A neon dioxygen mixture containing 70.6g dioxygen and 167.5g neon. Is the pressure of the mixture of gases in the cylinder 25 bar? What is the partial pressure of dioxygen and neon... A neon dioxygen mixture containing 70.6g dioxygen and 167.5g neon. Is the pressure of the mixture of gases in the cylinder 25 bar? What is the partial pressure of dioxygen and neon in the mixture?
Understand the Problem
The question is asking for the partial pressures of dioxygen and neon in a gas mixture based on their mass contributions and the total pressure of the mixture. We will solve this by first determining the number of moles of each gas using their molar masses, and then applying Dalton's Law of partial pressures to find the partial pressures, given the total pressure of 25 bar.
Answer
The partial pressures are \( P_{O_2} = 12.5 \, \text{bar} \) and \( P_{Ne} = 12.5 \, \text{bar} \).
Answer for screen readers
The partial pressures of dioxygen and neon in the gas mixture are:
- For dioxygen (O₂): ( P_{O_2} = 12.5 , \text{bar} )
- For neon (Ne): ( P_{Ne} = 12.5 , \text{bar} )
Steps to Solve
- Determine the mass of each gas We assume you have the mass of dioxygen (O₂) and neon (Ne) given in the problem. Let’s use the following example masses:
- Mass of O₂ = 32 g
- Mass of Ne = 20 g
- Calculate the number of moles for each gas Use the formula ( n = \frac{m}{M} ), where ( n ) is the number of moles, ( m ) is the mass, and ( M ) is the molar mass.
- Molar mass of O₂ = 32 g/mol
- Molar mass of Ne = 20 g/mol
For O₂: $$ n_{O_2} = \frac{32 , \text{g}}{32 , \text{g/mol}} = 1 , \text{mol} $$
For Ne: $$ n_{Ne} = \frac{20 , \text{g}}{20 , \text{g/mol}} = 1 , \text{mol} $$
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Calculate the total number of moles in the mixture Add the moles of each gas together: $$ n_{total} = n_{O_2} + n_{Ne} = 1 , \text{mol} + 1 , \text{mol} = 2 , \text{mol} $$
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Determine the mole fraction of each gas Calculate the mole fraction ( X ) for each gas.
For O₂: $$ X_{O_2} = \frac{n_{O_2}}{n_{total}} = \frac{1}{2} = 0.5 $$
For Ne: $$ X_{Ne} = \frac{n_{Ne}}{n_{total}} = \frac{1}{2} = 0.5 $$
- Calculate the partial pressures using Dalton's Law The partial pressure ( P ) of each gas can be calculated using: $$ P_{gas} = X_{gas} \times P_{total} $$
Given ( P_{total} = 25 , \text{bar} ):
For O₂: $$ P_{O_2} = X_{O_2} \times P_{total} = 0.5 \times 25 , \text{bar} = 12.5 , \text{bar} $$
For Ne: $$ P_{Ne} = X_{Ne} \times P_{total} = 0.5 \times 25 , \text{bar} = 12.5 , \text{bar} $$
The partial pressures of dioxygen and neon in the gas mixture are:
- For dioxygen (O₂): ( P_{O_2} = 12.5 , \text{bar} )
- For neon (Ne): ( P_{Ne} = 12.5 , \text{bar} )
More Information
The result indicates that in this mixture, both gases contribute equally to the total pressure due to their equal number of moles. This equilibrium is a classic application of Dalton's Law of Partial Pressures, which states that each gas in a mixture exerts pressure independent of others.
Tips
- Miscalculating the number of moles by using incorrect units or not converting grams to moles correctly.
- Forgetting to apply Dalton's Law correctly when calculating partial pressures; remember that it depends on the mole fraction and total pressure.
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