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Chapter 10: Problem 24

A sample of air occupies \(3.8 \mathrm{~L}\) when the pressure is \(1.2 \mathrm{~atm} .\) (a) What volume does it occupy at \(6.6 \mathrm{~atm}\) ? (b) What pressure is required to compress it to \(0.075 \mathrm{~L} ?\) (The temperature is kept constant.)

Short Answer

Expert verified
(a) 0.69 L; (b) 60.8 atm

Step by step solution

01

Understanding Boyle's Law

The problem involves changing pressure and volume while keeping the temperature constant, indicating Boyle's Law applies. Boyle's Law states that the product of the initial pressure and volume is equal to the product of the final pressure and volume, i.e., \( P_1V_1 = P_2V_2 \).
02

Setting Known Values for Part (a)

For part (a), we are given \( P_1 = 1.2 \text{ atm} \) and \( V_1 = 3.8 \text{ L} \), and we need to find the volume \( V_2 \) when the pressure \( P_2 = 6.6 \text{ atm} \). Using Boyle's Law, we set up the equation: \( 1.2 \times 3.8 = 6.6 \times V_2 \).
03

Solving for Volume in Part (a)

Rearrange the equation from part (a) to solve for \( V_2 \):\[ V_2 = \frac{1.2 \times 3.8}{6.6} \].Calculate \( V_2 \) to find the final volume when the pressure is increased to 6.6 atm.
04

Calculation for Part (a)

Calculate \( V_2 = \frac{1.2 \times 3.8}{6.6} \approx 0.69 \text{ L} \). Hence, the volume of the air sample at 6.6 atm is approximately 0.69 liters.
05

Setting Known Values for Part (b)

For part (b), we are given \( P_1 = 1.2 \text{ atm} \), \( V_1 = 3.8 \text{ L} \), and \( V_2 = 0.075 \text{ L} \). We need to find the required pressure \( P_2 \) to compress the sample to \( 0.075 \text{ L} \) using the equation: \( 1.2 \times 3.8 = P_2 \times 0.075 \).
06

Solving for Pressure in Part (b)

Rearrange the equation from part (b) to solve for \( P_2 \):\[ P_2 = \frac{1.2 \times 3.8}{0.075} \].Calculate \( P_2 \) to find the required pressure to compress the sample to 0.075 L.
07

Calculation for Part (b)

Calculate \( P_2 = \frac{1.2 \times 3.8}{0.075} \approx 60.8 \text{ atm} \). Therefore, the pressure required to compress the air sample to 0.075 L is approximately 60.8 atm.

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Key Concepts

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

Pressure and Volume Relationship
Boyle's Law beautifully illustrates the relationship between pressure and volume for a gas, provided the temperature remains constant. This fundamental principle of physics suggests that if you increase the pressure exerted on a gas, its volume will decrease, and vice versa. This is because gases are compressible; pressure pushes the gas molecules into a smaller space.

Here's how it works in simpler terms:
  • Imagine a balloon filled with air. If you squeeze it, the pressure on the air inside the balloon increases.
  • As the pressure increases, the air molecules are packed closer together 鈥� this means the balloon's volume decreases.
  • On the other hand, if you release some of that pressure, the balloon expands, showing that as pressure decreases, volume increases.
Mathematically, it's represented as: \( P_1V_1 = P_2V_2 \), where:
  • \( P_1 \) and \( V_1 \) are the initial pressure and volume.
  • \( P_2 \) and \( V_2 \) are the new pressure and volume after a change.
This relationship helps us predict the behavior of gases under different conditions, ensuring our calculations for volumes and pressures are accurate, as seen in exercises where this principle is applied.
Ideal Gas Law
The Ideal Gas Law combines several principles, including Boyle's Law, into one comprehensive equation: \[ PV = nRT \].This equation helps describe the state of an ideal gas by relating pressure \( P \), volume \( V \), the amount of gas \( n \), the ideal gas constant \( R \), and temperature \( T \).
  • "Ideal gas" refers to a theoretical gas that perfectly follows the equation, although real gases approximate this under many conditions.
  • The principle becomes particularly useful when we know how a gas will behave as conditions change, like in reacting gases or expanding balloons.
  • Boyle's Law can be derived from the Ideal Gas Law when the temperature and amount of gas are constant, leading to the same relationship \( PV = C \) (a constant).
It is crucial for advanced studies in chemistry and physics for predicting how gases will behave in various scenarios involving temperature, volume, and pressure changes.
Constant Temperature
Boyle's Law, and the related exercises, focus on scenarios where the temperature remains constant. This is also referred to as an isothermal process. When a process occurs at a constant temperature, we are specifically looking at an energy transfer where heat flows in or out, compensating for the work done, so the temperature does not change.
  • Isothermal processes allow gases to maintain their temperature even when the volume and pressure change.
  • For example, when compressing a gas in a piston slowly, the heat can escape or be absorbed, ensuring the temperature stays the same.
  • This constancy is crucial for applying Boyle's Law correctly; without it, the relationship \( P_1V_1 = P_2V_2 \) wouldn't hold.
In practical applications, maintaining a constant temperature might involve using a controlled environment or apparatus, such as a water bath, to ensure the changes in pressure and volume are studied under ideal conditions.

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Most popular questions from this chapter

At what temperature will He atoms have the same \(u_{\mathrm{rms}}\) value as \(\mathrm{N}_{2}\) molecules at \(25^{\circ} \mathrm{C} ?\)

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Dry ice is solid carbon dioxide. A \(0.050-\mathrm{g}\) sample of dry ice is placed in an evacuated 4.6-L vessel at \(30^{\circ} \mathrm{C}\). Calculate the pressure inside the vessel after all the dry ice has been converted to \(\mathrm{CO}_{2}\) gas.
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