# The Gas Laws Section 16.3 Pressure = force =N aream 2 1N/m 2 = 1 pascal 101.325 kPa = 760 mmHg = 1atm (normal atmospheric pressure) Because gas particles.

## Presentation on theme: "The Gas Laws Section 16.3 Pressure = force =N aream 2 1N/m 2 = 1 pascal 101.325 kPa = 760 mmHg = 1atm (normal atmospheric pressure) Because gas particles."— Presentation transcript:

The Gas Laws Section 16.3

Pressure = force =N aream 2 1N/m 2 = 1 pascal 101.325 kPa = 760 mmHg = 1atm (normal atmospheric pressure) Because gas particles are in constant motion, when they collide with objects they create pressure

Pressure - Volume Relationship Increasing the pressure on a gas decreases the volume of the gas Boyles Law

P 1 V 1 = P 2 V 2 P1 P1 = original pressure V1 V1 = original volume P2 P2 = new pressure V2 V2 = new volume

Example #1 - the gas in a balloon has a volume of 7.5L at 100. kPa. The balloon is released into the atmosphere, and the gas in it expands to 11 L. Assuming constant temperature what is the pressure on the balloon?

P 1 V 1 = P 2 V 2 (100.kPa)(7.5L) = P 2 (11L) 750 = P 2 (11) 750 = P 2 (11) (11)11 68kPa = P 2

Atmospheric pressure decreases with increase of altitude - less air = less pressure

This is why the balloon has to be partially filled when released into the atmosphere As the balloon goes up, the pressure decreases so the volume increases, i.e. the balloon expands

Temperature - Volume Relationship Charless Law -heating a gas causes the gas to expand - increasing the temperature of gas increases the volume of the gas

V 1 = V 2 T 1 T 2 V 1 = original volume T1 T1 = original temperature V2 V2 = new volume T2 T2 = new temperature Temp. must be in Kelvin!!! °C + 273 = K

Example #2 - A sample of gas occupies 24m 3 at 175.0K. What volume would the gas occupy at 400.0K? V 1 = V 2 T 1 T 2 24m 3 = V 2 175.0K 400.0K (24)(400.0) = V 2 175.0 55m 3 = V 2

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