Chapter 13

Gases

Make a Flip Book

Boyle’s Gas Law

How are pressure and volume related?

As the pressure increases, the volume decreases proportionally.

Boyle’s Gas Law

Identify and label 2 situations

A diver blows a 0.75 liter air bubble 10m under water. As it rises to the surface, the pressure goes from 2.25 atm to 1.03 atm. What will be the volume of the air in the bubble at the surface?

Identify and label 2 situations

A diver blows a 0.75 liter air bubble 10m under water. As it rises to the surface, the pressure goes from 2.25 atm to 1.03 atm. What will be the volume of the air in the bubble at the surface?

10m underwater Surface

V1 = 0.75 L V2 = ?

P1 = 2.25 atm P2 = 1.03 atm

Solve using Boyle’s Law

P1V1 = P2 V2

2.25 0.75 = 1.03 V2

Use magic Algebra both sides by 1.03

V2 = 2.25 0.75 1.03

V2 = 1.6 L

Check Answer

P1V1 = P2 V2

2.25 0.75 = 1.03 V2

V2 = 1.6 L

The pressure decreased by about half, 2.25 to 1.03.

So, the volume should double. Is 1.6 about twice 0.75?

STP

•Standard Pressure

•1 atm

•760 torr

•760 mmHg

Standard Temperature

• 273º K

• 0º C

Charles’s Law

How are temperature and volume related?

The volume of a given amount of gas is directly proportional to its kelvin temperature at constant pressure.

Temperature in kelvin units

Convert celsius degrees to kelvins

Tk = 273 + Tc

Charles’s Law

A helium balloon in a closed car occupies a volume of 2.23 L at 40.0C. If the car is parked on a hot day and the temperature inside rises to 75.0C, what is the new volume of the balloon, assuming the pressure remains constant.

Set up two situations

Car with balloon

T1 = 40C+ 273 = 313

V1 = 2.32 L

Car parked in the heat

T2 = 75.0C + 273 = 348

V2 = unknown

2.32 2

313 348348 2.32 313 2

2 2.58

V

V

V L

Assume the pressure and amount of gas remain constant

What volume will the gas in the balloon occupy at 250 K?

A gas at 89C occupies a volume of 0.67L. At what celsius temperature will the volume increase to 1.12L?

The celsius temperature of a 3.00L sample of gas is lowered from 80.0C to 30.0C. What will be the resulting volume of the gas?

Assume the pressure and amount of gas remain constant

What volume will the gas in the balloon occupy at 250 K?1.8LA gas at 89C occupies a volume of 0.67L. At what celsius temperature will the volume increase to 1.12L?330CThe celsius temperature of a 3.00L sample of gas is lowered from 80.0C to 30.0C. What will be the resulting volume of the gas?2.58L

Gay-Lussac’s Law

How are temperature and pressure of gas related?

Remember the temperature must be in kelvins.

Example

The pressure of the oxygen gas inside a canister is 5.00 atm at 25.0 C. The canister is located at a camp high on Mount Everest. If the temperature there falls to -10.0 C , what is the new pressure inside the canister?

Set up 2 situations

Warm

T1 = 25.0 C +273

P1 = 5.00 atm

Cold

T2 = -10.0 C + 273

P2 = unknown

5.00 2

298 2635

263 2298

2 4.41

P

P

P atm

Analyze the answer

The temperature decreased, so the pressure should decrease. The unit is atm matches the pressure unit given.

Examples

The pressure in an automobile tire is 1.88atm at 25.0 C. What will be the pressure if the temperature increases to 37.0 C.

Helium gas in a 2.00L cylinder is under 1.12 atm pressure. At 36.5 C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas in the cylinder?

Examples

The pressure in an automobile tire is 1.88atm at 25.0 C. What will be the pressure if the temperature increases to 37.0 C.

1.96 atm

Helium gas in a 2.00L cylinder is under 1.12 atm pressure. At 36.5 C, that same gas sample has a pressure of 2.56 atm. What was the initial temperature of the gas in the cylinder?

-138.0 C

The Combined Gas Law

Temperature measured in kelvins.

Example

A gas at 110 kPa and 30C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80.0C and the pressure increases to 440kPa, what is the new volume?

Set up 2 situations

Initial

P1 = 110 kPa

T1 = 30.0 C +273

V1 = 2.00L

Raised Temp

P2 = 440 kPa

T2 = 80.0 C + 273

V2 = unknownLV

V

V

58.02

2303

2110

440

353353

2440

303

00.2110