Question

A \(2.00-\mathrm{L}\) sample of \(\mathrm{O}_{2}(g)\) was collected over water at a total pressure of 785 torr and \(25^{\circ} \mathrm{C}\). When the \(\mathrm{O}_{2}(g)\) was dried (water vapor removed), the gas had a volume of \(1.94 \mathrm{~L}\) at \(25^{\circ} \mathrm{C}\) and 785 torr. Calculate the vapor pressure of water at \(25^{\circ} \mathrm{C}\).

Short answer

Insert the calculated values of \(n_{wet}\) and \(n_{dry}\), and solve for the vapor pressure of water at \(25^{\circ} C\): \(n_{wet} = \frac{785\,\text{torr} \cdot 2.00\,\text{L}}{62.36\,\frac{\text{L}\cdot\text{torr}}{\text{mol}\cdot\text{K}} \cdot 298.15\, \text{K}} \approx 0.0840\, \text{mol}\) \(n_{dry} = \frac{785\,\text{torr} \cdot 1.94\,\text{L}}{62.36\,\frac{\text{L}\cdot\text{torr}}{\text{mol}\cdot\text{K}} \cdot 298.15\, \text{K}} \approx 0.0813\, \text{mol}\) \(n_{water} = n_{wet} - n_{dry} \approx 0.0840\, \text{mol} - 0.0813\, \text{mol} \approx 0.0027\, \text{mol}\) Now, calculate the partial pressure of water vapor: \(P_{H_2O} = 785\,\text{torr} \times \frac{0.0027\, \text{mol}}{0.0840\, \text{mol}} \approx 26\, \text{torr}\) The vapor pressure of water at \(25^{\circ} C\) is approximately \(26\,\text{torr}\).

Step-by-step solution

Find the pressure of wet Oxygen gas

The pressure when the O_2(g) was collected over water is given as 785 torr. Let's denote this pressure as P_wet. \(P_{wet}= 785\,\text{torr}\)

Apply Ideal Gas Law to find moles of wet Oxygen gas

Now, we'll use the Ideal Gas Law, PV=nRT, to find the number of moles of the wet Oxygen gas, where: - P is pressure - V is volume - n is the number of moles - R is the ideal gas constant - T is temperature in Kelvin We need to convert temperature from Celsius to Kelvin first: \(T_{k} = 25^{\circ} C + 273.15 = 298.15\, K\) Now, we'll use the Ideal Gas Law and solve for moles of wet Oxygen gas (n_wet). The volume of wet Oxygen gas is given as 2.00 L. \(P_{wet}V_{wet} = n_{wet}RT\) \(785\,\text{torr} \cdot 2.00\,\text{L} = n_{wet}\left(62.36\,\frac{\text{L}\cdot\text{torr}}{\text{mol}\cdot\text{K}}\right) (298.15\, \text{K})\) Now, we should solve for \(n_{wet}\): \(n_{wet}= \frac{785\,\text{torr} \cdot 2.00\,\text{L}}{62.36\,\frac{\text{L}\cdot\text{torr}}{\text{mol}\cdot\text{K}} \cdot 298.15\, \text{K}}\)

Find the pressure of dry Oxygen gas

The pressure of dry Oxygen gas can also be denoted as P_dry, and it holds the same pressure as the wet oxygen gas because the pressure stays constant. This was given as 785 torr. \(P_{dry} = 785\,\text{torr}\)

Apply Ideal Gas Law to find moles of dry Oxygen gas

Now, we'll use the Ideal Gas Law again and solve for moles of dry Oxygen gas (n_dry). The volume of dry oxygen gas is given as 1.94 L. \(P_{dry}V_{dry} = n_{dry}RT\) \(785\,\text{torr} \cdot 1.94\,\text{L} = n_{dry}\left(62.36\,\frac{\text{L}\cdot\text{torr}}{\text{mol}\cdot\text{K}}\right)(298.15\, \text{K})\) Now, we should solve for \(n_{dry}\): \(n_{dry}= \frac{785\,\text{torr} \cdot 1.94\,\text{L}}{62.36\,\frac{\text{L}\cdot\text{torr}}{\text{mol}\cdot\text{K}} \cdot 298.15\, \text{K}}\)

