NOTES: 14.2 – Gas Laws

Pressure-Volume Relationship: (Boyle’s Law)

● Pressure and volume are inversely proportional

● As volume increases, pressure decreases● As volume decreases, pressure increases● P1V1 = P2V2

Boyle’s Law Example #1: A sample of gas occupies 12.0 L under a pressure of 1.2 atm. What would its volume be if the pressure were increased to 3.6 atm? (assume temp is constant)

● P1V1 = P2V2

● (1.2 atm)(12.0 L) = (3.6 atm)V2

● V2 = 4.0 L

Boyles’s Law Example #2:● For every 10 meters a scuba diver

descends into the water the pressure increases by 1 atmosphere (1 atm or 101.3 kPa). If a scuba diver fills a balloon with 5.00 L of air at the surface and descends to a depth of 100. meters, what will the new volume of the balloon be?

(assume pressure at the surface is 1.0 atm!!)

Boyles’s Law Example #2:● For every 10 meters a scuba diver descends into the water

the pressure increases by 1 atmosphere (1 atm or 101.3 kPa). If a scuba diver fills a balloon with 5.00 L of air at the surface and descends to a depth of 100. meters, what will the new volume of the balloon be?

● P1V1 = P2V2

● (1.0 atm)(5.0 L) = (11.0 atm)V2

● V2 = 0.455 L

Boyle’s Law Example #3: A high-altitude balloon contains 30.0 L of helium gas at 103.0 kPa. What is the volume when the balloon rises to an altitude where the pressure is only 0.247 atm? (assume constant temperature) **careful with pressure units!

● P1V1 = P2V2

● (103.0 kPa)(30.0 L) = (25.0 kPa)V2

● V2 = 124 L

Temperature-Volume Relationship: (Charles’s Law)

● Volume and temperature are directly proportional

● As temperature increases, volume increases● As temperature decreases, volume decreases● (V1/T1) = (V2/T2)

● ALWAYS USE KELVIN TEMPERATURES!!!!!

Charles’ Law Example #1: A sample of nitrogen gas occupies 117 mL at 100°C. At what temperature would it occupy 234 mL if the pressure does not change? (express answer in K and °C)

● V1 / T1= V2 / T2

● (117 mL) / (373 K) = (234 mL) / T2

● T2 = 746 K

● T2 = 473 ºC

Charles’s Law Example #2:● An 8.00 L capacity Ziplok bag is partially

filled with 2.50 L of CO2 at 0.50oC. If the bag is set in a sunny window and heated to 70.0oC, what will be the new volume of the CO2? Will the Ziplok bag burst?

Charles’s Law Example #2:● An 8.00 L capacity Ziplok bag is partially filled

with 2.50 L of CO2 at 0.50oC. If the bag is set in a sunny window and heated to 70.0oC, what will be the new volume of the CO2? Will the Ziplok bag burst?

● V1 / T1= V2 / T2

● (2.50 L) / (273.5 K) = (V2) / (343 K)

● V2 = 3.14 L● No, the bag won’t burst.

Charles’ Law Example #3: A 5.00 L sample of air at -50.0˚C is warmed to 100.0˚C. What is the new volume if the pressure remains constant?

● V1 / T1= V2 / T2

● (5.00 L) / (223 K) = (V2) / (373 K)

● V2 = 8.36 L

Temperature-Pressure Relationship: (Gay-Lussac’s Law)

● Pressure and temperature are directly proportional

● As temperature increases, pressure increases

● As temperature decreases, pressure decreases

● (P1/T1) = (P2/T2)

● USE KELVIN TEMPERATURES!!!!!

Gay-Lussac’s Example #1: The gas in a used aerosol can is at a pressure of 103 kPa at 25.0˚C. If this can is thrown onto a fire, what is the pressure of the gas when its temperature reaches 928˚C?

● P1 / T1= P2 / T2

● (103 kPa) / (298 K) = (P2) / (1201 K)

● P2 = 415 kPa

Gay-Lussac’s Example #2: The pressure in an automobile tire is 1.85 atm at 27.0˚C. At the end of a trip on a hot sunny day, the pressure has risen to 235 kPa. What is the temperature (in ˚C) of the air in the tire?

● P1 / T1= P2 / T2

● (1.85 atm) / (300. K) = (2.32 atm) / (T2)

● T2 = 376 K

● T2 = 103˚C