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A 10.0-L balloon contains helium gas at a pressure of \(655 \mathrm{mmHg}\). What is the final pressure, in millimeters of mercury, of the helium gas at each of the following volumes, if there is no change in temperature and amount of gas? a. \(20.0 \mathrm{~L}\) b. \(2.50 \mathrm{~L}\) c. \(13800 \mathrm{~mL}\) d. \(1250 \mathrm{~mL}\)

Short Answer

Expert verified
a. 327.5 mmHgb. 2620 mmHgc. 474.6 mmHgd. 5240 mmHg

Step by step solution

01

- Understand the gas law

The problem involves changing the volume of a gas, so use Boyle's Law, which states that for a given amount of gas at constant temperature, the product of pressure and volume is constant: \[ P_1 V_1 = P_2 V_2 \].
02

- Identify initial conditions

The initial conditions given are: Initial pressure, \( P_1 = 655 \, \text{mmHg} \) Initial volume, \( V_1 = 10.0 \, \text{L} \).
03

- Calculate final pressures for each volume

For each new volume, use the formula derived from Boyle's Law: \[ P_2 = \frac{P_1 V_1}{V_2} \].
04

- Solve for 20.0 L

For \( V_2 = 20.0 \, \text{L} \), plug in the values:\[ P_2 = \frac{655 \, \text{mmHg} \times 10.0 \, \text{L}}{20.0 \, \text{L}} = 327.5 \, \text{mmHg} \].
05

- Solve for 2.50 L

For \( V_2 = 2.50 \, \text{L} \), plug in the values:\[ P_2 = \frac{655 \, \text{mmHg} \times 10.0 \, \text{L}}{2.50 \, \text{L}} = 2620 \, \text{mmHg} \].
06

- Solve for 13800 mL

Convert 13800 mL to liters: \( 13800 \, \text{mL} = 13.8 \, \text{L} \).For \( V_2 = 13.8 \, \text{L} \), plug in the values:\[ P_2 = \frac{655 \, \text{mmHg} \times 10.0 \, \text{L}}{13.8 \, \text{L}} \approx 474.6 \, \text{mmHg} \].
07

- Solve for 1250 mL

Convert 1250 mL to liters: \( 1250 \, \text{mL} = 1.25 \, \text{L} \).For \( V_2 = 1.25 \, \text{L} \), plug in the values:\[ P_2 = \frac{655 \, \text{mmHg} \times 10.0 \, \text{L}}{1.25 \, \text{L}} = 5240 \, \text{mmHg} \].

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Key Concepts

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

Gas Laws
The **gas laws** are fundamental principles that describe the behavior of gases. They help us understand how variables such as pressure, volume, and temperature relate to each other in a gaseous state.
For instance, **Boyle's Law** is a key gas law that states that the pressure of a gas is inversely proportional to its volume when temperature and the amount of gas are kept constant.
In simpler terms, if the volume of gas decreases, its pressure increases, and vice versa.
This concept is crucial for solving problems where you need to find out how changes in one property (like volume) affect another (like pressure).

Some other important gas laws include:
  • **Charles's Law:** Describes how gas volume changes with temperature.
  • **Avogadro's Law:** Describes how gas volume changes with the number of gas particles.
Each of these laws can be combined to form the **Ideal Gas Law**, which we'll discuss in more detail later.
Pressure-Volume Relationship
The **pressure-volume relationship** in gases is foundational in understanding how gases behave under changing conditions.
This relationship is clearly explained by **Boyle's Law**, which can be mathematically represented as: \( P_1V_1 = P_2V_2 \).
Here, **\( P_1 \)** and **\( V_1 \)** represent the initial pressure and volume of a gas, respectively, while **\( P_2 \)** and **\( V_2 \)** represent the final pressure and volume after a change has occurred.

  • If the volume of a gas increases, the pressure decreases, provided the temperature and amount of gas stay the same.
  • Conversely, if the volume decreases, the pressure increases.
These principles are put into practice when solving problems involving changes in gas volume and pressure. For instance, given an initial pressure and volume, you can use Boyle's Law to find the new pressure if the volume changes:
  • For example, if you start with a volume of 10.0 L and a pressure of 655 mmHg, and the volume changes to 2.50 L, you can find the new pressure using the formula: \[ P_2 = \frac{P_1 \times V_1}{V_2} \ = \frac{655 \text{ mmHg} \times 10.0 \text{ L}}{2.50 \text{ L}} \ = 2620 \text{ mmHg}. \]
Ideal Gas Law
The **Ideal Gas Law** is a comprehensive equation that encapsulates the relationships described by other gas laws.
It combines Boyle's Law, Charles's Law, and Avogadro's Law into one formula:
  • \( PV = nRT \)
Here, **P** is the pressure, **V** is the volume, **n** is the number of moles of gas, **R** is the universal gas constant, and **T** is the temperature in Kelvin.

Using the Ideal Gas Law, you can predict how a gas will behave under different conditions:
  • It tells you how increasing the number of gas particles (n) will affect the pressure and volume.
  • It shows how changing the temperature (T) will influence the gas's pressure and volume.
The Ideal Gas Law is practical for many scenarios, from everyday applications (like understanding how a balloon inflates) to industrial processes (such as chemical reactions involving gases). While Boyle's Law focuses solely on the pressure-volume relationship at constant temperature, the Ideal Gas Law gives you a more complete picture by also considering the amount of gas and temperature.

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Most popular questions from this chapter

Calculate the final temperature, in degrees Celsius, \({ }^{\circ} \mathrm{C}\), for each of the following, with \(n\) and \(V\) constant: a. A sample of xenon at \(25^{\circ} \mathrm{C}\) and \(740 . \mathrm{mmHg}\) is cooled to give a pressure of \(620 . \mathrm{mmHg}\). b. A tank of argon gas with a pressure of \(0.950\) atm at \(-18^{\circ} \mathrm{C}\) is heated to give a pressure of 1250 torr.

Use the kinetic molecular theory of gases to explain each of the following: a. A container of nonstick cooking spray explodes when thrown into a fire. b. The air in a hot-air balloon is heated to make the balloon rise. c. You can smell the odor of cooking onions from far away.

A gas has a volume of \(4.00 \mathrm{~L}\) at \(0{ }^{\circ} \mathrm{C}\). What final temperature, in degrees Celsius, is needed to change the volume of the gas to each of the following, if \(n\) and \(P\) do not change? a. \(1.50 \mathrm{~L}\) b. \(1200 \mathrm{~mL}\) c. \(10.0 \mathrm{~L}\) d. \(50.0 \mathrm{~mL}\)

Identify the property of a gas that is measured in each of the following: a. \(425 \mathrm{~K}\) b. \(1.0 \mathrm{~atm}\) c. \(10.0 \mathrm{~L}\) d. \(0.50 \mathrm{~mol}\) of He

A scuba diver \(60 \mathrm{ft}\) below the ocean surface inhales \(50.0 \mathrm{~mL}\) of compressed air from a scuba tank at a pressure of \(3.00\) atm and a temperature of \(8{ }^{\circ} \mathrm{C}\). What is the final pressure of air, in atmospheres, in the lungs when the gas expands to \(150.0 \mathrm{~mL}\) at a body temperature of \(37^{\circ} \mathrm{C}\), and the amount of gas remains constant?

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