gases - 4 min at a constant temperature, the pressure exerted by a gas varies inversely with its...
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Gases - 4 min
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At a constant temperature, thepressure exerted by agas varies inverselywith its volume.
At a constant temperature, thepressure exerted by agas varies inverselywith its volume.
Used when only pressure changes.
Whatever pressure does,volume does the opposite.
Used when only pressure changes.
Whatever pressure does,volume does the opposite.
V2 = V1 P1
V2 = V1 P1 P2P2
Practice Problem
Practice Problem
A gas occupies 15.5 dm3 at a pressure of 30 mm Hg. What is its volume when the pressure is increased to 50 mm Hg. Assume temp does not change.
A gas occupies 15.5 dm3 at a pressure of 30 mm Hg. What is its volume when the pressure is increased to 50 mm Hg. Assume temp does not change.
V2 = V1 P1
V2 = V1 P1
P2P2
Write the proper equation.
V2 = 15.5 dm3 30 mm Hg
V2 = 15.5 dm3 30 mm Hg
50 mm Hg50 mm Hg
Substitute the given numbers.
V2 = 15.5 dm3 30 mm Hg
V2 = 15.5 dm3 30 mm Hg
50 mm Hg50 mm Hg
Since pressure increases,what will happen to volume?
V2 = 15.5 dm3 30 mm Hg
V2 = 15.5 dm3 30 mm Hg
50 mm Hg50 mm Hg
DECREASE V2 will be less than V1.
V2 = 15.5 dm3 30 mm Hg
V2 = 15.5 dm3 30 mm Hg
50 mm Hg50 mm Hg
Do the math.
V2 = 9.3 dm3
V2 = V1 P1 P2
All Boyle's Law problems should look like this:
= 15.5 dm3 30 mm Hg
50 mm Hg
= 9.3 dm3
stop
At a constant pressure,the volume of a gas varies directly with the
Kelvin temperature.
At a constant pressure,the volume of a gas varies directly with the
Kelvin temperature.
Used only when temperature changes.
Whatever temperature does,volume does the same.
Used only when temperature changes.
Whatever temperature does,volume does the same.
K e l v i n t e m p e r a t u r e
K = o C + 2 7 3
K e l v i n t e m p e r a t u r e
K = o C + 2 7 3
V2 = V1 T2V2 = V1 T2
T1T1
Practice Problem
Practice Problem
500 ml of air at 20 oC isheated to 60 oC with nopressure change. Whatis the new gas volume?
V2 = V1 T2
V2 = V1 T2
T1T1
Write the proper equation.
V2 = 500 ml 60 oC
V2 = 500 ml 60 oC 20 oC20 oC
Substitute the given numbers.
V2 = 500 ml 333 K
V2 = 500 ml 333 K 293 K293 K
Temps must be in Kelvins.
Add 273 to Celsius temps.
V2 = 500 ml 333 K
V2 = 500 ml 333 K 293 K293 K
Since temperature increases,what will happen to volume?
V2 = 500 ml 333 K
V2 = 500 ml 333 K 293 K293 K
INCREASE V2 will be greater than V1.
V2 = 500 ml 333 K
V2 = 500 ml 333 K 293 K293 K
Do the math.
V2 = 600 ml
V2 = V1 T2 T1
All Charles' Law problems should look like this:
= 500 ml 333 K
293 K
= 600 mlstop
In the real world,pressure andtemperature BOTHwill change.
In the real world,pressure andtemperature BOTHwill change.
What do we do then??
What do we do then??
Used when both temperature and pressure change
Used when both temperature and pressure change
V2 = V1 P1 T
2
V2 = V1 P1 T
2 P2 T1P2 T1
Practice Problem
Practice Problem
A gas has a volume of 810 mlat 44 oC and 325 mm Hg. Whatwould be the gas volume at227 oC and 650 mm Hg?
V2 = V1 P1 T 2
V2 = V1 P1 T 2
P2 T1P2 T1
Write the proper equation.
