Question

A sample of oxygen collected over water at 25°C (vapor pressure H2O 5 23.8 mm Hg). The wet gas occupies a volume of 7.28 L at a total pressure of 1.25 bar. If all the water is removed, what volume will the dry oxygen occupy at a pressure of 1.10 atm and a temperature of 27°C?

Short answer

The volume of dry oxygen will be 7.632 L.

Step-by-step solution

Convert all the given quantities to the appropriate units

We need to use the SI units for temperature and pressure. Convert the given quantities in the problem: - Temperature: 25°C = 25 + 273.15 = 298.15 K - Vapor pressure of water (H2O) at 25°C: 23.8 mm Hg = (23.8 / 760) atm = 0.0313 atm - Total pressure of wet gas: 1.25 bar = 1.25 * 0.9869 atm = 1.2336 atm - The new temperature for dry oxygen: 27°C = 27 + 273.15 = 300.15 K - The new pressure for dry oxygen: 1.10 atm (already given in the appropriate units)

Determine the partial pressure of oxygen in the wet gas

Use Dalton's Law of Partial Pressures to find the partial pressure of oxygen: Partial pressure of oxygen (O2) = Total pressure of wet gas - Vapor pressure of water Partial pressure of O2 = 1.2336 atm - 0.0313 atm = 1.2023 atm

Use the Ideal Gas Law to find the number of moles of oxygen

Apply Ideal Gas Law (PV = nRT) for the wet gas to find the number of moles of oxygen (O2), where P is the partial pressure of oxygen, V is the volume of wet gas and T is the temperature: - R (ideal gas constant) = 0.0821 L atm / (mol K) - n = number of moles of oxygen 1.2023 atm * 7.28 L = n * 0.0821 L atm / (mol K) * 298.15 K Solve for n: n = (1.2023 atm * 7.28 L) / (0.0821 L atm / (mol K) * 298.15 K) = 0.3064 mol

Use the Ideal Gas Law to find the volume of dry oxygen under the new conditions

Now, we have the number of moles of oxygen, along with the new pressure and temperature. Apply the Ideal Gas Law (PV = nRT) again for the dry oxygen: - P (new pressure) = 1.10 atm - T (new temperature) = 300.15 K 1.10 atm * V = 0.3064 mol * 0.0821 L atm / (mol K) * 300.15 K Solve for the new volume V of dry oxygen: V = (0.3064 mol * 0.0821 L atm / (mol K) * 300.15 K) / 1.10 atm = 7.632 L The volume of dry oxygen at a pressure of 1.10 atm and a temperature of 27°C will be 7.632 L.

Key concepts

Partial Pressure

Imagine a group of different gases sharing the same room; each one of them exerts a certain pressure as if they were alone in the room. This pressure is what we call 'partial pressure'. It's an essential concept in gas laws because it helps us understand how individual gases within a mixture behave.

In the context of the textbook exercise, oxygen gas is collected over water, and both the oxygen and the water vapor exert partial pressures. The total pressure recorded is a sum of both these partial pressures. When we say 'the partial pressure of oxygen', we're referring to the contribution to the total pressure made by the oxygen alone, excluding the influence of the water vapor.

To calculate this, we subtract the vapor pressure of water from the total pressure measured. If you find this concept a bit tricky, a helpful analogy might be thinking of it as splitting a bill among friends. Just like you'd figure out your share of a restaurant bill, you determine the partial pressure for each gas in a mixture.

Vapor Pressure

Have you ever noticed that a closed bottle of water left in a car on a hot day seems to have more water droplets on the inside surface than on a cooler day? That's a glimpse of 'vapor pressure' in action. Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature.

In simpler terms, it represents how much a liquid tends to evaporate given a set of conditions. At 25°C, water has a vapor pressure, which is a measure of water molecules escaping into the gaseous phase. In our exercise, the vapor pressure of water at 25°C is given as 23.8 mmHg. This value is important because it impacts the total pressure inside the container that also houses oxygen.

Dalton's Law of Partial Pressures

To master mixtures of gases, remember this guy: John Dalton. He's the one who said when gases hang out together, they act independently. This is precisely what Dalton's Law of Partial Pressures is all about. It states that the total pressure exerted by a mixture of non-reactive gases is equal to the sum of the partial pressures of each gas.

It's like every gas in the mix has a personal space, and the pressures they exert don't mess with each other. Hence, in our exercise scenario, to find the pressure of just the oxygen (O2) in the wet gas, we use Dalton’s law to subtract the vapor pressure from the total pressure. Once we get the partial pressure of oxygen, we can use it to figure out how much space the oxygen would need (its volume) under different conditions, using the Ideal Gas Law.