Pressure, Volume and Temperature

Problems, problems, problems

Make a group graph that compares your data, to that which was calculated.

Temperature in Kelvin

Volume in Liters

Remember the BIG equation.

P1 • V1 P2 • V2

---------- = ------------

T1 T2

Boyles’s Law Formula P1 • V1 = P2 • V2

You have a tire with a volume of 1 liter. It has a pressure of 101 kPa (which is what ordinary atmosphere is). You reduce the volume by one half. What is the new pressure?

Boyles’s Law Formula P1 • V1 = P2 • V2

You have a tire with a volume of 1 liter. It has a pressure of 101 k Pa (which is what ordinary atmosphere is). You reduce the volume by one half. What is the new pressure?

101 kPa • 1 liter = P2 • .5 litera. Divide both sides by .5 literb. Answer is 202 kPa

You have a set of lungs that have a volume of 1.5 liters. You dive down, and have a new volume of .5 liters. What is the pressure on your body due to water?

You have a set of lungs that have a volume of 1.5 liters. You dive down, and have a new volume of .5 liters. What is the pressure on your body due to water?P1 – 101 k PaV1 – 1.5 liters 101 kPa • 1.5 liters = P2 • .5 litersP2 -- unknownV2 -- .5 liters Divide both sides by .5 liters, answer is 303 kPa

You have a balloon, and it has a volume of 2.5 liters and a pressure of 101 kPa. You release it, and it rises until the air pressure is 50 kPa. What is the new volume of the balloon? Assume the helium in the balloon is free to expand.

You have a balloon, and it has a volume of 2.5 liters and a pressure of 101 kPa. You release it, and it rises until the air pressure is 50 kPa. What is the new volume of the balloon? Assume the helium in the balloon is free to expand.

P1 – 101 kPaV1 – 2.5 litersP2 – 50 kPaV2 – unknown 101 kPa • 2.5 liters = 50 kPa • V2

Divide both sides by 50 k Pa, answer is 5.05 liters.

You have a container that has been squeezed to a volume of 3.25 liters, and a pressure of 450 kPa. What was the pressure of the container when the container had a volume of 10 liters?

You have a container that has been squeezed to a volume of 3.25 liters, and a pressure of 450 k Pa. What was the pressure of the container when the container had a volume of 10 liters?

P1 – unknown kPaV1 – 10 liters P1 • 10 liters = 450 kPa • 3.25 litersP2 – 450 kPaV2 – 3.25 liters divide both sides by 10 liters, answer is 146.25 kPa

Charles’s Law Formula V1/T1 = V2/T2Kelvin Temperature Scale -- add 273˚ to the Centigrade scale.You MUST use Kelvin scale with this formula

You have a tire with a volume of 1 liter. It has a temperature of 25˚C (which is what room temperature is). You reduce the volume (by chilling) by one half, to .5 liter. What is the new temperature? Remember to change to Kelvin, not centigrade.

Charles’s Law Formula V1/T1 = V2/T2Kelvin Temperature Scale -- add 273˚ to the Centigrade scale.You MUST use Kelvin scale with this formula

You have a tire with a volume of 1 liter. It has a temperature of 25˚C (which is what room temperature is). You reduce the volume (by chilling) by one half, to .5 liter. What is the new temperature? Remember to change to Kelvin, not centigrade.

V1 – 1 literT1 – 298 degrees KelvinV2 -- .5 litersT2 – unknown degrees Kelvin(1 liter ÷ 298 K) = (.5 liters ÷ T2)

Cross multiply and divide, answer is 149 K

You have a balloon, with a temperature of 25˚C, and a volume of .75 liters. You chill it to a volume of .5 liters. Pressure remains the same. What is its new temperature?

You have a balloon, with a temperature of 25˚C, and a volume of .75 liters. You chill it to a volume of .5 liters. Pressure remains the same. What is its new temperature?

V1 – .75 litersT1 – 298 degrees Kelvin V2 -- .5 litersT2 – unknown degrees Kelvin

(.75 liter ÷ 298 K) = (.5 liter ÷ T2)

Cross multiply and divide, answer is 199 K

You have a container that has been chilled to a volume of 3.25 liters, and a temperature of 400˚C. What was the temperature of the container when the container had a volume of 10 liters?

You have a container that has been chilled to a volume of 3.25 liters, and a temperature of 400˚C. What was the temperature of the container when the container had a volume of 10 liters?

V1 – 10 litersT1 – unknown degrees KelvinV2 – 3.25 litersT2 – 673 degrees Kelvin

(10 liter ÷ T2 K) = (3.25 liter ÷ 673 K)

Cross multiply and divide, answer is 2,071 K