Chapter 5 Gases

Barometer The tube is sealed at the top and open at the bottom in the mercury pool. Measures atmospheric pressure.

As the air pressure rises, it pushes down more on the pool, forcing mercury up higher in the column. As air pressure falls, it allows more mercury to drain out of the column.

1 mm Hg = 1 torr so all also = 760 torr Recall… Standard pressure in the three units are 760 mm Hg = 1 atm = 101.325 kPa 1 mm Hg = 1 torr so all also = 760 torr NEW: = 14.7 psi

Manometer Measures pressure of a gas in a closed container

Boyle’s Law Robert Boyle described the relationship between pressure (P) and volume (V) of a gas. As V increases, P decreases. As V decreases, P increases. This is an inversely proportional relationship. When one variable goes up, the other goes down.

Boyle’s Law P1V1 = P2V2 Boyle Visual

Boyle’s Law - Example The gas in a balloon has a volume of 7.5L at 100kPa. The balloon is released into the atmosphere, and the gas expands to a volume of 11L. Assuming temperature is constant, what is the new pressure? Answer: 68 kPa

Charles’ Law Jacques Charles observed the relationship between volume (V) and temperature (T). As T up, V also up. As T down, V down. This is a directly proportional relationship. When one variable goes up, the other goes up too.

V1/T1 = V2/T2 Charles Visual

Charles’s Law - Example A gas sample occupies 2.5L at 300.0K. What volume will the gas occupy at 80.0K? Answer: 0.67L

Guy-Lussac’s Law Joseph Guy-Lussac studied the relationship between pressure (P) and temperature (T). As T up, P also up. As T down, so did P. This is a directly proportional relationship. As one variable goes one way, so does the other. P1/T1 = P2/T2

Guy-Lussac’s Law - Example A gas has a pressure of 6.58kPa at 539K. If the temperature is decreased to 211K at a constant volume, what will the resulting pressure be? Answer: 2.58 kPa

Combined Gas Law Often volume, temperature, and pressure all change at the same time. Deals with a fixed amount of gas. T represents temperature in Kelvin Units must match on both sides P1V1 = P2V2 T1 T2

Example A gas at 155 kPa and 25ºC occupies a container with initial volume of 1.00 L. By changing the volume, the pressure of the gas increases to 605 kPa and the temperature is raised to 125ºC. What is the new volume? Answer: 0.342 L

Ideal Gas Law PV=nRT NEED KELVIN R = 0.08206 L(atm)/K(mol) NEED ATM TO USE THIS R

Example You fill a rigid steel container (volume = 20.0 L) with nitrogen gas to a final pressure of 2.00 * 104 kPa at 28ºC. How many moles of nitrogen gas does the cylinder contain? Answer: 1.60 * 102 mol N2(g)

Another Example What pressure will be exerted by 0.450 mol of a gas at 25ºC if it is contained in a 0.650 L vessel? Answer: 16.9atm

Molar Volume One mole of an ideal gas takes up 22.42 liters of space (the last branch of the mole roadmap)