Chemistry: An Introduction to General, Organic, and Biological Chemistry

Chapter 8: Gases

Introduction

Gases play a crucial role in numerous biological processes and industrial applications. This chapter introduces the fundamental properties of gases, the laws governing their behavior, and practical applications in fields such as respiratory therapy. Respiratory therapists assess and treat patients' respiratory issues by conducting diagnostic tests that include measuring breathing capacity and blood gas concentrations.

8.1 Properties of Gases

The atmosphere, or the gases surrounding Earth, contains a mixture of gases, including:

  • Oxygen (O₂): 21% of the atmosphere, essential for life processes in both plants and animals.
  • Nitrogen (N₂): 78% of the atmosphere.
  • Ozone (O₃): Formed in the upper atmosphere when oxygen interacts with ultraviolet light.
  • Other gases such as argon, carbon dioxide (CO₂), and water vapor.
Learning Goal
  • Describe the kinetic molecular theory of gases and the units of measurement used for gases.

Kinetic Molecular Theory

The kinetic molecular theory explains the behavior of gases by suggesting that:

  • Gases consist of tiny particles that move rapidly in straight lines.
  • There are no significant attractive or repulsive forces between the particles.
  • The particles are spaced far apart compared to their size.
  • The volume occupied by the gas particles is negligible compared to the volume of the container.
  • The average kinetic energy of the particles increases with temperature.

Properties That Describe a Gas

Gases are characterized by four primary properties:

  • Pressure (P): The force exerted by gas particles colliding with the walls of their container.
  • Volume (V): The space occupied by the gas, directly correlated with the container's volume.
  • Temperature (T): A measure related to the average kinetic energy of the gas particles, measured in Kelvin (K).
  • Amount (n): The quantity of gas generally expressed in moles (mol).

Volume of a Gas

  • The volume of a gas matches the volume of its container.
  • It is normally measured in liters (L) or milliliters (mL).
  • Volume increases as temperature increases, maintaining constant pressure.

Temperature of a Gas

  • Temperature correlates with the average kinetic energy of gas molecules and is measured using the Kelvin scale.
  • When a gas's temperature decreases, its molecules collide less frequently; conversely, an increase in temperature leads to more collisions.

Pressure of a Gas

  • Pressure measures the collisions of gas particles with the container walls, expressed in various units:
    • Millimeters of mercury (mmHg) or Torr
    • Atmospheres (atm)
    • Pascals (Pa) or kilopascals (kPa)
    • Pounds per square inch (psi)
  • Atmospheric pressure is the pressure exerted by gases in the atmosphere, measured as 1 atm at sea level.

Barometers Measure Pressure

  • A barometer measures atmospheric pressure, represented as the height of a mercury column: 760 mmHg = 1 atm = 760 Torr.
  • Invented by Evangelista Torricelli, the barometer is 760 mm high at sea level.

Atmospheric Pressure

  • Atmospheric pressure diminishes with increasing altitude.
  • At sea level, atmospheric pressure equals 1 atm.

Units for Measuring Pressure

  • Standard units for measuring pressure include mmHg, Torr, atm, Pa, kPa, and psi.

Altitude and Atmospheric Pressure

  • Atmospheric pressure is subject to fluctuation based on weather and altitude.
    • High atmospheric pressure on sunny days causes the mercury column in barometers to rise.
    • On rainy days, the lower pressure results in a drop in the mercury column.

Learning Check

  1. What is 475 mmHg expressed in atmospheres?
    • A. 475 atm
    • B. 0.625 atm
    • C. 3.61 × 105 atm
  2. The pressure in a tire is 2.00 atm. What is this pressure in millimeters of mercury?
    • A. 2.00 mmHg
    • B. 1520 mmHg
    • C. 22,300 mmHg
Solution
  1. 475 mmHg = 0.625 atm (Answer B).
  2. 2.00 atm is 1520 mmHg (Answer B).

8.2 Pressure and Volume: Boyle's Law

Boyle's Law states that the pressure of a gas is inversely related to its volume when temperature and amount of gas are held constant.

Learning Goal

Use the pressure–volume relationship (Boyle’s law) to calculate the unknown pressure or volume when temperature and amount of gas do not change.

Boyle's Law Formula
  • Boyle's law can be mathematically expressed as:
    P1V1=P2V2P_1V_1 = P_2V_2
    Where:
  • $P_1$ and $V_1$ are the initial pressure and volume.
  • $P_2$ and $V_2$ are the final pressure and volume.
Example of Boyle’s Law
  • If $P_1 = 8.0$ atm and $V_1 = 2.0$ L, then
    P1V1=16extatmLP_1V_1 = 16 ext{ atm L}
  • For $P_2 = 4.0$ atm and $V_2 = 4.0$ L,
    P2V2=16extatmLP_2V_2 = 16 ext{ atm L}
  • For $P_3 = 2.0$ atm and $V_3 = 8.0$ L,
    P3V3=16extatmLP_3V_3 = 16 ext{ atm L}
Chemistry Link to Health: Boyle's Law and Breathing
  • During inhalation:
    • Lungs expand, reducing pressure in the lungs, letting air flow in.
  • During exhalation:
    • Lung volume decreases, increasing pressure, expelling air outward.
Calculations Using Boyle’s Law

To find the new volume of an 8.0-L sample of Freon gas initially at 550 mmHg after its pressure changes to 2200 mmHg:

  1. Given and Needed Quantities:
    • $P_1 = 550$ mmHg, $V_1 = 8.0$ L, $P_2 = 2200$ mmHg, find $V_2$.
  2. Problem Analysis:
    • Predict: As pressure increases, volume decreases.
  3. Rearranging Boyle’s Law:
    • V2=P1V1P2V_2 = \frac{P_1V_1}{P_2}
  4. Substituting Values: (actual calculation should show various mathematical steps).
Learning Check

If a sample of oxygen gas has a volume of 12.0 L at 600 mmHg, what is the new pressure when the volume changes to 36.0 L?

