gas laws what to do when conditions are ideal. boyle’s law what was the relationship between...
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Gas Laws
What to do when conditions are ideal
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Boyle’s LawWhat was the relationship between pressure and
volume?
When P Then V
When P Then V
Algebraically this is written as P=k/V
When you solved for k; PV=k
Therefore P1V1=P2V2
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Gas Laws – Boyle’s LawConstant: Temperature
Relationship: Pressure is inversely proportional to volume
Pressure 1/volume
Written As: P1V1 = P2V2
Pressure is typically in atm or torr
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Charles’ LawWhat was the relationship between Temperature and
Volume?
When T Then V
When T Then V
Algebraically this is written as V=kT
When you solved for k; V/T=k
Therefore V1/T1=V2/T2
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Gas Laws – Charles’ LawConstant: Pressure
Relationship: Temperature is directly proportional to volume
Temp Volume
Written As: V1/T1 = V2/T2
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Charles’ LawWhat unit of measure is needed for these
calculations? C or K?
Temperature is in K (K = 273 + C)
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Gas Laws – Gay-Lussac’s LawConstant: Volume
Relationship: Pressure is directly proportional to temperature
Pressure Temperature
Written As: P1/T1 = P2/T2
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The IDEAL GAS Law –this is what we will useWhen we put all three laws together:
PV nT (n= number of moles)
PV = nRT (R= ideal gas law constant)
R=62.4 L torr/K mol or .08206 L atm/K mol
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Lecture PLUS Timberlake 2000 9
Ideal Gases
Behave as described by the ideal gas equation; no real gas is actually ideal
Within a few %, ideal gas equation describes most real gases at room temperature and pressures of 1 atm or less
In real gases, particles attract each other reducing the pressure
Real gases behave more like ideal gases as pressure approaches zero.
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Lecture PLUS Timberlake 2000 10
PV = nRT
R is known as the universal gas constant
Using STP conditions P V
R = PV = (1.00 atm)(22.4 L) nT (1mol) (273K)
n T
= 0.0821 L-atm
mol-K
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LecturePLUS Timberlake 11
Combined Gas Law
P1V1 = P2V2
T1 T2
Isolate V2
P1V1T2 = P2V2T1
V2 = P1V1T2
P2T1
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Lecture PLUS Timberlake 2000 12
Learning Check G15What is the value of R when the STP value for P is 760 mmHg?
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Lecture PLUS Timberlake 2000 13
Solution G15What is the value of R when the STP value for P is 760 mmHg?
R = PV = (760 mm Hg) (22.4 L)
nT (1mol) (273K)
= 62.4 L-mm Hg mol-K
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Lecture PLUS Timberlake 2000 14
Learning Check G16
Dinitrogen monoxide (N2O), laughing gas, is used by dentists as an anesthetic. If 2.86 mol of gas occupies a 20.0 L tank at 23°C, what is the pressure (mmHg) in the tank in the dentist office?
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Lecture PLUS Timberlake 2000 15
Solution G16
Set up data for 3 of the 4 gas variables
Adjust to match the units of R
V = 20.0 L 20.0 L
T = 23°C + 273 296 K
n = 2.86 mol 2.86 mol
P = ? ?
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Lecture PLUS Timberlake 2000 16
Rearrange ideal gas law for unknown P
P = nRT
V
Substitute values of n, R, T and V and solve for P
P = (2.86 mol)(62.4L-mmHg)(296 K)
(20.0 L) (K-mol)
= 2.64 x 103 mm Hg
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Lecture PLUS Timberlake 2000 17
Learning Check G17
A 5.0 L cylinder contains oxygen gas at 20.0°C and 735 mm Hg. How many grams of oxygen are in the cylinder?
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Lecture PLUS Timberlake 2000 18
Solution G17
Solve ideal gas equation for n (moles)n = PV
RT
= (735 mmHg)(5.0 L)(mol K) (62.4 mmHg L)(293 K)
= 0. 20 mol O2 x 32.0 g O2 = 6.4 g O2
1 mol O2
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LecturePLUS Timberlake 19
Learning Check C1
Solve the combined gas laws for T2.
