Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 5: Problem 47

An aerosol can contains 400. mL of compressed gas at 5.20 atm. When all of the gas is sprayed into a large plastic bag, the bag inflates to a volume of 2.14 L. What is the pressure of gas in the plastic bag? Assume a constant temperature.

Short Answer

Expert verified
The pressure of the gas inside the plastic bag is approximately 0.97 atm.

Step by step solution

01

Understand the problem and the given data

We have an initial volume (V1) of 400 mL, which is the volume of compressed gas inside the aerosol can. The initial pressure (P1) is 5.20 atm. When the gas is transferred to the plastic bag, it inflates to a final volume (V2) of 2.14 L. We need to find the final pressure (P2) of the gas inside the plastic bag. Note: It is important to convert all given volumes to the same units.
02

Convert initial volume to liters

To convert the initial volume from mL to L, we can use the following conversion factor: 1 L = 1000 mL Therefore, the initial volume (V1) in liters is: V1 = 400 mL × \( \frac{1 L}{1000 mL} \) = 0.4 L
03

Apply the ideal gas law

Since the temperature and the number of moles of gas remain constant, we can use the following equation, which is derived from the ideal gas law: \( \frac{P1 \times V1}{T1} = \frac{P2 \times V2}{T2} \) Since T1 = T2, the equation can be simplified to: P1 × V1 = P2 × V2 We know the values of P1, V1, and V2, so we can solve for P2: P2 = \( \frac{P1 \times V1}{V2} \)
04

Calculate the final pressure

Now, we can substitute the given values into the equation to find P2: P2 = \( \frac{5.20 \; atm \times 0.4 \; L}{2.14 \; L} \) ≈ 0.97 atm So, the pressure of the gas inside the plastic bag is approximately 0.97 atm.

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!

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

A \(2.50-\mathrm{L}\) flask contains 0.60 \(\mathrm{g} \mathrm{O}_{2}\) at a temperature of \(22^{\circ} \mathrm{C}\) . What is the pressure (in atm) inside the flask?

Freon- 12\(\left(\mathrm{CF}_{2} \mathrm{Cl}_{2}\right)\) is commonly used as the refrigerant in central home air conditioners. The system is initially charged to a pressure of 4.8 atm. Express this pressure in each of the fol lowing units \((1 \mathrm{atm}=14.7 \mathrm{psi}).\) a. mm Hg b. torr c. Pa d. psi

Suppose two \(200.0-\mathrm{L}\) tanks are to be filled separately with the gases helium and hydrogen. What mass of each gas is needed to produce a pressure of 2.70 atm in its respective tank at \(24^{\circ} \mathrm{C} ?\)

A 1.00 -L gas sample at \(100 .^{\circ} \mathrm{C}\) and 600 . torr contains 50.0\(\%\) helium and 50.0\(\%\) xenon by mass. What are the partial pressures of the individual gases?

Ideal gas particles are assumed to be volume less and to neither attract nor repel each other. Why are these assumptions crucial to the validity of Dalton’s law of partial pressures?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

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