A sample of a gas occupies a volume of \(725 \mathrm{~mL}\) at \(825 \mathrm{torr}\). At constant temperature, what will be the new pressure (torr) when the volume changes to the following: (a) \(283 \mathrm{~mL}\) (b) \(2.87 \mathrm{~L}\)

Short Answer

Expert verified
The new pressures are (a) approximately 2111.2 torr and (b) approximately 208.2 torr.

Step by step solution

01

Use Boyle’s Law to set up the equation

Boyle’s Law states that for a given mass of gas at constant temperature, the pressure of the gas is inversely proportional to its volume. The equation is given by: \( P_1V_1 = P_2V_2 \). In this case, \( P_1 = 825 \) torr, \( V_1 = 725 \) mL, \( V_2 \) will change as per each scenario and \( P_2 \) is what we need to find.
02

Solve for the new pressure (a) \( P_{2a} \) when \( V_2 = 283 \) mL

Plugging the values into Boyle's Law for scenario (a), we get: \( 825 \) torr * \( 725 \) mL = \( P_{2a} \) * \( 283 \) mL. Solving for \( P_{2a} \), we have \( P_{2a} = \frac{825 \text{ torr} \times 725 \text{ mL}}{283 \text{ mL}} \).
03

Calculate the new pressure (a) \( P_{2a} \)

Perform the calculation for \( P_{2a} \): \( P_{2a} = \frac{825 \text{ torr} \times 725 \text{ mL}}{283 \text{ mL}} = \frac{597375 \text{ torr} \cdot \text{ mL}}{283 \text{ mL}} \approx 2111.2 \text{ torr} \).
04

Solve for the new pressure (b) \( P_{2b} \) when \( V_2 = 2.87 \) L

First, convert \( V_2 \) from liters to milliliters: \( V_2 = 2.87 \) L * \( 1000 \) mL/L = \( 2870 \) mL. Use Boyle's Law for scenario (b): \( 825 \) torr * \( 725 \) mL = \( P_{2b} \) * \( 2870 \) mL and solve for \( P_{2b} \).
05

Calculate the new pressure (b) \( P_{2b} \)

Perform the calculation for \( P_{2b} \): \( P_{2b} = \frac{825 \text{ torr} \times 725 \text{ mL}}{2870 \text{ mL}} = \frac{597375 \text{ torr} \cdot \text{ mL}}{2870 \text{ mL}} \approx 208.2 \text{ torr} \).

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 Vaia!

Key Concepts

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

Gas Laws
Understanding the behavior of gases is pivotal in the field of chemistry. The gas laws are a set of generalized equations which scientists have derived to describe how gases react to changes in pressure, volume, and temperature. One of the fundamental aspects of these laws is that they help predict the properties of a gas after it undergoes a change in one or more of these states. Boyle's Law, Charles's Law, and Gay-Lussac's Law are the three primary laws that describe the bulk behavior of a gas. When combined, they form the combined gas law, and they can also be derived from the ideal gas law, which incorporates all the variables into a single equation. To truly grasp how gases will behave under various conditions, one must comprehend these laws, which can also unveil the molecular level interactions within a given sample of gas.
Pressure-Volume Relationship
Boyle's Law is particularly focused on the pressure-volume relationship of a gas. It states that at constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume. This means that when the volume of a gas decreases, the pressure increases, as long as the temperature remains unchanged. The inverse nature of this relationship is crucial for various applications, including breathing mechanisms, syringes, and even the automotive engines. The mathematical representation of Boyle's Law is expressed as P_1V_1 = P_2V_2, where P_1 and V_1 are the initial pressure and volume, and P_2 and V_2 are the pressure and volume after the change. By understanding and applying this basic principle, one can solve practical problems involving the compression or expansion of gases.
Chemistry Education
In the field of chemistry education, simplifying complex concepts such as Boyle's Law is essential for effective learning. Through the step-by-step demonstration of how to manipulate and apply this law, students can build a solid foundation in understanding gas behavior. Learning to solve problems using Boyle's Law is more than just learning to rearrange an equation; it's about developing a conceptual framework for how particles of gas will respond to changes in their environment. Exercises involving this law not only enhance students' problem-solving skills but also encourage a deeper comprehension of the theoretical knowledge. Thus, teaching approaches need to focus both on solving equations and on fostering a practical understanding of the concepts at play—preparing students for real-world applications in various scientific fields.

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

The tires on an automobile were filled with air to 30. psi at \(71.0^{\circ} \mathrm{F}\). When driving at high speeds, the tires become hot. If the tires have a bursting pressure of 44 psi, at what temperature \(\left({ }^{\circ} \mathrm{F}\right)\) will the tires "blow out"?

Three gases were added to the same 2.0-I. container. The total pressure of the gases was 790 torr at room temperature \(\left(25.0^{\circ} \mathrm{C}\right)\). If the mixture contained \(0.65 \mathrm{~g}\) of oxygen gas, \(0.58 \mathrm{~g}\) of carbon dioxide, and an unknown amount of nitrogen gas, determine the following: (a) the total number of moles of gas in the container (b) the number of grams of nitrogen in the container (c) the partial pressure of each gas in the mixture

A baker is making strawberry cupcakes and wants to ensure they are very light. To do this, he must add enough baking soda to increase the volume of the cupcakes by \(55.0 \%\). All of the carbon dioxide produced by the decomposition of the baking soda is incorporated into the cupcakes. The baking soda or sodium bicarbonate will react with citric acid in the batter according to the reaction \(3 \mathrm{NaHCO}_{3}+\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7} \longrightarrow \mathrm{Na}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}+3 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{CO}_{2}\) which occurs with a \(63.7 \%\) yield. How many grams of baking soda must be added to \(1.32 \mathrm{~L}\) of cupcake batter at a baking temperature of \(325^{\circ} \mathrm{F}\) and a pressure of 738 torr to achieve the correct consistency?

You have a \(10-\mathrm{L}\) container filled with \(0.5 \mathrm{~mol}\) of \(\mathrm{O}_{2}\) gas at a temperature of \(30 .{ }^{\circ} \mathrm{C}\) with a pressure of 945 torr. (a) What will happen to the pressure if the container size is doubled while keeping the temperature and number of moles constant? (b) What will happen to the pressure when the temperature is doubled while keeping the size of the container and the number of moles constant? (c) What will happen to the pressure when the amount of \(\mathrm{O}_{2}\) gas is cut in half while keeping the size of the container and the temperature constant? (d) What will happen to the pressure if 1 mole of \(\mathrm{N}_{2}\) gas is added to the container while keeping the temperature and size of the container the same?

A soccer ball of constant volume \(2.24 \mathrm{~L}\) is pumped up with air to a gauge pressure of \(13 \mathrm{lb} /\) in. \({ }^{2}\) at \(20.0^{\circ} \mathrm{C}\). The molar mass of air is about \(29 \mathrm{~g} / \mathrm{mol}\). (a) How many moles of air are in the ball? (b) What mass of air is in the ball? (c) During the game, the temperature rises to \(30.0^{\circ} \mathrm{C}\). What mass of air must be allowed to escape to bring the gauge pressure back to its original value?

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.

Sign-up for free