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Gases
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Pressure
The force per unit area of a surface.
Units: N/cm2
N, Newton: SI unit of force
As area of contact changes, force changes
500 N = 1.7 N
500 N = 83.3 N
300 cm2
cm2
6.0 cm2
cm2
Christina,
Will you go to the prom with me?
Steve
Barometer
• Used to measure the pressure of gases.
barometerplanet.com
Units of Pressure for Gases
• Millimeters of mercury (mm Hg)
• Torricelli or 1 torr = 1mm Hg
• 760 mm Hg = 1 atmosphere at sea level
when temp is 0oC
• Pascal = pressure exerted by a force of one
newton acting on an area of one square meter.
Pa = N/m2
• 1.013 x 105 kPa = 1 atmosphere= 10.1N/cm2
Pressure conversion sample:
Express 0.725 atm in a) mm Hg and b)
kilopascals (kPa)
A) 0.725 atm x 760 mm Hg = 551 mm Hg
1 atm
B) 0.725 atm x 101.325 kPa = 73.5 kPa
1 atm
Standard Conditions (STP)
Standard Temperature is
o
0 C or 273K
Standard Pressure is 1 atm
or 760 mm of Hg
You must have done by next meeting:
A list of the units of pressure (see page 364) and
A list of the gas laws:
Dalton’s Law of Partial Pressures: PT = P1+P2+…
Boyle’s: P1V1 = P2V2
T constant
Charles’s: V1 = V2
P constant
T1 T2
Gay-Lussac’s: P1 = P2
V constant
T1 T2
Combined Gas: P1V1 = P2V2
T1
T2
Ideal Gas: PV = nRT
Dalton’s Law of Partial Pressures
The total pressure of a gas mixture is the sum of
the partial pressures of the component gases.
Gas collected through water picks up water
vapor, so allow inside and outside water levels
in a gas collection device to stabilize and:
Patm = Pgas + PH20
Dalton sample problem:
Oxygen gas is collected by water displacement.
The barometric pressure and the temperature
during the experiment are 731.0 torr and
25.0oC. What was the partial pressure of the
oxygen collected?
PT = Patm = 731.0 torr
PH20 = 23.8 torr (see vapor pressure of water at
25.0 oC from table in handout or book – Table
A-8)
PT = Patm = 731.0 torr
PH20 = 23.8 torr (see vapor pressure of water at
25.0 oC from table in handout or book – Table
A-8)
Patm = PO2 + PH20
So PO2 = Patm – PH20
PO2 = 731.0 torr – 23.8 torr = 707.2 torr
Boyle’s Law – at constant temperature, volume
of a fixed gas varies inversely with the pressure.
If 100.0 mL of a gas, originally at 760 torr, is
compressed to a pressure of 800 torr, at a
constant temperature, what would be its final
volume?
P1V1 = P2V2 --> V2 = P1V1
P2
V2 = 100mL(760 torr) = 95.0 mL
800 torr
Charles’ Law – the volume of a fixed mass of gas
at constant pressure varies directly with the
Kelvin temperature.
A sample of neon gas has a volume of 752 mL at
25.0oC. What will the volume at 100.0oC be if
pressure is constant?
V1 = V2 --> V2 = V1T2
T1 T2
T1
V2 = 752 mL (100.0oC) = 300.8 mL
25.0oC
Gay-Lussac’s Law – the pressure of a fixed mass
of gas at constant volume varies directly with
the Kelvin temperature.
At 122oC the pressure of a sample of nitrogen
gas is 1.07 atm. What will the pressure be at
205oC, assuming constant volume?
P1/T1 = P2/T2 --> P2 = P1T2
T1
P2 = 1.07 atm(205&273) = 1.29 atm
122+273
Gay-Lussac’s Law of combining
volumes
At constant temperature and pressure, the
volumes of gaseous reactants and products
can be expressed as ratios of small whole
numbers.
H2 + Cl2 -->
2HCl
1L
1L
2L
H:Cl:HCl = 1:1:2
Formulas must be written correctly and
chemical equation balanced.
Combined Gas Law – expresses the relationship
between pressure, volume, and temperature of
a fixed amount of a gas. PV = k
T
P1V1 = P2V2 --> To find V2: V2 = P1V1T2
T1
T2
P2T1
Problem: The volume of a gas is 27.5mL at
22.0oC and 0.974 atm. What will be the
volume at 15.OoC and 0.993 atm?
Temps to K:
22+273 = 295K and 15+273=288K
The volume of a gas is 27.5mL at 22.0oC and 0.974
atm. What will be the volume at 15.OoC and 0.993
atm?
V2 = P1V1T2
V1 = 27.5ml
P2T1
T1 = 295K
V2 = 0.974atm(27.5ml)(288K)
P1 = 0.974atm
0.993atm(295K)
V2 = ?
V2 = 26.3 mL
T2 = 288K
P2 = 0.993atm
Hint: in solving gas law problems, use the combined
gas law and quantities that don’t change will cancel
out.
Avogadro’s Law – equal volumes of gases at the
same temperature and pressure contain equal
numbers of molecules.
Ratios apply here also.
2H2
+
O2
-->
2H2O
2molecules
1molecule
2molecules
2mol
1mol
2mol
2volumes
1volume
2volumes
Standard molar volume of a gas is the volume
occupied by one mole of a gas at STP.
Standard molar volume = 22.4 L/mol
Steve,
Yes, I will go to the prom with you.
Christina
At STP, what is the volume of 7.08 mol
of nitrogen gas?
7.08 mol (22.4L) = 158 L
1 mol
A sample of gas occupies 11.9 L at STP. How
many moles of the gas are present?
11.9L (1 mol) = 0.531 mol
22.4L
Assuming all volume measurements are made at the
same temperature and pressure, what volume of
hydrogen gas is needed to react completely with 4.55 L
of oxygen gas to produce water vapor?
Write and balance the equation.
Label known and unknown.
Do unit analysis.
Gas Stoichiometry – dealing with proportional
relationships between reactants and products in
a chemical reaction.
2CO2(g)
+ O2(g)
-->
2CO2(g)
2molecules
1 molecule
2 molecules
2 mol
1 mol
2 mol
2 volumes
1 volume
2 volumes
Assuming all volume measurements are made at
the same temperature and pressure, what
volume of hydrogen gas is needed to react
completely with 4.55 L of oxygen gas to
produce water vapor?
Write the correct chemical reaction first:
2H2 + O2 --> 2H2O
Indicate known and unknown:
Solve:
Ideal Gas Law – the mathematical relationship
among pressure, volume, temperature, and the
number of moles of a gas.
PV = nRT --> R = PV
R is the ideal gas
nT
constant.
At STP, R = 1 atm(22.414 L) = 0.0821 L atm
1 mol(273.15K)
mol K
R is the ideal gas constant.
What pressure, in atmospheres, is exerted by
0.325 mol of hydrogen gas in a 4.08 L container
at 35oC?
PV = nRT --> P = nRT
V
T = 35 + 273 = 308K
P = 0.325 mol 0.0821L atm) (308K)
4.08 L
mol K
P = 2.01 atm
Gases spread out in a container – diffusion
Gases can randomly pass through a tiny opening
in a container (leak out) - effusion