Properties of Gases and The Ideal Gas Law

When tackling gas law problems, it's crucial to identify the given variables, what's being asked, and which variables remain constant. This will guide you to the correct equation.

Boyle's Law (Pressure-Volume Relationship)

  • Describes: The inverse relationship between pressure (P) and volume (V) of a gas when temperature (T) and the number of moles (n) are kept constant.

  • Formula: P1V1 = P2V2

  • Approach: Use this law when you have initial and final conditions for pressure and volume, and the temperature and moles of gas don't change. Remember if one increases, the other must decrease proportionally.

Worked Example:

Problem: A gas occupies 12.0 \, L at 1.20 \, atm. What is its volume if the pressure increases to 2.40 \, atm while temperature and moles remain constant?

Step-by-step Explanation:

  1. List Knowns and Unknowns:

    • Initial Pressure (P_1) = 1.20 \, atm

    • Initial Volume (V_1) = 12.0 \, L

    • Final Pressure (P_2) = 2.40 \, atm

    • Final Volume (V_2) = ?

  2. Choose the Right Formula: Since pressure and volume change while temperature and moles are constant, use Boyle's Law: P1V1 = P2V2

  3. Rearrange and Solve: Solve for V_2

    • V2 = (P1V1)/P2

    • V_2 = \frac{(1.20 \, atm)(12.0 \, L)}{2.40 \, atm}

    • V_2 = 6.00 \, L

  4. Check Your Answer: The pressure doubled, so the volume should be halved, which is consistent with the answer.

Charles's Law (Volume-Temperature Relationship)

  • Describes: The direct relationship between the volume (V) and absolute temperature (T) of a gas when pressure (P) and the number of moles (n) are kept constant.

  • Formula: V1/T1 = V2/T2

  • Approach: Apply this law when volume and temperature change, but pressure and moles are constant. Important: Temperature must always be in Kelvin (K = ^\circ C + 273.15).

Worked Example:

Problem: A balloon has a volume of 2.50 \, L at 25 ^\circ C. If the temperature is increased to 50 ^\circ C at constant pressure and moles, what is the new volume?

Step-by-step Explanation:

  1. List Knowns and Unknowns:

    • Initial Volume (V_1) = 2.50 \, L

    • Initial Temperature (T_1) = 25 ^\circ C

    • Final Temperature (T_2) = 50 ^\circ C

    • Final Volume (V_2) = ?

  2. Convert Units: Convert temperatures to Kelvin.

    • T_1 = 25 ^\circ C + 273.15 = 298.15 \, K

    • T_2 = 50 ^\circ C + 273.15 = 323.15 \, K

  3. Choose the Right Formula: Since volume and temperature change at constant pressure and moles, use Charles's Law: V1/T1 = V2/T2

  4. Rearrange and Solve: Solve for V_2

    • V2 = (V1T2)/T1

    • V_2 = \frac{(2.50 \, L)(323.15 \, K)}{298.15 \, K}

    • V_2 \approx 2.71 \, L

  5. Check Your Answer: The temperature increased, so the volume also increased, which is consistent with the direct relationship of Charles's Law.

Gay-Lussac's Law (Pressure-Temperature Relationship)

  • Describes: The direct relationship between the pressure (P) and absolute temperature (T) of a gas when volume (V) and the number of moles (n) are kept constant.

  • Formula: P1/T1 = P2/T2

  • Approach: Use this when pressure and temperature change under constant volume and moles. Again, temperature must be in Kelvin.

Combined Gas Law

  • Describes: A combination of Boyle's, Charles's, and Gay-Lussac's laws, applicable when only the number of moles (n) is constant.

  • Formula: (P1V1)/T1 = (P2V2)/T2

  • Approach: This is your go-to when pressure, volume, and temperature all change (or some combination of two while the third is unknown) but the amount of gas remains the same. Make sure temperatures are in Kelvin.

Avogadro's Law (Volume-Moles Relationship)

  • Describes: The direct relationship between the volume (V) and the number of moles (n) of a gas when pressure (P) and temperature (T) are kept constant.

  • Formula: V1/n1 = V2/n2

  • Approach: Use this when the volume of a gas changes due to a change in the amount (moles) of gas, at constant temperature and pressure. It's often used with stoichiometry problems involving gases.

Ideal Gas Law

  • Describes: Describes the state of a hypothetical ideal gas, relating pressure, volume, number of moles, and absolute temperature.

  • Formula: PV = nRT

  • Where:

    • P = Pressure

    • V = Volume

    • n = Number of moles

    • R = Ideal gas constant (e.g., 0.0821 \, L \cdot atm / (mol \cdot K) or 8.314 \, J / (mol \cdot K), choose based on units of P and V)

    • T = Absolute temperature (in Kelvin)

  • Approach: Use this law when you're given three of the four properties (P, V, n, T) of a gas at a single point in time and need to find the fourth. It's not for comparing initial and final states.

Worked Example:

Problem: What is the pressure exerted by 0.50 \, mol of a gas at 27 ^\circ C in a 10.0 \, L container? Use R = 0.0821 \, L \cdot atm / (mol \cdot K).

Step-by-step Explanation:

  1. List Knowns and Unknowns:

    • Number of moles (n) = 0.50 \, mol

    • Temperature (T) = 27 ^\circ C

    • Volume (V) = 10.0 \, L

    • Ideal gas constant (R) = 0.0821 \, L \cdot atm / (mol \cdot K)

    • Pressure (P) = ?

