The Gas Laws Section 16.3 Pressure = force =N aream 2 1N/m 2 = 1 pascal 101.325 kPa = 760 mmHg =...

The Gas LawsSection 16.3

Pressure = force = N area m2

1N/m2 = 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

Boyle’s Law

P1V1 = P2V2

P1 = original pressureV1 = original volumeP2 = new pressureV2 = 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?

P1V1 = P2V2

(100.kPa)(7.5L) = P2(11L)750 = P2(11)750 = P2(11)(11) 11

68kPa = P2

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

Charles’s Law-heating a gas causes the gas to expand- increasing the temperature of gas increases the volume of the gas

V1 = V2

T1 T2

V1 = original volumeT1 = original temperatureV2 = new volumeT2 = new temperature

Temp. must be in Kelvin!!! °C + 273 = K

Example #2 - A sample of gas occupies 24m3

at 175.0K. What volume would the gas occupy at 400.0K?

V1 = V2

T1 T2

24m3 = V2

175.0K 400.0K(24)(400.0) = V2

175.055m3 = V2