_{tot} = n_{water} + n_{benzene} + n_{mercury}, and

x_{water} + x_{benzene} + x_{mercury} = 1.

Even though it's an open vessel, we are told there is "no ventilation". This means we can assume the vessel contains no air, only the gas phases of water, benzene, and mercury.

 

[Bacat]

rbsaway, I didn't realize that I could assume there was no air in the room. pV=nRT makes a lot more sense now.

Thank you!

 

[Borek]

Air has nothing to do with the question - you are given volume, temperature and pressure. That's perfectly enough to calculate number of moles. Presence of other gases doesn't change it.