Boyle's Law

Boyle's Law

• Pressure and volume are inversely proportional.
• As pressure increases, volume decreases, and vice versa.
• The equation for Boyle's Law is P x V = constant.
• This law assumes a constant number of moles and temperature.
• If pressure doubles, the volume is halved to maintain the constant.
• An alternative way to express Boyle's Law is: PᵢVᵢ = PfVf, comparing initial and final states of a gas.

Boyle's Law Example

• An initial volume of 10 liters at a pressure of 1 atm results in a product of 10 atm-liters.
• Doubling the pressure to 2 atm decreases the volume to 5 liters, maintaining the 10 atm-liters product.
• Further doubling the pressure to 4 atm reduces the volume to 2.5 liters, keeping the product constant.

Practice Problem 1

• A 5-liter gas sample at 25°C and 3.0 atm is compressed at constant temperature to 1 liter.
• Since temperature and the amount of gas are constant, Boyle's Law applies.
• The initial volume (Vi) is 5 liters, and the initial pressure (Pi) is 3.0 atm.
• The final volume (Vf) is 1 liter, and the goal is to find the final pressure (Pf).
• Using the equation P1V1 = P2V2: (3 atm) (5 liters) = Pf(1 liter).
• Solving for Pf: Pf = 15 atm.
• Since the volume decreased by a factor of 5, the pressure increased by the same factor.

Practice Problem 2

• A 3.5-liter gas sample expands at constant temperature until the pressure is 0.10 atm.
• Initial volume (Vi) is 3.5 liters, and initial pressure (Pi) is 1 atm.
• Final pressure (Pf) is 0.10 atm, and the goal is to find the final volume (Vf).
• Using Boyle's Law: (1 atm)(3.5 liters) = (0.10 atm)Vf.
• Solving for Vf: Vf = 35 liters.
• As the pressure decreased by a factor of 10, the volume increased by the same factor.

Charles' Law

• Charles' Law relates temperature and volume, stating they are directly proportional.
• If temperature and volume are changing, number of moles and pressure must be constant.
• The equation is V/T = constant.
• Temperature and volume grow or decrease together to maintain a constant ratio.

Charles' Law Example

• Starting with a gas at 10 liters and 273 Kelvin.
• Doubling the temperature to 546 Kelvin doubles the volume to 20 liters, when pressure is kept constant.
• This principle is used in hot air balloons, where heated air expands, decreasing density and causing the balloon to rise.

Practice Problem

• A 2.5-liter gas sample at 25°C is heated to 50°C at constant pressure.
• Converting Celsius to Kelvin: 25°C = 298 K and 50°C = 323 K.
• The volume will not double because the temperature in Kelvin did not double.
• Initial volume (Vi) is 2.5 liters, initial temperature (Ti) is 298 K, and final temperature (Tf) is 323 K.
• Using Charles' Law: (2.5 liters) / (298 K) = Vf / (323 K).
• Solving for Vf: Vf = 2.7 liters.
• To double the volume, the temperature must double in Kelvin: 298 K * 2 = 596 K.
• 596 K is equal to 323°C.

Combined Gas Law

• The Combined Gas Law merges Boyle's Law and Charles' Law.
• The amount of gas is kept constant with pressure, temperature, and volume varying.
• The equation is PᵢVᵢ/Tᵢ = PfVf/Tf.
• An example considers a sample of nitrogen with an initial volume of 0.100 liters at 300K and 1.00 atm increasing to 350K.

Combined Gas Law Applications

• If initial and final pressures are the same, they cancel out in the combined gas law equation, simplifying the problem to Charles's Law.
• If pressure changes, the combined gas law must be used.

Example Calculation using Combined Gas Law

• The problem involves calculating the final temperature of a gas sample after compression, given initial and final volumes and pressures.
• Initial volume: 0.5 liters; Initial pressure: 1.0 atm; Initial temperature: 25°C (298 K).
• Final volume: 0.05 liters; Final pressure: 5.0 atm; Unknown: Final temperature.
• The combined gas law formula is P₁V₁/T₁ = P₂V₂/T₂.
• Substitute to solve for T₂: T₂ = (P₂V₂T₁) / (P₁V₁)
• The calculated final temperature is 149 K.
• The result is logical because as volume decreases, temperature should also decrease according to Charles's Law.

Avogadro's Law

• Avogadro's Law states that equal volumes of any gas contain the same number of particles (or moles) at the same temperature and pressure.
• Formally defined as V/n = constant (where V is volume and n is the number of moles).
• Initial and final states can be related: V₁/n₁ = V₂/n₂.

Example Calculation Using Avogadro's Law

• Problem: Determine the final volume occupied by 16.5 moles of carbon monoxide, given that 5.5 moles occupy 20.6 liters at the same temperature and pressure.
• Initial moles: 5.5 moles; Initial volume: 20.6 liters.
• Final moles: 16.5 moles; Unknown: Final volume.
• The new volume should be approximately three times the initial volume, given the mole ratio.
• Solving for the final volume yields 61.8 liters.

Molar Volume of Gas

• Molar volume refers to the volume occupied by one mole of a gas.

