The Gas Laws

Boyle’s Law: Pressure and Volume

• If the temperature is constant, as the pressure of a gas increases, the volume decreases.

• In turn, as the pressure decreases, the volume increases.

Boyle’s Law: Pressure and Volume

• Boyle’s law states that for a given mass of gas at constant temperature, the volume of the gas varies inversely with pressure.

• P1 x V1 = P2 x V2

• P = pressure• V = volume

Using Boyle’s Law

• A balloon contains 30L of helium gas at 103 kPa. What is the volume of the helium when the balloon rises to an altitude where the pressure is only 25 kPa?

• Step 1: use Boyle’s law: P1 x V1 = P2 x V2

Using Boyle’s Law

• Step 2: set up equation to solve for V2

o V2 = (V1 x P1) / P2

• Step 3: plug and chugo V2 = (30L x 103 kPa) / 25 kPao V2 = 124 L

Using Boyle’s Law Practice

• 1. Nitrous oxide (N2O) is used as an anesthetic. The pressure on 2.5 L of N2O changes from 105 kPa to 40.5 kPa. If the temperature does not change, what will the new volume be?

Using Boyle’s Law Practice

• 2. A gas with a volume of 4 L at a pressure of 205 kPa is allowed to expand to a volume of 12 L. What is the pressure in the container if the temperature remains constant?

Charles’s Law: Temperature and Volume

• As the temperature of an enclosed gas increases, the volume increases, if the pressure is constant.

Charles’s Law: Temperature and Volume

• Charles’s law states that the volume of a fixed mass of gas is directly proportional to its Kelvin temperature if the pressure is kept constant.

• V1/T1 = V2/T2

Using Charles’s Law

• A balloon inflated in a room at 24 °C has a volume of 4 L. The balloon is then heated to a temperature of 58 °C. What is the new volume if the pressure remains constant?

• Step 1: Turn Celsius into Kelvin K = C + 273o 24 + 273 = 297 K for T1o 58 + 273 = 331 K for T2

Using Charles’s Law

• Step 2: rearrange Charles’s law to isolate V2o V2 = (V1 x T2) / T1

o Step 3: plug and chugo V2 = (4L x 331K) / 297K = 4.46 L

Using Charles’s Law Practice

• 3. If a sample of gas occupies 6.8 L at 325°C, what will its volume be at 25°C if the pressure does not change?

Using Charles’s Law Practice

• 4. Exactly 5L of air at -50 °C is warmed to 100 °C. What is the new volume if the pressure remains constant?

Gay-Lussac’s Law: Pressure and Temperature

• As the temperature of an enclosed gas increases, the pressure increases, if the volume is constant.

Gay-Lussac’s Law: Pressure and Temperature

• Gay-Lussac’s law states that the pressure of a gas is directly proportional to the Kelvin temperature if the volume remains constant.

• P1/T1 = P2/T2

Using Gay-Lussac’s Law

• The gas in a used aerosol can is at a pressure of 103kPac at 25°C. If the can is thrown onto a fire, what will the pressure be when the temperature reaches 928°C?

• Step 1: turn celsius into kelvino 25 + 273 = 298 Ko 928 + 273 = 1201 K

Using Gay-Lussac’s Law

• Step 2: Use Gay-Lussac’s law and isolate P2o P2 = (P1 x T2) / T1

o Step 3: Plug and chugo P2 = (103 kPa x 1201 K) / 298 K = 415 kPa

Using Gay-Lussac’s Law Practice

• 5. The pressure in a car tire is 198 kPa at 27°C. After a long drive, the pressure is 225 kPa. What is the temperature of the air in the tire?

Using Gay-Lussac’s Law Practice

• 6. a sample of nitrogen gas has a pressure of 6.58 kPa at 539 K. If the volume does not change, what will the pressure be at 211 K?

The Combined Gas Law

• The combined gas law describes the relationship among the pressure, temperature, and volume of an enclosed gas.

The Combined Gas Law

• The combined gas law allows you to do calculations for situations in which only the amount of gas is constant.

• P1 x (V1/T1) = P2 x (V2/T2)

Using The Combined Gas Law

• The volume of a gas filled balloon is 30L at 313 K and 153 kPa pressure. What would the volume be at standard temperature and pressure (STP)?

• Step 1: Realize that T2 = 273K and P2 = 101.3 kPa because they are at STP.

Using The Combined Gas Law

• Step 2: Rearrange the combined gas law to isolate V2:o V2 = (V1 x P1 x T2) / (P2 x T1)

• Step 3: Plug and Chugo V2 = (30L x 153kPa x 273K) / (101.3kPa x 313K)o V2 = 39.5 L

Combined Gas Law Practice

• 7. A gas at 155 kPa and 25°C has an initial volume of 1L. The pressure of the gas increases to 605 kPa as the temperature is raised to 125°C. What is the new volume?

Combined Gas Law Practice

• 8. A 5L air sample has a pressure of 107kPa at a temperature of 250°C. If the temperature is raised to 102°C and the volume expands to 7L, what will the new pressure be?