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Chapter 5: Problem 129

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

Expert verified
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

01

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. \(P_{wet}= 785\,\text{torr}\)
02

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 \(n_{wet}\): \(n_{wet}= \frac{785\,\text{torr} \cdot 2.00\,\text{L}}{62.36\,\frac{\text{L}\cdot\text{torr}}{\text{mol}\cdot\text{K}} \cdot 298.15\, \text{K}}\)
03

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. \(P_{dry} = 785\,\text{torr}\)
04

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 \(n_{dry}\): \(n_{dry}= \frac{785\,\text{torr} \cdot 1.94\,\text{L}}{62.36\,\frac{\text{L}\cdot\text{torr}}{\text{mol}\cdot\text{K}} \cdot 298.15\, \text{K}}\)
05

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): \(n_{water} = n_{wet} - n_{dry}\)
06

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掳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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

A spherical glass container of unknown volume contains helium gas at \(25^{\circ} \mathrm{C}\) and \(1.960 \mathrm{~atm}\). When a portion of the helium is withdrawn and adjusted to \(1.00\) atm at \(25^{\circ} \mathrm{C}\), it is found to have a volume of \(1.75 \mathrm{~cm}^{3}\). The gas remaining in the first container shows a pressure of \(1.710\) atm. Calculate the volume of the spherical container.

A mixture of cyclopropane and oxygen is sometimes used as a general anesthetic. Consider a balloon filled with an anesthetic mixture of cyclopropane and oxygen at \(170 .\) torr and 570 . torr. respectively. Calculate the ratio of the moles \(\mathrm{O}_{2}\) to moles cyclopropane in this mixture.

Small quantities of hydrogen gas can be prepared in the laboratory by the addition of aqueous hydrochloric acid to metallic zinc. $$ \mathrm{Zn}(s)+2 \mathrm{HCl}(a q) \longrightarrow \mathrm{ZnCl}_{2}(a q)+\mathrm{H}_{2}(g) $$ Typically, the hydrogen gas is bubbled through water for collection and becomes saturated with water vapor. Suppose \(240 . \mathrm{mL}\) of hydrogen gas is collected at \(30 .{ }^{\circ} \mathrm{C}\) and has a total pressure of \(1.032\) atm by this process. What is the partial pressure of hydrogen gas in the sample? How many grams of zinc must have reacted to produce this quantity of hydrogen? (The vapor pressure of water is 32 torr at \(30^{\circ} \mathrm{C}\).)

A container is filled with an ideal gas to a pressure of \(40.0\) atm at \(0^{\circ} \mathrm{C}\). a. What will be the pressure in the container if it is heated to \(45^{\circ} \mathrm{C}\) ? b. At what temperature would the pressure be \(1.50 \times 10^{2}\) atm? c. At what temperature would the pressure be \(25.0 \mathrm{~atm} ?\)

One way of separating oxygen isotopes is by gaseous diffusion of carbon monoxide. The gaseous diffusion process behaves like an effusion process. Calculate the relative rates of effusion of \({ }^{12} \mathrm{C}^{16} \mathrm{O},{ }^{12} \mathrm{C}^{17} \mathrm{O}\), and \({ }^{12} \mathrm{C}^{18} \mathrm{O}\). Name some advantages and disad- vantages of separating oxygen isotopes by gaseous diffusion of carbon dioxide instead of carbon monoxide.
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