Standard Temperature and Pressure (STP)

• Standard Temperature and Pressure (STP) conditions are defined as 1 atm pressure and 273 K (0°C) temperature.
• Memorizing STP conditions is essential.
• At STP, one mole of any ideal gas occupies 22.4 liters.
• Applies regardless of the gas's identity, as long as it is an ideal gas.

Density Calculation at STP

• Example: Calculate the density of 4 grams of helium at STP.
• 4 grams of helium is one mole (molar mass of helium is approximately 4 g/mol).
• Density is mass divided by volume.
• The volume of one mole of helium at STP is 22.4 liters.
• Density = 4 grams / 22.4 liters = 0.178 grams per liter.
• Gases' densities are typically measured in grams per liter.

Ideal Gas Law

• Combines Boyle's, Charles's, and Avogadro's Laws into a single equation.
• The Ideal Gas Law is expressed as PV = nRT (Pressure × Volume = moles × Ideal Gas Constant × Temperature).
• The ideal gas constant is denoted as 'R'.
• Memorizing the Ideal Gas Law is crucial.

Ideal Gas Constant (R)

• The value of R depends on the units used for pressure, volume, and temperature.
• The most common units for 'R' in gas law problems are atm·L / (mol·K).
• R = 0.0821 atm·L / (mol·K) when pressure is in atm, volume in liters, amount in moles, and temperature in Kelvin
• The value of R can be derived from STP conditions: at 1 atm and 273 K, one mole of gas occupies 22.4 liters.

Example Problem Using Ideal Gas Law

• Determine the volume occupied by 5 grams of methane at 25°C and 1 atm using PV = nRT.
• The problem does not suggest that the gas is undergoing any change.
• Convert 5 grams of methane to moles using its molar mass (16.05 g/mol): 5 grams / 16.05 g/mol = 0.31 moles.
• Ensure all parameters are in the correct units: Pressure in atm, Volume in Liters, n in moles, T in Kelvin
• The temperature is 25°C, which converts to 298 K.
• Volume V = nRT / P = (0.31 moles × 0.0821 atm·L/mol·K × 298 K) / 1 atm = 7.6 liters.

Units for Gas Law Equations

• For PV = nRT, it is essential to use the specified units: Pressure (atm), Volume (liters), amount (moles), and Temperature (Kelvin).
• Boyle's Law, Charles's Law, or the combined gas law only requires the temperature to be in Kelvin.
• With the ideal gas law, you must use the same units because of the units present in R.
• You can convert millimeters of mercury to atm by dividing by 760

Gas Laws and Calculations

• When converting units in gas law problems, use Kelvin for temperature because the Celsius to Kelvin conversion involves addition, not multiplication or division.
• For pressure, volume, and moles, any unit is acceptable if it's the same on both sides of the equation.
• When solving for an unknown variable using the ideal gas law (PV=nRT), ensure all units match the gas constant R (0.0821 atm⋅L/mol⋅K).

Problem-Solving Example: Mass of Nitrogen Gas

• Problem: Determine the mass of nitrogen gas required to occupy 3 liters at 100°C and 700 mm Hg.
• Solution involves two steps: first calculate the moles, then find the mass using the molar mass.
• Convert temperature from Celsius to Kelvin by adding 273 (100°C + 273 = 373 K).
• Convert pressure from millimeters of mercury (mm Hg) to atmospheres (atm) using the conversion factor 760 mm Hg = 1 atm.
• The pressure of 700 mm Hg is equal to 0.921 atm.
• Rearrange the ideal gas law to solve for moles (n = PV/RT).
• Substitute the values P = 0.921 atm, V = 3.0 L, R = 0.0821 atm⋅L/mol⋅K, and T = 373 K into the equation.
• Calculate the moles of nitrogen gas to be 0.0902 moles.
• Multiply the moles of nitrogen gas by the molar mass of nitrogen (28.02 g/mol) to find the mass which is 2.5 grams.
• These kinds of problems involve determining an unknown variable given the other three in the ideal gas law equation.
• The units used in the ideal gas law need to match the units in the gas constant R.

Dalton's Law of Partial Pressures

• Dalton's Law states that the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of each individual gas.
• The partial pressure of each gas is the pressure that the gas would exert if it occupied the container alone.
• The total pressure is simply the sum of the individual pressures.
• Example: If air contains 0.2 atm of oxygen and 0.8 atm of nitrogen, the total pressure is 1 atm.
• This law is based on the kinetic theory of gases, which assumes that gases do not interact with each other.

Ideal vs. Real Gases

• Ideal gases are a theoretical concept, while real gases deviate from ideal behavior under certain conditions.
• Most gases behave close to ideally at room temperature and pressure.
• Gases deviate from ideal behavior at low temperatures and high pressures.
• Nonpolar gases tend to behave more ideally than polar gases.
• Nonpolar substances interact very little with each other.
• Polar gases experience stronger attractive forces (dipole-dipole interactions, hydrogen bonds), causing deviation from ideal behavior.
• For example, nitrogen (N2), a nonpolar diatomic gas, behaves more ideally than water vapor (H2O), which is polar and forms hydrogen bonds.