the gas laws section 16.3 pressure = force =n aream 2 1n/m 2 = 1 pascal 101.325 kpa = 760 mmhg =...
TRANSCRIPT
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