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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
The vapor pressure of water at 25°C is 21.8 torr.

Step by step solution

01

Write down the given values.

We are given the following information: - Volume of the gas mixture (O₂ and H₂O vapor): V_total = 2.00 L - Total pressure: P_total = 785 torr - Temperature: T = 25°C = 298K - Volume of dry O₂ gas: V_O₂ = 1.94 L
02

Determine the moles of dry Oâ‚‚ gas using the Ideal Gas Law.

First, we need to calculate the number of moles of dry Oâ‚‚ gas using the Ideal Gas Law: \(PV = nRT\) Here, we are looking for n (number of moles), and we are given the pressure (P), volume (V), and temperature (T). The gas constant R = 62.363 L torr / K mol for pressure given in torr. Rearranging the equation to find n: \(n = \frac{PV}{RT}\) Using the values provided for dry Oâ‚‚ gas, we have: P = 785 torr, V = 1.94 L, and T = 298K Plugging these values into the equation: \(n = \frac{(785\,\text{torr})(1.94\,\text{L})}{(62.363\,\text{L torr / K mol})(298\,\text{K})}\)
03

Calculate the number of moles of dry Oâ‚‚ gas.

Now, we can calculate the number of moles of dry Oâ‚‚ gas: \(n = \frac{(785\,\text{torr})(1.94\,\text{L})}{(62.363\,\text{L torr / K mol})(298\,\text{K})} = 0.0815\,\text{mol}\)
04

Calculate the partial pressure of Oâ‚‚ in the total pressure.

Next, we can use the ideal gas law again to find the partial pressure of Oâ‚‚ gas in the total pressure: \(P_{\text{O}_2} = \frac{nRT}{V}\) Using the values we calculated previously for dry Oâ‚‚ gas: \(P_{\text{O}_2} = \frac{(0.0815\,\text{mol})(62.363\,\text{L torr / K mol})(298\,\text{K})}{2.00\,\text{L}}\)
05

Calculate the partial pressure of Oâ‚‚ gas.

Now, we can calculate the partial pressure of Oâ‚‚ gas: \(P_{\text{O}_2} = \frac{(0.0815\,\text{mol})(62.363\,\text{L torr / K mol})(298\,\text{K})}{2.00\,\text{L}} = 763.2\,\text{torr}\)
06

Calculate the vapor pressure of water using Dalton's Law of Partial Pressures.

According to Dalton's Law of Partial Pressures, the total pressure is equal to the sum of the partial pressures of the gases: \(P_{\text{total}} = P_{\text{O}_2} + P_{\text{H}_2\text{O}}\) Now, we can solve for the vapor pressure of water, \(P_{\text{H}_2\text{O}}\): \(P_{\text{H}_2\text{O}} = P_{\text{total}} - P_{\text{O}_2}\) Using the values we know: \(P_{\text{H}_2\text{O}} = 785\,\text{torr} - 763.2\,\text{torr}\)
07

Calculate the vapor pressure of water at 25°C.

Finally, we can calculate the vapor pressure of water at 25°C: \(P_{\text{H}_2\text{O}} = 785\,\text{torr} - 763.2\,\text{torr} = 21.8\,\text{torr}\) The vapor pressure of water at 25°C is 21.8 torr.

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Key Concepts

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

Dalton's Law of Partial Pressures
Dalton's Law of Partial Pressures is a fundamental concept when dealing with gas mixtures. This law states that in a mixture of non-reacting gases, the total pressure exerted is equal to the sum of the partial pressures of individual gases. Each gas in the mixture behaves independently, exerting pressure as if it alone occupied the entire volume. So, for our exercise:
  • Total Pressure (\(P_{\text{total}}\)) = Pressure of Oxygen (\(P_{\text{O}_2}\)) + Pressure of Water Vapor (\(P_{\text{H}_2\text{O}}\))
By knowing the total pressure and calculating the pressure of the dry oxygen gas, we can easily find the vapor pressure of water by using a simple subtraction. This is a practical example of applying Dalton's law to determine unknown pressures in gas mixtures over water or other solvents.
Vapor Pressure
Vapor pressure is an important concept when you have a liquid in equilibrium with its vapor. It is the pressure exerted by the vapor in equilibrium with its liquid phase at a given temperature. In our problem, the oxygen gas was collected over water, which means some water evaporated into the gas. At a specific temperature, such as 25°C in our exercise, water has a characteristic vapor pressure, which we determined using Dalton’s Law.