Calculate moles of water vapor

We can find the moles of water vapor (n_water) by subtracting moles of dry oxygen (n_dry) from moles of wet oxygen (n_wet): \(n_{water} = n_{wet} - n_{dry}\)

Calculate the partial pressure of water vapor

Next, we'll use Dalton's Law of Partial Pressures to find the vapor pressure of water at 25°C. According to Dalton's Law: \(P_{total} = P_{O_2} + P_{H_2O}\) We have the total pressure and partial pressure of the dry Oxygen gas. Thus, we can find the partial pressure of water vapor (P_H2O): \(P_{H_2O} = P_{total} - P_{O_2}\) \(P_{H_2O} = 785\,\text{torr} - 785\,\text{torr}\) #Conclusion#Insert the calculated values of \(n_{wet}\) and \(n_{dry}\), and solve for the vapor pressure of water at \(25^{\circ}\, C\). This will give you the final answer.

Key concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, and temperature of a gas with its amount in moles. It's expressed as \(PV = nRT\), where:

• \(P\) is the pressure of the gas.
• \(V\) is the volume.
• \(n\) represents the number of moles.
• \(R\) is the universal gas constant, often given as \(62.36\, \text{L} \cdot \text{torr}/\text{mol} \cdot \text{K}\).
• \(T\) is the temperature in Kelvin.

To solve the Ideal Gas Law, it is crucial to ensure consistent units to avoid errors. Temperature must be in Kelvin, so you often need to convert from Celsius by adding 273.15. This equation allows you to solve for any one of these variables if you have the other three.
As seen in the exercise, the Ideal Gas Law is used to calculate the number of moles of oxygen gas before and after drying it to determine various properties like vapor pressure.

Partial Pressure

Partial pressure refers to the pressure exerted by a single type of gas in a mixture of gases. It's a concept particularly useful when dealing with gases collected over water.
When gases are collected over water, they are mixed with water vapor. Each gas in the mixture behaves independently, and the partial pressure of each gas is the pressure it would exert if it were alone.
In the given exercise, the partial pressure of the dry oxygen gas remains the same as the total pressure when wet. However, it helps highlight the individual contributions to the total pressure when using Dalton's Law. Calculating partial pressures is a vital step toward understanding interactions between gases in a mixture.

Dalton's Law of Partial Pressures

Dalton's Law of Partial Pressures is key when dealing with mixtures of gases. It states that the total pressure exerted by a mixture of non-reacting gases is the sum of the partial pressures of individual gases.
This can be mathematically expressed as:\[P_{\text{total}} = P_1 + P_2 + P_3 + ... + P_n\]Where \(P_{\text{total}}\) is the total pressure and \(P_1, P_2, \ldots\) are the partial pressures of each gas present.
In the exercise, Dalton's Law helps calculate the vapor pressure of water by determining the part of the total pressure attributable to water vapor. By subtracting the pressure of dry oxygen from the total pressure, you isolate the vapor pressure of water at the experiment's temperature.

Gas Collection over Water

In scientific experiments, gases are often collected over water. This involves capturing a gas in a container already filled with water, allowing the gas to displace the water. The pressure inside the container is due to both the gas and the water vapor it contains.
The challenge is to determine the exact pressure of just the gas collected. This requires accounting for the water vapor pressure, which varies with temperature.
By using the Ideal Gas Law alongside Dalton's Law, and after determining the partial pressure of the gas, you can subtract it to find the vapor pressure from the total pressure. This method is exemplified in the exercise problem, demonstrating how these laws are practically applied to derive the vapor pressure of water from the given data.