Problem 1
Question
A mixture of one mole each of \(\mathrm{H}_{2}, \mathrm{He}\) and \(\mathrm{O}_{2}\) each are enclosed in a cylinder of volume \(V\) at temperature \(T\). If the partial pressure of \(\mathrm{H}_{2}\) is \(2 \mathrm{~atm}\), the total pressure of the gases in the cylinder is : (a) 6 atm (b) 38 atm (c) \(14 \mathrm{~atm}\) (d) \(22 \mathrm{~atm}\)
Step-by-Step Solution
Verified Answer
The total pressure of the gases in the cylinder is 6 atm.
1Step 1: Understand Dalton's Law of Partial Pressures
Dalton's Law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of individual gases. Mathematically, this can be expressed as \( P_{\text{total}} = P_{1} + P_{2} + P_{3} + \ldots \), where each \( P_{i} \) is the partial pressure of a gas in the mixture.
2Step 2: Identify Given Values
We are given that each gas in the mixture (\(\text{H}_2\),\(\text{He}\), and \(\text{O}_2\)) occupies 1 mole. The partial pressure of\(\text{H}_2\) is provided as \(2 \text{ atm}\). This suggests each gas has an equal contribution due to equal moles.
3Step 3: Calculate Partial Pressures of Other Gases
Since the gases are in equal moles and considering ideal gas behavior in similar conditions, the partial pressures of \(\text{He}\) and \(\text{O}_2\) will also be equal to that of \(\text{H}_2\), which is \(\text{2 atm}\). Thus, the partial pressures for \(\text{He}\) and \(\text{O}_2\) are also \(2 \text{ atm}\) each.
4Step 4: Sum Up Partial Pressures to Find Total Pressure
Using the partial pressures calculated in the previous steps, we now find the total pressure:\[ P_{\text{total}} = P_{\text{H}_2} + P_{\text{He}} + P_{\text{O}_2} = 2 \text{ atm} + 2 \text{ atm} + 2 \text{ atm} = 6 \text{ atm} \]
Key Concepts
Partial PressureIdeal Gas BehaviorNon-reacting GasesMixture of Gases
Partial Pressure
Partial pressure is a concept introduced by John Dalton that helps us understand how individual gases behave in a mixture. Imagine you have a box filled with several gases. Each gas exerts its own pressure as if the other gases weren't present at all. This is known as the gas's partial pressure.
- The partial pressure is directly proportional to the amount of gas in the mixture.
- Partial pressures add up to give the total pressure of the gas mixture.
- For example, in our exercise, \( \text{H}_2 \) has a partial pressure of \( 2 \, \text{atm} \).
Ideal Gas Behavior
Ideal gas behavior is a useful assumption about how gases act under certain conditions. It simplifies the math and allows us to make predictions about the properties of gases.
For our problem, gases like \( \text{H}_2 \), \( \text{He} \), and \( \text{O}_2 \) are treated as ideal, helping us easily determine their partial pressures based on moles.
- In ideal conditions, gases are assumed to have no interactions between them.
- The gas particles themselves are considered to take up no volume—they're point particles.
For our problem, gases like \( \text{H}_2 \), \( \text{He} \), and \( \text{O}_2 \) are treated as ideal, helping us easily determine their partial pressures based on moles.
Non-reacting Gases
It is crucial that the gases in our mixture do not react with one another. If they did, their behavior would deviate from what's expected in simple mixtures.
- Non-reacting gases keep their chemical identities in a mixture.
- No chemical reactions mean their partial pressures are unaffected by changes in chemical composition.
Mixture of Gases
A mixture of gases is exactly what it sounds like—different gases hanging out in the same space. It's vital to understand that each gas keeps its original pressure contribution, which brings us back to partial pressures.
The mixture's total pressure can only be found when all individual pressures are summed.
The mixture's total pressure can only be found when all individual pressures are summed.
- Each gas contributes based solely on its own number of moles and conditions.
- The mixture's properties depend on the collective contribution of its components.
Other exercises in this chapter
Problem 2
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The predominant intermolecular forces present in ethyl acetate, a liquid, are: (a) London dispersion and dipole-dipole (b) hydrogen bonding and London dispersio
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At a given temperature \(\mathrm{T}\), gases \(\mathrm{Ne}, \mathrm{Ar}, \mathrm{Xe}\) and \(\mathrm{Kr}\) are found to deviate from ideal gas behaviour. Their
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