# Calculating mass of vapor from vapor pressure

• Bacat
You can have a whole bunch of different gases and mixtures in the same volume, and you will still use the same equation to calculate number of moles of a specific gas.In summary, the problem involves an open vessel containing water, benzene, and mercury at room temperature and pressure. The vapor pressures of the substances are given and the goal is to calculate the mass of each substance in the air if there is no ventilation. The equations used are PV=nRT and Raoult's Law. The three partial pressures are considered separately and the mole fractions are calculated using the total pressure and the individual vapor pressures. It is also possible to solve the problem without using Raoult's Law by considering the total number of moles of all
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?

This is simple application of PV=nRT.

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.

Last edited:
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!

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.

Khaled_84

## 1. What is vapor pressure?

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid or solid phase at a given temperature. It is a measure of the tendency of a substance to escape from its liquid or solid state and enter the gas phase.

## 2. How is vapor pressure related to temperature?

Vapor pressure increases as temperature increases. This is because at higher temperatures, more molecules have enough energy to escape from the liquid or solid phase and enter the gas phase.

## 3. What is the ideal gas law and how is it used to calculate the mass of vapor?

The ideal gas law is PV=nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. By rearranging this equation, we can solve for n, which represents the number of moles of gas. Multiplying this number by the molar mass of the gas will give us the mass of vapor.

## 4. How does the type of substance affect the calculation of vapor mass?

The type of substance does affect the calculation of vapor mass. Each substance has a unique molar mass, which will impact the calculation. Additionally, some substances may have different vapor pressure equations that need to be used in the calculation.

## 5. Why is it important to calculate the mass of vapor from vapor pressure?

Calculating the mass of vapor from vapor pressure is important in many scientific and industrial processes. It can help determine the amount of a substance present in a mixture, the concentration of a vapor in a confined space, and the potential for evaporation or condensation in a given system.

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