  2. Convert Units: Convert temperature to Kelvin.

    • T = 27 ^\circ C + 273.15 = 300.15 \, K

  3. Choose the Right Formula: Since you have n, T, V, and R for a single state, use the Ideal Gas Law: PV = nRT

  4. Rearrange and Solve: Solve for P

    • P = (nRT)/V

    • P = \frac{(0.50 \, mol)(0.0821 \, L \cdot atm / (mol \cdot K))(300.15 \, K)}{10.0 \, L}

    • P \approx 1.23 \, atm

Graham's Law (Effusion and Diffusion)

  • Describes: How the rate at which gas particles escape (effuse) through a small hole or spread out (diffuse) depends on their molar mass. Lighter gases effuse/diffuse faster than heavier gases.

  • Effusion: The process where a gas escapes through a tiny hole into a vacuum.

  • Diffusion: The process where gas particles spread out from an area of higher concentration to an area of lower concentration.

  • Difference: Effusion involves escaping through a small opening, while diffusion involves spreading out within a volume.

  • Formula: Rate1/Rate2 = sqrt(M2/M1)

  • Where:

    • Rate = Rate of effusion or diffusion

    • M = Molar mass of the gas

  • Approach: Use this law to compare the rates of effusion or diffusion for two different gases, or to determine an unknown molar mass if rates are known. The rates are inversely proportional to the square root of their molar masses.

Worked Example:

Problem: If oxygen gas (O2, molar mass 32.0 \, g/mol) effuses at a rate of 1.00 \, mol/min, what is the effusion rate of hydrogen gas (H2, molar mass 2.02 \, g/mol) under the same conditions?

Step-by-step Explanation:

  1. List Knowns and Unknowns:

    • Rate (O_2) = 1.00 \, mol/min (Let's call this Rate1)

    • Molar mass (O2), M1 = 32.0 \, g/mol

    • Molar mass (H2), M2 = 2.02 \, g/mol

    • Rate (H_2) = ? (Let's call this Rate2)

  2. Choose the Right Formula: Use Graham's Law: Rate1/Rate2 = sqrt(M2/M1)

  3. Rearrange and Solve: Solve for Rate_2

    • Rate2 = Rate1 * sqrt(M1/M2)

    • Rate_2 = (1.00 \, mol/min) \times \sqrt{\frac{32.0 \, g/mol}{2.02 \, g/mol}}

    • Rate_2 = (1.00 \, mol/min) \times \sqrt{15.84}

    • Rate_2 \approx (1.00 \, mol/min) \times 3.98

    • Rate_2 \approx 3.98 \, mol/min

  4. Check Your Answer: Hydrogen is much lighter than oxygen, so it should effuse much faster, which is consistent with the answer.

Gas Laws with Density

  • Describes: How to incorporate the density of a gas into gas law calculations, often derived from the Ideal Gas Law.

  • Formula: D = (PM_mol)/(RT)

  • Where:

    • D = Density (typically in g/L)

    • P = Pressure

    • M_{mol} = Molar mass of the gas

    • R = Ideal gas constant

    • T = Absolute temperature (in Kelvin)

  • Approach: Use this formula when you are given or asked for the density of a gas, along with pressure, temperature, and/or molar mass. It's a rearrangement of the Ideal Gas Law: PV = nRT where n = m/M{mol} and D = m/V. Thus, P (m/D) = (m/M{mol}) RT \Rightarrow P/D = RT/M{mol} \Rightarrow D = PM{mol}/RT

Worked Example:

Problem: What is the density of carbon dioxide (CO_2, molar mass 44.01 \, g/mol) at 1.50 \, atm and 25 ^\circ C? Use R = 0.08206 \, L \cdot atm / (mol \cdot K).

Step-by-step Explanation:

  1. List Knowns and Unknowns:

    • Pressure (P) = 1.50 \, atm

    • Molar mass (M_{mol}) = 44.01 \, g/mol

    • Temperature (T) = 25 ^\circ C

    • Gas constant (R) = 0.08206 \, L \cdot atm / (mol \cdot K)

    • Density (D) = ?

  2. Convert Units: Convert temperature to Kelvin.

    • T = 25 ^\circ C + 273.15 = 298.15 \, K

  3. Choose the Right Formula: Use the density form of the Ideal Gas Law: D = (PM_mol)/(RT)

  4. Rearrange and Solve: The formula is already arranged to solve for D.

    • D = \frac{(1.50 \, atm)(44.01 \, g/mol)}{(0.08206 \, L \cdot atm / (mol \cdot K))(298.15 \, K)}

    • D \approx \frac{66.015 \, g \cdot atm / mol}{24.466 \, L \cdot atm / mol}

    • D \approx 2.70 \, g/L

  5. Check Your Answer: Units cancel to g/L, which is appropriate for density.

General Tips for Solving Gas Law Problems:

  1. Convert Units: Always ensure all quantities are in consistent units. For temperature, always convert to Kelvin (K = ^\circ C + 273.15).

  2. Identify Constants: Determine which variables remain constant. This helps narrow down the applicable law.

  3. List Knowns and Unknowns: Write down all given values and what you need to find. This clarifies the problem.

  4. Choose the Right Formula: Select the gas law that relates your knowns and unknowns, considering what's constant.

  5. Rearrange and Solve: Algebraically rearrange the formula to solve for the unknown, then plug in your values.

  6. Check Your Answer: Does your answer make sense? For example, if temperature increases and volume is constant, you'd expect pressure to increase (Gay-Lussac's Law).