Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 25

A sample of methane is placed in a \(10.0-\mathrm{L}\) container at \(25^{\circ} \mathrm{C}\) and \(725 \mathrm{mmHg}\). The gas sample is then moved to a \(7.50-\mathrm{L}\) container at \(25^{\circ} \mathrm{C}\). What is the gas pressure in the second container?

Short Answer

Expert verified
966.67 mmHg

Step by step solution

01

Understand the Gas Law

This exercise involves gas laws. Since temperature and the amount of gas are constant, apply Boyle's Law which states that for a fixed amount of gas at constant temperature, the pressure and volume of a gas are inversely proportional: \[ P_1 V_1 = P_2 V_2 \] where \( P_1 \) and \( P_2 \) are the initial and final pressures, and \( V_1 \) and \( V_2 \) are the initial and final volumes.
02

Identify known values

From the problem, the known values are: \( P_1 = 725 \text{ mmHg} \), \( V_1 = 10.0 \text{ L} \), \( V_2 = 7.50 \text{ L} \).
03

Rearrange Boyle's Law

To find the final pressure \( P_2 \), rearrange the Boyle's Law formula to solve for \( P_2 \): \[ P_2 = \frac{P_1 V_1}{V_2} \]
04

Substitute the known values

Substitute the known values into the rearranged equation: \[ P_2 = \frac{725 \text{ mmHg} \times 10.0 \text{ L}}{7.50 \text{ L}} \]
05

Calculate the final pressure

Perform the calculation to determine \( P_2 \): \[ P_2 = \frac{7250 \text{ mmHg} \text{ L}}{7.50 \text{ L}} = 966.67 \text{ mmHg} \]
06

Interpret the result

The final pressure of the methane gas in the smaller container is approximately 966.67 mmHg.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Gas Laws
Gas laws describe the behavior of gases under different conditions of pressure, volume, and temperature. Several laws combined make up the gas laws, which include Boyle's Law, Charles's Law, and Avogadro's Law, among others. These laws help us predict how a gas will behave as we change one or more of its properties while keeping others constant.

In the given exercise, Boyle's Law is applied. This specific law is straightforward and relates pressure and volume of a gas at a constant temperature and amount of gas. Understanding Boyle's Law is fundamental for solving problems like the one presented.
Pressure-Volume Relationship
The pressure-volume relationship is a critical aspect of gas behavior. According to Boyle's Law, the volume of a gas is inversely proportional to its pressure when temperature and quantity of gas are held constant.

This relationship is expressed mathematically as: P_1 V_1 = P_2 V_2n . This means if you increase the volume of the container, the pressure decreases, and if you decrease the volume, the pressure increases, provided the temperature and amount of gas do not change.
Inverse Proportionality of Pressure and Volume
Inverse proportionality is a key concept within Boyle's Law. It means that as one variable increases, the other decreases in such a way that their product remains constant. In the context of gases, if the volume of the gas is reduced, the pressure increases proportionally, and vice versa.

For instance, in our exercise, the methane gas was initially in a 10.0 L container at 725 mmHg. When moved to a smaller 7.50 L container, the pressure increased to approximately 966.67 mmHg, illustrating the inverse proportionality between pressure and volume. This predictable behavior is crucial in various scientific and practical applications, such as in understanding how gases will behave in different scenarios.

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

White phosphorus melts and then vaporizes at high temperatures. The gas effuses at a rate that is 0.404 times that of neon in the same apparatus under the same conditions. How many atoms are in a molecule of gaseous white phosphorus?

What is the effect of the following on the volume of \(1 \mathrm{~mol}\) of an ideal gas? (a) The pressure is tripled (at constant \(T\) ). (b) The absolute temperature is increased by a factor of 3.0 (at constant \(P\) ). (c) Three more moles of the gas is added (at constant \(P\) and \(T\) ).

When gaseous \(\mathrm{F}_{2}\) and solid \(\mathrm{I}_{2}\) are heated to high temperatures, the \(\mathrm{I}_{2}\) sublimes and gaseous iodine heptafluoride forms. If 350. torr of \(\mathrm{F}_{2}\) and \(2.50 \mathrm{~g}\) of solid \(\mathrm{I}_{2}\) are put into a 2.50 - \(\mathrm{L}\) container at \(250 . \mathrm{K}\) and the container is heated to \(550 . \mathrm{K},\) what is the final pressure (in torr)? What is the partial pressure of \(\mathrm{I}_{2}\) gas?

For each of the following, which shows the greater deviation from ideal behavior at the same set of conditions? Explain. (a) Argon or xenon (b) Water vapor or neon (c) Mercury vapor or radon (d) Water vapor or methane

Nitrogen dioxide is used industrially to produce nitric acid, but it contributes to acid rain and photochemical smog. What volume \((\) in \(\mathrm{L})\) of nitrogen dioxide is formed at 735 torr and \(28.2^{\circ} \mathrm{C}\) by reacting \(4.95 \mathrm{~cm}^{3}\) of copper \(\left(d=8.95 \mathrm{~g} / \mathrm{cm}^{3}\right)\) with \(230.0 \mathrm{~mL}\) of nitric acid \(\left(d=1.42 \mathrm{~g} / \mathrm{cm}^{3}, 68.0 \% \mathrm{HNO}_{3}\right.\) by mass \() ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemistry Branches

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation
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.

Sign-up for free