When a gas is collected over water, the vapor pressure needs to be accounted for to isolate the pressure exerted by the gas you're interested in. Therefore, understanding the concept of vapor pressure is essential in exact calculations when gases are involved.
Moles Calculation
Calculating the number of moles in a given gas situation is crucial and often involves the Ideal Gas Law. The Ideal Gas Law is depicted as:
  • \( PV = nRT \)
where:
  • \(P\) is the pressure
  • \(V\) is the volume
  • \(n\) is the number of moles
  • \(R\) is the gas constant
  • \(T\) is the temperature in Kelvin
Rearranging this equation to solve for \(n\) gives us the number of moles. This step is crucial in determining the partial pressures of gas mixtures. Once you know the moles, you can plug them back into the gas law to find other parameters like pressure or volume. In this exercise, knowing the moles of oxygen helped us calculate its partial pressure in the mixture.
Gas Mixtures
Gas mixtures are a combination of different gases occupying the same volume. They're an essential part of understanding real-world gas behavior because gases are rarely found in isolation. Each component in a gas mixture contributes to the overall pressure based on its proportion or mole fraction in the mix.

In our exercise, the gas mixture consisted of oxygen and water vapor. When the oxygen gas was initially gathered, the water vapor contributed to the total pressure inside the container. Adjusting calculations for this enables a more precise understanding of the behavior of the target gas (oxygen in this instance). Studying gas mixtures helps in various fields, including environmental science, chemistry, and engineering, where gas interactions often take place.

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Most popular questions from this chapter

A \(20.0\) -\(\mathrm{L}\) stainless steel container at \(25^{\circ} \mathrm{C}\) was charged with \(2.00\) atm of hydrogen gas and \(3.00\) atm of oxygen gas. A spark ignited the mixture, producing water. What is the pressure in the tank at \(25^{\circ} \mathrm{C} ?\) If the exact same experiment were performed, but the temperature was \(125^{\circ} \mathrm{C}\) instead of \(25^{\circ} \mathrm{C},\) what would be the pressure in the tank?

Consider two gases, \(A\) and \(B\), each in a \(1.0\) -\(\mathrm{L}\) container with both gases at the same temperature and pressure. The mass of gas \(A\) in the container is \(0.34\) \(\mathrm{g}\) and the mass of gas \(B\) in the container is \(0.48 \mathrm{g}\). a. Which gas sample has the most molecules present? Explain. b. Which gas sample has the largest average kinetic energy? Explain. c. Which gas sample has the fastest average velocity? Explain. d. How can the pressure in the two containers be equal to each other since the larger gas \(B\) molecules collide with the container walls more forcefully?

A tank contains a mixture of \(52.5 \mathrm{g}\) oxygen gas and \(65.1 \mathrm{g}\) carbon dioxide gas at \(27^{\circ} \mathrm{C}\). The total pressure in the tank is \(9.21\) atm. Calculate the partial pressures of each gas in the container.

Consider the following reaction: $$4 \mathrm{Al}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{Al}_{2} \mathrm{O}_{3}(s)$$ It takes \(2.00 \mathrm{L}\) of pure oxygen gas at \(STP\) to react completely with a certain sample of aluminum. What is the mass of aluminum reacted?

A steel cylinder contains \(5.00\) moles of graphite (pure carbon) and \(5.00\) moles of \(\mathrm{O}_{2}\). The mixture is ignited and all the graphite reacts. Combustion produces a mixture of \(\mathrm{CO}\) gas and \(\mathrm{CO}_{2}\) gas. After the cylinder has cooled to its original temperature, it is found that the pressure of the cylinder has increased by \(17.0 \% .\) Calculate the mole fractions of \(\mathrm{CO}, \mathrm{CO}_{2},\) and \(\mathrm{O}_{2}\) in the final gaseous mixture.

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