/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Consider the apparatus shown in ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider the apparatus shown in the following drawing. (a) When the valve between the two containers is opened and the gases allowed to mix, how does the volume occupied by the \(\mathrm{N}_{2}\) gas change? What is the partial pressure of \(\mathrm{N}_{2}\) after mixing? (b) How does the volume of the \(\mathrm{O}_{2}\) gas change when the gases mix? What is the partial pressure of \(\mathrm{O}_{2}\) in the mixture? (c) What is the total pressure in the container after the gases mix?

Short Answer

Expert verified
When the valve between the two containers is opened and the gases mix, the volume occupied by the N2 gas remains the same, as the total volume of the container is unchanged. The partial pressure of N2 after mixing is \(P_{N_2}' = x_{N_2} P\), where \(x_{N_2}\) is the mole fraction of N2 and P is the total pressure of the mixture. Similarly, the volume occupied by the O2 gas remains the same when the gases mix. The partial pressure of O2 after mixing is \(P_{O_2}' = x_{O_2}P\), where \(x_{O_2}\) is the mole fraction of O2. The total pressure in the container after the gases mix is given by the sum of the partial pressures of N2 and O2: \(P_{total} = P_{N_2}' + P_{O_2}'\).

Step by step solution

01

Analyze the initial conditions for both gases

Initially, both the gases are in separate containers. We'll need to find the volume occupied by each gas before mixing. Let's denote the initial volume of N2 gas as \(V_{N_2}\) and the initial volume of O2 gas as \(V_{O_2}\). The initial volume can be found using the ideal gas law: \[ PV = nRT \] where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.
02

Calculate the initial volumes of N2 and O2

Using the ideal gas law, we can find the initial volumes occupied by each gas: \(V_{N_2} = \frac{n_{N_2}RT}{P_{N_2}}\) \(V_{O_2} = \frac{n_{O_2}RT}{P_{O_2}}\)
03

Analyze the gas mixture

When the two gases are mixed, the total volume V of the container remains the same. So, the volume occupied by each gas remains the same. The partial pressures of the gases will change, based on the mole fractions of the gases in the mixture. We can use the following formula for partial pressures: \(P_i = x_iP\) where \(P_i\) is the partial pressure of the gas i, \(x_i\) is the mole fraction of the gas i, and P is the total pressure.
04

Calculate the partial pressure of N2 after mixing

The mole fraction of N2 gas, \(x_{N_2}\), is given by: \(x_{N_2} = \frac{n_{N_2}}{n_{N_2} + n_{O_2}}\) The total pressure of the mixture P can be found using the ideal gas law for the overall mixture: \(P = \frac{(n_{N_2} + n_{O_2})RT}{V}\) The partial pressure of N2 gas, \(P_{N_2}'\), is given by: \(P_{N_2}' = x_{N_2} P\)
05

Calculate the partial pressure of O2 after mixing

The mole fraction of O2 gas, \(x_{O_2}\), is given by: \(x_{O_2} = \frac{n_{O_2}}{n_{N_2} + n_{O_2}}\) The partial pressure of O2 gas, \(P_{O_2}'\), is given by: \(P_{O_2}' = x_{O_2}P\)
06

Calculate the total pressure in the container after the gases mix

The total pressure of the mixture after mixing can be found by summing the partial pressures of N2 and O2: \(P_{total} = P_{N_2}' + P_{O_2}'\) This is the final answer for the total pressure in the container after the gases mix.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Partial Pressure
When multiple gases are mixed in a container, each gas exerts its own pressure as if it were alone. This is what we call partial pressure.

For example, if you mix nitrogen (\(N_2\)) and oxygen (\(O_2\)) in a container, the gases don't change their volume, but each exerts individual pressure. We calculate this using the formula:

\[P_i = x_iP\]
where \\(P_i\)\ is the partial pressure of a specific gas, \\(x_i\)\ is the mole fraction, and \(P\)\ is the total pressure of the mixture.

