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Calculate the total pressure (in atm) of a mixture of \(0.0300 \mathrm{~mol}\) of helium, He, and \(0.0400 \mathrm{~mol}\) of oxygen, \(\mathrm{O}_{2}\) in a 4.00-L flask at \(20^{\circ} \mathrm{C}\). Assume ideal gas behavior.

Short Answer

Expert verified
The total pressure of the gas mixture is 0.421 atm.

Step by step solution

01

Identify the Known Variables

We are given the number of moles of helium (\(n_{He} = 0.0300 \text{ mol}\)), the number of moles of oxygen (\(n_{O_2} = 0.0400 \text{ mol}\)), the volume of the container (\(V = 4.00 \text{ L}\)), and the temperature (\(T = 20^{\circ} \text{C} = 293 \text{ K}\)). We also need the universal gas constant \(R = 0.0821 \text{ L}\cdot \text{atm}/(\text{mol}\cdot \text{K})\).
02

Calculate Partial Pressure of Helium

Use the ideal gas law, \(PV = nRT\), to find the partial pressure of helium. Substituting in the known values for helium:\[ P_{He} = \frac{n_{He}RT}{V} = \frac{0.0300 \times 0.0821 \times 293}{4.00} = 0.180 \text{ atm} \]
03

Calculate Partial Pressure of Oxygen

Similarly, use the ideal gas law to find the partial pressure of oxygen. Substituting the known values for oxygen:\[ P_{O_2} = \frac{n_{O_2}RT}{V} = \frac{0.0400 \times 0.0821 \times 293}{4.00} = 0.241 \text{ atm} \]
04

Calculate Total Pressure

The total pressure is the sum of the partial pressures of helium and oxygen due to Dalton's Law of Partial Pressures. Thus:\[ P_{total} = P_{He} + P_{O_2} = 0.180 + 0.241 = 0.421 \text{ atm} \]

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Key Concepts

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

Partial Pressure
Partial pressure is a concept used in chemistry to describe the pressure exerted by a specific gas within a mixture of gases. Each gas in a mixture, even though they are mixed together, behaves independently of the others. This concept is crucial when dealing with environments where multiple gases are present, such as inside the human lungs or in a chemical reaction involving gas mixtures.

To calculate the partial pressure of a gas, we use the Ideal Gas Law, which is expressed as:
  • P (pressure)
  • V (volume)
  • n (number of moles)
  • R (universal gas constant)
  • T (temperature in Kelvin)
The formula is given by: \[ P = \frac{nRT}{V} \]Where
  • P is the partial pressure we want to calculate
  • n is the number of moles of the specific gas
Using this formula, you can determine the partial pressure contributed by each gas in the mixture.
Dalton's Law of Partial Pressures
Dalton's Law of Partial Pressures is a fundamental principle in gases, particularly when dealing with mixtures. This law states that the total pressure exerted by a mixture of gases is equal to the sum of the partial pressures of each individual gas in the mixture.

Mathematically, this is expressed as:
  • \[P_{total} = P_1 + P_2 + P_3 + ...\]
Where
  • Ptotal is the total pressure of the mixture
  • P1, P2, P3, ... are the partial pressures of the individual gases
This law is particularly useful in scenarios like the mixture of helium and oxygen in a flask, where the total pressure is needed. By individually calculating the partial pressures of each gas, as shown in the Ideal Gas Law calculations, we can simply add them up to find the total pressure. This is why understanding both the concept of partial pressure and Dalton's Law is essential.
Universal Gas Constant
The Universal Gas Constant, denoted by the symbol R, is a key term in the Ideal Gas Law equation, linking the properties of pressure, volume, and temperature for ideal gases. The value of R is determined to be 0.0821 L·atm/(mol·K) when using pressure in atmospheres and volume in liters.

This constant is crucial as it serves as a bridge in calculations involving gases, enabling us to quantify the relationship between moles and pressure under ideal conditions. In real-world applications, knowing R helps you perform accurate calculations of gas behavior where the Ideal Gas Law is applicable.

Understanding the Universal Gas Constant helps in a wide range of applications:
  • Predicting the behavior of gases at different temperatures or pressures
  • Determining how gases will react in chemical reactions
Thus, mastering the use of this constant is essential for anyone studying or working with the properties of gases.

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Most popular questions from this chapter

What is the density of ammonia gas, \(\mathrm{NH}_{3}\), at \(31^{\circ} \mathrm{C}\) and \(751 \mathrm{mmHg}\) ? Obtain the density in grams per liter.

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5.117 Liquid oxygen was first prepared by heating potassium chlorate, \(\mathrm{KClO}_{3},\) in a closed vessel to obtain oxygen at high pressure. The oxygen was cooled until it liquefied. $$ 2 \mathrm{KClO}_{3}(s) \longrightarrow 2 \mathrm{KCl}(s)+3 \mathrm{O}_{2}(g) $$ If \(171 \mathrm{~g}\) of potassium chlorate reacts in a \(2.70-\mathrm{L}\) vessel, which was initially evacuated, what pressure of oxygen will be attained when the temperature is finally cooled to \(25^{\circ} \mathrm{C} ?\) Use the preceding chemical equation and ignore the volume of solid product.

Gas Laws and Kinetic Theory of Gases I Shown here are two identical containers labeled \(\mathrm{A}\) and \(\mathrm{B}\). Container A contains a molecule of an ideal gas, and container B contains two molecules of an ideal gas. Both containers are at the same temperature. (Note that small numbers of molecules and atoms are being represented in these examples in order that you can easily compare the amounts. Real containers with so few molecules and atoms would be unlikely.) How do the pressures in the two containers compare? Be sure to explain your answer. Shown below are four different containers \((\mathrm{C}, \mathrm{D}, \mathrm{E}\) and \(\mathrm{F}\) ), each with the same volume and at the same temperature. How do the pressures of the gases in the containers compare? Container \(\mathrm{H}\) below has twice the volume of container G. How will the pressure in the containers compare? Explain your reasoning. How will the pressure of containers \(\mathrm{G}\) and \(\mathrm{H}\) compare if you add two more gas molecules to container \(\mathrm{H}\) ? Consider containers I and J below. Container J has twice the volume of container \(\mathrm{I}\). Container \(\mathrm{I}\) is at a temperature of \(100 \mathrm{~K},\) and container \(\mathrm{J}\) is at \(200 \mathrm{~K}\). How does the pressure in container I compare with that in container \(\mathrm{J} ?\) Include an explanation as part of your answer.

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