  • A. 200 mmHg
  • B. 400 mmHg
  • C. 1200 mmHg
Solution
  1. Analyzing Given:
    • $P_1 = 600$ mmHg, $V_1 = 12.0$ L, $V_2 = 36.0$ L.
  2. Rearranging to find $P_2$:
    P2=P1V1V2P_2 = \frac{P_1V_1}{V_2}
  3. Substituting values gives:
    • The answer is A, 200 mmHg.

8.3 Temperature and Volume: Charles's Law

Increasing the temperature of a gas sample causes its kinetic energy and volume to rise when pressure and amount of gas are constant.

Learning Goal

Use the temperature–volume relationship (Charles's law) to calculate unknown temperature or volume when pressure and amount of gas do not change.

Charles's Law Formula
  • Expressed mathematically as:
    V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2}
  • Where:
    • $T$ is in Kelvin.
Example of Charles’s Law
  • Initial Volume at 21 °C (294 K) of a balloon is 785 mL and changes to 0 °C (273 K).
  1. Given Quantities:
    • $T_1 = 294$ K, $V_1 = 785$ mL, $T_2 = 273$ K.
  2. Rearranging the Equation:
    • Solve for $V_2$:
      V2=V1T2T1V_2 = \frac{V_1T_2}{T_1}
  3. Final Calculation: (detailed calculation of substitution).
Learning Check

A sample of oxygen gas that has a volume of 420 mL at 18 °C, what is the temperature when the volume is 640 mL?

  • A. 443 °C
  • B. 170 °C
  • C. −82 °C
Solution
  1. Analyze Given:
    • $T_1 = 291$ K, $V_1 = 420$ mL, $V_2 = 640$ mL.
  2. Rearranging for $T_2$:
    T2=V2T1V1T_2 = \frac{V_2 T_1}{V_1}
  3. Substituting: (provide calculations). The answer is B.

8.4 Temperature and Pressure: Gay-Lussac’s Law

Gay-Lussac’s Law states that the pressure of a gas is directly proportional to its Kelvin temperature when volume and amount are constant.

Learning Goal

Use the temperature–pressure relationship (Gay-Lussac’s law) to calculate unknown temperature or pressure when volume and amount do not change.

Gay-Lussac’s Law Formula
  • Expressed mathematically as:
    P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}
Example of Gay-Lussac's Law
  • If $P_1 = 2.0$ atm at a temperature of 18 °C, what will be $P_2$ at 62 °C?
  1. Given Quantities:
    • $T_1 = 291$ K, $T_2 = 335$ K.
  2. Calculating Pressure:
    • Substitute into the equation for $P_2$ and calculate.
Learning Check

What is the pressure of gas at 645 Torr at 128 °C if raised to 824 Torr?

Solution
  1. Identify Given and Needed Quantities.
  2. Rearranging: (Provide equation alterations for use).

8.5 The Combined Gas Law

The Combined Gas Law integrates Boyle's, Charles's, and Gay-Lussac’s laws, allowing calculations when considering changes in pressure, volume, and temperature simultaneously while keeping n (amount of gas) constant.

Formula

P1V1T1=P2V2T2\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}

Learning Check

Calculate the volume when pressure changes from 646 mmHg to 802 mmHg and temperature alters from 35 °C to −95 °C:

  1. Identify Variables
  2. Rearranging and Calculating.

8.6 Volume and Moles: Avogadro’s Law

  • Avogadro's Law states that the volume of a gas (V) is directly proportional to the number of moles (n) of gas at constant temperature and pressure. In standard temperature and pressure (STP), one mole occupies 22.4 L.
Molar Volume at STP
  • Standard Temperature: 0 °C (273 K)
  • Standard Pressure: 1 atm (760 mmHg)
  • Molar Volume: 22.4 L/mole

8.7 Partial Pressure: Dalton’s Law

  • Dalton’s law of partial pressures states that the total pressure of a gas mixture is the sum of the partial pressures of the individual gases present in the mixture.
  • Formula:
    Ptotal=P1+P2+P3+…P_{total} = P_1 + P_2 + P_3 +…
Example Calculation for Partial Pressure
  • A scuba tank contains O₂ at 0.450 atm and He at 855 mmHg:
  1. Equation Setup:
    • Calculate total pressure using given equations.
  2. Final Calculation: (Substitute and calculate).
Learning Check

For a deep dive, if a mixture of gases totals 8.00 atm, determine the partial pressure of the helium given the partial pressure of oxygen is 1280 mmHg.