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LecturePLUS Timberlake 20
Solution C1
Solve the combined gas law for T2.
(Hint: cross-multiply first.)
P1V1 = P2V2
T1 T2
P1V1T2 = P2V2T1
T2 = P2V2T1
P1V1
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LecturePLUS Timberlake 21
Combined Gas Law Problem
A sample of helium gas has a volume of 0.180 L, a pressure of 0.800 atm and a temperature of 29°C. What is the new temperature(°C) of the gas at a volume of 90.0 mL and a pressure of 3.20 atm?
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LecturePLUS Timberlake 22
Data Table
Set up Data Table
P1 = 0.800 atm V1 = 0.180 L T1 = 302 K
P2 = 3.20 atm V2= 90.0 mL T2 = ????
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LecturePLUS Timberlake 23
Solution
Solve for T2
Enter data
T2 = 302 K x atm x mL = K
atm mL
T2 = K - 273 = °C
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LecturePLUS Timberlake 24
Calculation
Solve for T2
T2 = 302 K x 3.20 atm x 90.0 mL = 604 K
0.800 atm 180.0 mL
T2 = 604 K - 273 = 331 °C
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LecturePLUS Timberlake 25
Learning Check C2
A gas has a volume of 675 mL at 35°C and 0.850 atm pressure. What is the temperature in °C when the gas has a volume of 0.315 L and a pressure of 802 mm Hg?
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LecturePLUS Timberlake 26
Solution G9
T1 = 308 K T2 = ?
V1 = 675 mL V2 = 0.315 L = 315 mL
P1 = 0.850 atm P2 = 802 mm Hg = 646 mm Hg
T2 = 308 K x 802 mm Hg x 315 mL
646 mm Hg 675 mL
P inc, T inc V dec, T dec
= 178 K - 273 = - 95°C
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LecturePLUS Timberlake 27
Volume and Moles
How does adding more molecules of a gas change the volume of the air in a tire?
If a tire has a leak, how does the loss of air (gas) molecules change the volume?
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LecturePLUS Timberlake 28
Learning Check C3
True (1) or False(2)
1.___The P exerted by a gas at constant V is not affected by the T of the gas.
2.___ At constant P, the V of a gas is directly proportional to the absolute T
3.___ At constant T, doubling the P will cause the V of
the gas sample to decrease to one-half its original V.
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LecturePLUS Timberlake 29
Solution C3
True (1) or False(2)
1. (2)The P exerted by a gas at constant V is not affected by the T of the gas.
2. (1) At constant P, the V of a gas is directly proportional to the absolute T
3. (1) At constant T, doubling the P will cause the V of
the gas sample to decrease to one-half its original V.
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LecturePLUS Timberlake 30
Avogadro’s Law
When a gas is at constant T and P, the V is directly proportional to the number of moles (n) of gas
V1 = V2
n1 n2
initial final
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LecturePLUS Timberlake 31
STP
The volumes of gases can be compared when they have the same temperature and pressure (STP).
Standard temperature 0°C or 273 K
Standard pressure 1 atm (760 mm Hg)
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LecturePLUS Timberlake 32
Learning Check C4
A sample of neon gas used in a neon sign has a volume of 15 L at STP. What is the volume (L) of the neon gas at 2.0 atm and –25°C?
P1 = V1 = T1 = K
P2 = V2 = ?? T2 = K
V2 = 15 L x atm x K = 6.8 L
atm K
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LecturePLUS Timberlake 33
Solution C4
P1 = 1.0 atm V1 = 15 L T1 = 273 K
P2 = 2.0 atm V2 = ?? T2 = 248 K
V2 = 15 L x 1.0 atm x 248 K = 6.8 L
2.0 atm 273 K
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LecturePLUS Timberlake 34
Molar Volume
At STP
4.0 g He 16.0 g CH4 44.0 g CO2
1 mole 1 mole 1mole (STP) (STP) (STP)
V = 22.4 L V = 22.4 L V = 22.4 L
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LecturePLUS Timberlake 35
Molar Volume Factor
1 mole of a gas at STP = 22.4 L
22.4 L and 1 mole
1 mole 22.4 L
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LecturePLUS Timberlake 36
Learning Check C5A.What is the volume at STP of 4.00 g of CH4?
