Homework 5 Gay Lussacs Law 1 Data units

Homework # 5 Gay Lussac’s Law 1. Data + units 2. Set up the problem + units 3. Answer + units

Aim 5: How can you describe Gay- Lussac’s Law?

Gay-Lussac’s Law • Discovered in 1802 by Joseph Gay-Lussac

The tank contains a gas ……. .

How much propane is in the tank? Propane tanks are widely used with barbeque grills. But it’s not fun to find out half-way through your grilling that you’ve run out of gas. You can buy gauges that measure the pressure inside the tank to see how much is left. The gauge measures pressure and will register a higher pressure on a hot day than it will on a cold day. So you need to take the air temperature into account when you decide whether or not to refill the tank before your next cook-out.

At a Constant Volume

Volume and number of particles are constant

What happens when you increase the temperature of the container? u Increased temperature leads to increase average velocity of the particles u. Increased average force per collision u. Increased rate of collision u. Increased pressure Increase Pressure

At a Constant Volume

At a Constant Volume

Think about a SCUBA tank where volume is constant. What happens to the molecular kinetics inside of it when you heat the tank? Since the particles are moving faster the number of collisions with the side of the tank increases, increasing the pressure.

1. Gay-Lussac Law • For a gas confined at a constant volume, the pressure and temperature are directly related.

2. Directly Proportional: P R E S S U R E Temperature

“At constant Volume” P 1 V 1 = P 2 V 2 T 1 T 2 P 1 = P 2 T 1 T 2

P 1 = P 2 T 1 T 2 P R E S S U R E P TEMPERATURE

1. Determine the pressure change when a constant volume of gas at 1. 00 atm is heated from 20. 0 °C to 30. 0 °C. P 1 = P 2 T 1 T 2 Data P 1= 1. 00 atm T 1 = 20. 0 °C = 293. 0 K P 2 = ? T 2 = 30. 0 °C = 303. 0 K

P 1 = P 2 T 1 T 2 P 2 = P 1 T 2 T 1 P 2 = 1. 00 atm x 303. 0 K 293. 0 K P 2 = 1. 03 atm

2. A gas has a pressure of 0. 370 atm at 50. 0 °C. What is the pressure at standard temperature? Data P 1= 0. 370 atm T 1 = 50. 0 °C = 323. 0 K P 2 = ? T 2 = 273 K (at standard temperature : table A)

P 2 = P 1 T 2 T 1 P 2 = 0. 370 atm x 273 K 323 K P 2 =0. 313 atm

3. If a gas in a closed container is pressurized from 15. 0 atmospheres to 16. 0 atmospheres and its original temperature was 25. 0 °C, what would the final temperature of the gas be? Data P 1= 15. 0 atm T 1 = 25 °C = 298 K P 2 = 16. 0 atm T 2 =

T 2 = P 2 T 1 P 1 T 2 = 16. 0 atm x 298 K 15 atm T 2 = 317. 87 K

• 760 mm. Hg = 101. 3 KPa

4. A gas has a pressure of 699. 0 mm Hg at 40. 0 °C. What is the temperature at standard pressure? Data P 1= 699. 0 mm Hg T 1 = 40. 0 °C = 313 K P 2 = 760. 0 mm Hg (at standard pressure) T 2 = ? *1 atm= 101. 3 Kpa= 760 mm Hg

P 1 = P 2 T 1 T 2 = P 2 T 1 P 1 T 2 = 760 mm Hg x 313 K 2 sf 3 sf 699. 0 mm Hg 4 sf T 2 = 340 K (2 sf)

5. If a gas is cooled from 323. 0 K to 273. 15 K and the volume is kept constant what final pressure would result if the original pressure was 750. 0 mm Hg? Data P 1= 750. 0 mm Hg T 1 = 320 K P 2 = ? T 2 = 273. 15 K

P 2 = P 1 T 2 T 1 P 2 = 750. 0 mm Hg x 273. 15 K 320 K P 2 = 640 mm Hg