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A sealed, rigid container contains three gases: 28.0 \(\mathrm{g}\) of nitrogen, 40.0 \(\mathrm{g}\) of argon, and 36.0 g of water vapor. If the total pressure exerted by the gases is \(2.0 \mathrm{atm},\) what is the partial pressure of the nitrogen? (A) 0.33 atm (B) 0.40 atm (C) 0.50 \(\mathrm{atm}\) (D) 2.0 \(\mathrm{atm}\)

Short Answer

Expert verified
The partial pressure of nitrogen is 0.5 atm.

Step by step solution

01

Determine the number of moles

First, convert the mass of each gas to moles using their respective molar masses. For nitrogen, Ar(N) = 14, thus molar mass = 28 g/mol. Therefore, number of moles = mass/molar mass = 28.0g / 28 g/mol = 1 mol. Similarly, for argon, Ar(Ar) = 40, thus molar mass = 40 g/mol and number of moles = mass/molar mass = 40.0 g / 40 g/mol = 1 mol. For water vapor, molar mass = 18 g/mol and number of moles = mass/molar mass = 36.0 g / 18 g/mol = 2 mol.
02

Calculate the total number of moles

The total number of moles, \(n_{total}\), is the sum of moles of all three gases = 1 mol N2 + 1 mol Ar + 2 mol H2O = 4 mol.
03

Calculate the partial pressure of nitrogen

Using the formula for partial pressure \(P_{gas} = P_{total} * (n_{gas}/n_{total})\), we substitute the values: \(P_{N2} = 2.0 atm * (1 mol / 4 mol) = 0.5 atm\). Therefore, the partial pressure of nitrogen is 0.5 atm.

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Key Concepts

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

Ideal Gas Law
The Ideal Gas Law is a fundamental concept in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is typically given by the equation \( PV = nRT \), where:
  • \( P \) is the pressure of the gas
  • \( V \) is the volume
  • \( n \) is the number of moles
  • \( R \) is the ideal gas constant
  • \( T \) is the temperature in Kelvin
In this exercise, while the Ideal Gas Law itself isn't directly used to solve the problem, understanding it helps grasp how gases behave under different conditions. The law assumes gases behave ideally, meaning they have point-sized particles with no intermolecular forces, and the gas particles are constantly in random motion. This simplifies calculations and provides a baseline for understanding gas mixtures in a container. It's important to remember that the Ideal Gas Law applies best under conditions of low pressure and high temperature, where real gases behave more ideally.
Mole Fraction
Mole fraction is a way to express the composition of a gas mixture. It describes the ratio of the number of moles of one component to the total number of moles in the mixture. The formula is \( X_{gas} = \frac{n_{gas}}{n_{total}} \), where
  • \( X_{gas} \) is the mole fraction of the gas
  • \( n_{gas} \) is the number of moles of the particular gas
  • \( n_{total} \) is the total number of moles in the mixture
For example, in this problem, we have 1 mole of nitrogen, 1 mole of argon, and 2 moles of water vapor, making a total of 4 moles. The mole fraction of nitrogen would be \( \frac{1}{4} \) or 0.25. Mole fractions give us insight into the contribution of each gas to the mixture's behavior, and they are particularly useful when calculating partial pressures using Dalton’s Law.
Dalton's Law of Partial Pressures
Dalton’s Law of Partial Pressures 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. The partial pressure of a gas is the pressure it would exert if it were alone in a container of the same volume. The formula is \( P_{total} = P_1 + P_2 + P_3 + \ldots \) where each \( P \) represents the partial pressure of each gas. To find the partial pressure of a gas, use \( P_{gas} = P_{total} \times X_{gas} \). Here, \( X_{gas} \) is the mole fraction, as calculated through dividing the moles of the specific gas by the total moles in the mixture.For instance, if the total pressure in the container is 2.0 atm, and the mole fraction of nitrogen is 0.25, the partial pressure for nitrogen is \( 2.0 \text{ atm} \times 0.25 = 0.5 \text{ atm}\). Dalton's Law simplifies understanding how each component contributes to the overall pressure, assisting in problems like our exercise where the individual gas pressures need to be calculated.

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