CH 10 (11/6) (PG 1-9)

Chapter 10: Gases and Their Properties

Properties of Gases

• Compressibility: Gases can be compressed, allowing their volumes to decrease under pressure.

• Pressure Exertion: Gases exert pressure on any surrounding surface due to collisions with that surface.

• Expansion: Gases expand to fill the available volume of a container.

• Mixing: Gases mix homogeneously with one another.

Importance of Physical Properties

• Physical properties of gases are dependent on:

  • Pressure (P)

  • Temperature (T)

  • Volume (V)

  • Amount of substance (n)

• Example: When stating "oxygen is a gas", it refers to conditions of normal atmospheric pressure and room temperature.

Understanding Pressure

• Pressure is measured using a barometer, invented by Evangelista Torricelli.

• Pressure is defined through the relationship between the height of a liquid column (e.g., mercury) and the exerted atmospheric pressure.

Measurement of Pressure Units

• 1 atm is defined as:

  • 760 mm Hg = 1 atm

  • 760 torr = 1 atm

  • 101,325 Pa = 1 atm

  • 1.01325 bar = 1 atm

  • 14.696 lb/in² = 1 atm

• Note: Pa (Pascal) is often noted in kilopascals (kPa) where 1 kPa = 10³ Pa. Bar is also noted in millibars (mbar) where 1 mbar = 10⁻³ bar.

Example Calculation

• Conversion of pressure from mm Hg to atmospheres:

  • Given: 610 mm Hg

  • Calculation:
    $610 ext{ mm Hg} imes rac{1 ext{ atm}}{760 ext{ mm Hg}} = 0.803 ext{ atm}$
    $0.803 ext{ atm} = 0.803 ext{ atm} imes rac{101325 ext{ Pa}}{1 ext{ atm}} = 81300 ext{ Pa} = 81.3 ext{ kPa}$

Fundamental Gas Laws

• Gas Laws: Explore the relationships among Pressure (P), Volume (V), Temperature (T), and the number of moles (n).

Boyle’s Law

• Definition: The pressure of a gas is inversely proportional to the volume at a constant number of moles and constant temperature.

  • Mathematically: $PV = ext{constant}$

• Illustrative Example: Using a bicycle pump. Reducing the volume increases gas pressure, forcing air into a tire.

Example Application of Boyle’s Law

• Given a nitrogen gas sample with:

  • Pressure: 67.5 mm Hg

  • Initial Volume: 500.0 mL

  • Final Volume: 125 mL

• Find new pressure: $P<em>1V</em>1 = P<em>2V</em>2$ $67.5 ext{ mm Hg} imes 500.0 ext{ mL} = P_2 imes 125 ext{ mL}$

  • Solve for $P<em>2$ resulting in: $P</em>2 = 270. ext{ mm Hg}$

Charles’s Law

• Definition: The volume of a gas is directly proportional to its absolute temperature at a constant pressure and number of moles.

  • Mathematically: $rac{V<em>1}{T</em>1} = rac{V<em>2}{T</em>2}$

• Absolute Temperature Conversion:

  • $T ( ext{K}) = T (°C) + 273.15$

Example Application of Charles’s Law

• Scenario: A 5.0 mL sample of CO₂ gas at 22 °C is placed in an ice bath (0 °C).

• Assume the pressure remains constant:

  • Find new volume using temperature conversions:

  • React accordingly with given conditions.

General Gas Law

• Combination of Boyle’s and Charles’s Laws:

• Applicable for conditions where both temperature and pressure change.

  • $rac{P<em>1V</em>1}{T<em>1} = rac{P</em>2V<em>2}{T</em>2}$

Avogadro’s Hypothesis

• States that equal volumes of gases, at the same temperature and pressure contain equal numbers of molecules (or moles).

• This implies that volume is directly proportional to the number of moles when T and P are held constant:
  $V ext{} ext{is proportional to} ext{ } n$

Ideal Gas Law

• Definition: Combines the relationships defined in previous laws:
  $PV = nRT$

• Where:

  • P = Pressure

  • V = Volume

  • n = Number of moles

  • T = Absolute temperature in Kelvin

  • R = Ideal gas constant ($R = 0.08206 rac{L ext{atm}}{mol ext{K}}$)

Example Ideal Gas Calculation

• Find moles of gas present in a 250 mL flask with an oxygen pressure of 1.3 atm at 31 °C:

• Convert the temperature to Kelvin (K): $T = 31 °C + 273.15 = 304.15 ext{ K}$

  • Use ideal gas law:
    $n = rac{PV}{RT}$
    $n = rac{1.3 ext{ atm} imes 0.250 ext{ L}}{(0.08206 rac{L ext{atm}}{mol ext{K}}) imes 304.15 ext{ K}} = 0.013 mol$

Gas Density Calculation

• The density of a gas can be calculated from the ideal gas law by rearranging:

  • $d = rac{m}{V} = rac{PM}{RT}$

• Example: To find the density of oxygen at STP with a molar mass of O₂ = 32.00 g/mol.

Standard Temperature and Pressure (STP)

• Defined as:

  • 1 atm (760 torr) and 0 °C (273.15 K).

• Used for comparing gases under consistent conditions.

Example Problem at STP

• Calculate the volume occupied by 43.7 g of hydrogen gas at STP:

  • $43.7 g imes rac{1 mol}{2.02 g/mol} = 21.6 mol$

  • Using Ideal Gas Law:
    $PV = nRT <br>ightarrow V = rac{nRT}{P}$

Stoichiometry with Gases

• Gas reactions can often be related in terms of volumes.

• Example problem: For the reaction: 2 CO(g) + O₂(g) → 2 CO₂(g).

  • If 0.5 L of O₂ is consumed, then using stoichiometry:
    $0.5 L O₂ imes rac{2 L CO₂}{1 L O₂} = 1.00 L CO₂$.

Dalton’s Law of Partial Pressures

• Definition: The total pressure of a gas mixture is equal to the sum of the partial pressures of individual gases:
  $P<em>{total} = P</em>A + P<em>B + P</em>C + …$

Mole Fraction

• The mole fraction $X<em>A$ for gas A is given by: $X</em>A = rac{n<em>A}{n</em>{total}}$

  • Where nA is the mole of gas A and ntotal is the total moles in the mixture.

Practical Example**

• Given a gas mixture, calculate the partial pressures and the mole fractions of different gases in a flask.

This structured approach allows for systematic study, calculations and application under real-world scenarios in the field of chemistry.