1) 5.60 L 2) 11.2 L 3) 44.8 L
B. How many grams of He are present in 8.0 L of
gas at STP?
1) 25.6 g 2) 0.357 g 3) 1.43 g
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LecturePLUS Timberlake 37
Solution C5A.What is the volume at STP of 4.00 g of CH4?
4.00 g CH4 x 1 mole CH4 x 22.4 L (STP) = 5.60 L
16.0 g CH4 1 mole CH4
B. How many grams of He are present in 8.0 L of gas at STP?
8.00 L x 1 mole He x 4.00 g He = 1.43 g He
22.4 He 1 mole He
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LecturePLUS Timberlake 38
Daltons’ Law of Partial PressuresPartial Pressure
Pressure each gas in a mixture would exert if it were the only gas in the container
Dalton's Law of Partial Pressures
The total pressure exerted by a gas mixture is the sum of the partial pressures of the gases in that mixture.
PT = P1 + P2 + P3 + .....
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LecturePLUS Timberlake 39
Gases in the AirThe % of gases in air Partial pressure (STP)
78.08% N2 593.4 mmHg
20.95% O2 159.2 mmHg
0.94% Ar 7.1 mmHg
0.03% CO2 0.2 mmHg
PAIR = PN + PO + PAr + PCO = 760 mmHg 2 2 2
Total Pressure 760 mm Hg
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LecturePLUS Timberlake 40
Learning Check C6
A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in
the air?
1) 35.6 2) 156 3) 760
B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?
1) 557 2) 9.14 3) 0.109
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LecturePLUS Timberlake 41
Solution C6
A.If the atmospheric pressure today is 745 mm Hg, what is the partial pressure (mm Hg) of O2 in
the air?
2) 156
B. At an atmospheric pressure of 714, what is the partial pressure (mm Hg) N2 in the air?
1) 557
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LecturePLUS Timberlake 42
Partial Pressures
The total pressure of a gas mixture depends
on the total number of gas particles, not on
the types of particles.
P = 1.00 atm P = 1.00 atm
0.5 mole O2
+ 0.3 mole He+ 0.2 mole Ar
1 mole H2
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LecturePLUS Timberlake 43
Health NoteWhen a scuba diver is several hundred feet
under water, the high pressures cause N2 from the
tank air to dissolve in the blood. If the diver rises too fast, the dissolved N2 will form bubbles in the
blood, a dangerous and painful condition called "the bends". Helium, which is inert, less dense, and does not dissolve in the blood, is mixed with O2 in
scuba tanks used for deep descents.
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LecturePLUS Timberlake 44
Learning Check C7
A 5.00 L scuba tank contains 1.05 mole of O2
and 0.418 mole He at 25°C. What is the partial pressure of each gas, and what is the total pressure in the tank?
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LecturePLUS Timberlake 45
Solution C7
P = nRT PT = PO + PHe
V 2
PT = 1.47 mol x 0.0821 L-atm x 298 K
5.00 L (K mol)
= 7.19 atm
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Lecture PLUS Timberlake 2000 46
Molar Mass of a gas
What is the molar mass of a gas if 0.250 g of the gas occupy 215 mL at 0.813 atm and 30.0°C?
n = PV = (0.813 atm) (0.215 L) = 0.00703 mol
RT (0.0821 L-atm/molK) (303K)
Molar mass = g = 0.250 g = 35.6 g/mol
mol 0.00703 mol
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Lecture PLUS Timberlake 2000 47
Density of a Gas
Calculate the density in g/L of O2 gas at STP. From STP, we know the P and T.
P = 1.00 atm T = 273 K
Rearrange the ideal gas equation for moles/L
PV = nRT PV = nRT P = n
RTV RTV RT V
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Lecture PLUS Timberlake 2000 48
Substitute
(1.00 atm ) mol-K = 0.0446 mol O2/L
(0.0821 L-atm) (273 K)
Change moles/L to g/L
0.0446 mol O2 x 32.0 g O2 = 1.43 g/L
1 L 1 mol O2
Therefore the density of O2 gas at STP is
1.43 grams per liter