This means each gas's pressure depends on its proportion in the mixture. Partial pressure is crucial for understanding how different gases contribute to the total pressure in a system.
Mole Fraction
The mole fraction is a way to express the concentration of a specific gas in a mixture. It's calculated by comparing the moles of one gas to the total moles in the mixture.
It's expressed as \[x_i = \frac{n_i}{n_{total}}\]
where \\(n_i\)\ is the moles of a particular gas, and \\(n_{total}\)\ is the total moles of all gases combined in the mixture.

The mole fraction has no units, making it a simple ratio.
Using mole fraction makes it easy to determine the partial pressure when combining gases, since it represents each gas's share of the total number of particles in the mixture.
  • Mole fraction allows us to focus on the relative quantities of gases.
  • This concept helps in further calculations, like partial pressures and reactions involving gases.
Gas Mixture
A gas mixture is created when different gases are combined in the same container, like our nitrogen and oxygen example. When mixed, each gas occupies the total container volume, but they do not influence each other’s behavior. This happens because gas particles are far apart, allowing free movement.

The gas mixture's behavior is predicted by combining the properties of the individual gases. When mixing gases:
  • The temperature remains constant.
  • Each gas acts as if it were alone.
  • The volume occupied by each gas is the volume of the container.

Mixing gases doesn't change their individual characteristics, such as volume or pressure exerted. Instead, it allows us to study their collective properties and impacts.
Dalton's Law of Partial Pressures
Dalton's Law of Partial Pressures helps explain how gas mixtures behave. It states that the total pressure of a gas mixture equals the sum of the partial pressures of each individual gas.
Expressed as \[P_{total} = P_1 + P_2 + P_3 + ... + P_n\]
where each \(P\)\ is the partial pressure of a gas.

Dalton’s Law is immensely helpful when calculating how different gases contribute to a mixture’s total pressure in a container. Each gas's pressure remains unaffected by the presence of other gases. This principle allows chemists to predict and balance how gases will interact in mixtures.

By understanding Dalton's Law, one can efficiently calculate the pressures in any gas system, which is crucial in both experimental and theoretical chemistry contexts.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Ammonia and hydrogen chloride react to form solid ammonium chloride: $$ \mathrm{NH}_{3}(g)+\mathrm{HCl}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s) $$ Two \(2.00\) - \(\mathrm{L}\) flasks at \(25^{\circ} \mathrm{C}\) are connected by a valve, as shown in the drawing on the next page. One flask contains \(5.00 \mathrm{~g}\) of \(\mathrm{NH}_{3}(g)\), and the other contains \(5.00 \mathrm{~g}\) of \(\mathrm{HCl}(g)\). When the valve is opened, the gases react until one is completely consumed. (a) Which gas will remain in the system after the reaction is complete? (b) What will be the final pressure of the system after the reaction is complete? (Neglect the volume of the ammonium chloride formed.) (c) What mass of ammo-

Hydrogen gas is produced when zinc reacts with sulfuric acid: $$ \mathrm{Zn}(s)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q) \longrightarrow \mathrm{ZnSO}_{4}(a q)+\mathrm{H}_{2}(g) $$ If \(159 \mathrm{~mL}\) of wet \(\mathrm{H}_{2}\) is collected over water at \(24^{\circ} \mathrm{C}\) and a barometric pressure of 738 torr, how many grams of \(\mathrm{Zn}\) have been consumed? (The vapor pressure of water is tabulated in Appendix B.)

Consider the sample of gas depicted here. What would the drawing look like if the volume and temperature remained constant while you removed enough of the gas to decrease the pressure by a factor of 2\(?[\) Section 10.3\(]\) \begin{equation}\begin{array}{l}{\text { (a) It would contain the same number of molecules. }} \\ {\text { (b) It would contain half as many molecules. }} \\\ {\text { (c) It would contain twice as many molecules. }} \\ {\text { (d) There is insufficient data to say. }}\end{array}\end{equation}

The typical atmospheric pressure on top of Mt. Everest \((29,028 \mathrm{ft})\) is about 265 torr. Convert this pressure to (a) atm, (b) \(\mathrm{mm} \mathrm{Hg}\), (c) pascals, (d) bars, (e) psi.

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and has a boiling point at atmospheric pressure of \(-164^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) is oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\) ? (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane just described and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL}\); the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.