V2 = 810 ml 325 mm Hg 500 K V2 = 810 ml 325 mm Hg 500 K
650 mm Hg 317 K 650 mm Hg 317 K
Substitute the given numbers.
V2 = 810 ml 325 mm Hg 500 K V2 = 810 ml 325 mm Hg 500 K
650 mm Hg 317 K 650 mm Hg 317 K
Do the math.
= 640 ml
Calculate the volume of a gas at STP if 5.05 dm3 of the gas are collected at 27.5 oC and 95.0 kPa.
V2 = V1 P1 T2
P2 T1
5.05 dm3 95.0 kPa 273 K
101 kPa 300.5 K
=
= 4.32 dm3
The total pressure
in a container is the sum of the partial pressures of all the gases in the container
The total pressure
in a container is the sum of the partial pressures of all the gases in the container
This law is used most
often when doing
calculations with gases collected over water.
This law is used most
often when doing
calculations with gases collected over water.
Practice Problem #1
Practice Problem #1
A container holds three gases:
O2, CO2, and N2. The partial
pressures are 2 atm, 3 atm,
and 4 atm respectively. What
is the total pressure inside the
container?
A container holds three gases:
O2, CO2, and N2. The partial
pressures are 2 atm, 3 atm,
and 4 atm respectively. What
is the total pressure inside the
container?
O2 - 2 atm
CO2 - 3 atm
N2 - 4 atm
O2 - 2 atm
CO2 - 3 atm
N2 - 4 atm
Total Pressure = 9 atm
Collecting a GasOver Water
As gas bubblesthroughthe water ...
As gas bubblesthroughthe water ...
water vapor mixeswith thegas beingcollected.
water vapor mixeswith thegas beingcollected.
Gas pressure incontaineris equalto airpressure ...
Gas pressure incontaineris equalto airpressure ...
when water levelsare theSAME.
when water levelsare theSAME.
Practice Problem #2
Practice Problem #2
If 60 liters of nitrogen gas is
collected over water when the
temperature is 40 oC and
atmospheric pressure is 760
mm Hg, what is the partial
pressure of the nitrogen?
If 60 liters of nitrogen gas is
collected over water when the
temperature is 40 oC and
atmospheric pressure is 760
mm Hg, what is the partial
pressure of the nitrogen?
From research, the vapor
pressure of H2O at 40 oC
is 7.4 kPa.
From research, the vapor
pressure of H2O at 40 oC
is 7.4 kPa. 101.0 kPa Total - 7.4 kPa H2O
93.6 kPa N2
At equal temperature andpressure, equal volumes of gases contain the same
number of molecules.
At equal temperature andpressure, equal volumes of gases contain the same
number of molecules.
At STP, 22.4 dm3 of
any gas contains one mole of molecules,6.02 X 1023
At STP, 22.4 dm3 of
any gas contains one mole of molecules,6.02 X 1023
Theoretical gas
molecules that
have mass,
but no volume
Theoretical gas
molecules that
have mass,
but no volume
P V = n R TP V = n R T
P V = n R T
P - standard pressure
P V = n R T
P - standard pressure
P V = n R T
V - molar volume
P V = n R T
V - molar volume
P V = n R T
n - number of moles
P V = n R T
n - number of moles
P V = n R T
R - 8.31 dm3 . kPa
P V = n R T
R - 8.31 dm3 . kPa
mole . K
P V = n R T
T - standard temp (K)
P V = n R T
T - standard temp (K)
How many moles of gas are found in a 500 dm3 container if the conditions inside the container are 25 oC and 200 kPa?
PV = nRT n = PVRT
200 kPa 500 dm3 mole K
298 K 8.31 dm3 kPa
= 40.4 moles
Substituting for n:
moles = mass (m)
Substituting for n:
moles = mass (m)
molecular mass (M)
P V = m R TP V = m R T
M
The relative rates at which
two gases diffuse under
identical conditions vary
inversely as the square roots
of their molecular masses.
The relative rates at which
two gases diffuse under
identical conditions vary
inversely as the square roots
of their molecular masses.
Fuel in a Butane Lighter
