GASES: Calculating partial pressure of a gas from sketch Some N gas is mixed with some O gas, and the sketch below shows a representative sample of the mixture. The total pressure is found to be [50.10] kev. carbon hydrogen sulfur nitrogen oxygen chlorine Calculate the mole fraction and partial pressure of each gas in this mixture. Round your answers to significant digits. You are assuming an ideal gas. gas mole fraction partial pressure OA ton ton
The Correct Answer and Explanation is:
Step 1: Count the Molecules
From the sketch:
- Red spheres = oxygen (O₂)
- Blue spheres = nitrogen (N₂)
Let’s count:
- Nitrogen (blue): 14 molecules
- Oxygen (red): 6 molecules
Step 2: Mole Fraction Calculation
The mole fraction of a gas is calculated as:Mole fraction of gas A=Number of molecules of ATotal number of molecules\text{Mole fraction of gas A} = \frac{\text{Number of molecules of A}}{\text{Total number of molecules}}Mole fraction of gas A=Total number of moleculesNumber of molecules of A
Total molecules = 14 (N₂) + 6 (O₂) = 20
- Mole fraction of N₂ = 14 / 20 = 0.7000
- Mole fraction of O₂ = 6 / 20 = 0.3000
Step 3: Use Dalton’s Law to Find Partial Pressures
Dalton’s Law:Ppartial=Xgas×PtotalP_{\text{partial}} = X_{\text{gas}} \times P_{\text{total}}Ppartial=Xgas×Ptotal
Where:
- Ptotal=1150P_{\text{total}} = 1150Ptotal=1150 torr
- XgasX_{\text{gas}}Xgas is the mole fraction
For N₂:PN2=0.7000×1150=805.0 torrP_{N_2} = 0.7000 \times 1150 = 805.0 \, \text{torr}PN2=0.7000×1150=805.0torr
For O₂:PO2=0.3000×1150=345.0 torrP_{O_2} = 0.3000 \times 1150 = 345.0 \, \text{torr}PO2=0.3000×1150=345.0torr
Final Answers
| Gas | Mole Fraction | Partial Pressure (torr) |
|---|---|---|
| N₂ | 0.7000 | 805.0 |
| O₂ | 0.3000 | 345.0 |
Explanation
In a gas mixture, the total pressure exerted is the sum of the partial pressures of each individual gas. This is explained by Dalton’s Law of Partial Pressures, which assumes gases in a mixture do not chemically react and each gas behaves ideally. Ideal gases exert pressure independently of one another, proportional to their mole fraction in the mixture.
To solve this problem, we visually counted the number of N₂ and O₂ molecules shown in the representative diagram. Since the ideal gas law relates pressure to the number of moles (or particles), the ratio of molecule counts accurately reflects the mole ratio. We found there were 14 nitrogen molecules and 6 oxygen molecules, totaling 20.
Using the formula for mole fraction (moles of gas/total moles\text{moles of gas} / \text{total moles}moles of gas/total moles), we calculated mole fractions of 0.7000 for N₂ and 0.3000 for O₂. This means nitrogen makes up 70% of the gas mixture by particle count.
Applying Dalton’s Law, we multiplied these mole fractions by the total pressure (1150 torr) to find the partial pressures of each gas. This yielded 805.0 torr for nitrogen and 345.0 torr for oxygen. These values also sum to 1150 torr, verifying the accuracy.
This exercise reinforces the principle that in an ideal gas mixture, the individual contributions to pressure are directly proportional to mole count, regardless of the chemical identity of the gases involved.
