Transcript Gas Laws - Schoolwires

Gas Laws
What to do when conditions are ideal
Boyle’s Law
What 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
Gas Laws – Boyle’s Law
Constant: Temperature
Relationship: Pressure is inversely
proportional to volume
Pressure a 1/volume
Written As: P1V1 = P2V2
Pressure is typically in atm or torr
Charles’ Law
What 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
Gas Laws – Charles’ Law
Constant: Pressure
Relationship: Temperature is directly
proportional to volume
Temp a Volume
Written As: V1/T1 = V2/T2
Charles’ Law
What unit of measure is needed for these
calculations? C or K?
Temperature is in K (K = 273 + C)
Gas Laws – Gay-Lussac’s Law
Constant: Volume
Relationship: Pressure is directly proportional
to temperature
Pressure a Temperature
Written As: P1/T1 = P2/T2
The IDEAL GAS Law –this is
what we will use
When we put all three laws together:
PV a 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
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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PV = nRT
R is known as the universal gas constant
Using STP conditions
P
V
R
=
PV
nT
= (1.00 atm)(22.4 L)
(1mol) (273K)
n
T
= 0.0821 L-atm
mol-K
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Combined Gas Law
P1 V 1
T1
= P2V2
T2
Isolate V2
P1 V 1 T2
V2 =
=
P 2 V 2 T1
P1V1T2
P2 T 1
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Learning Check G15
What is the value of R when the STP value for
P is 760 mmHg?
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Solution G15
What 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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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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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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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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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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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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Learning Check C1
Solve the combined gas laws for T2.
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Solution C1
Solve the combined gas law for T2.
(Hint: cross-multiply first.)
P1 V 1
T1
=
P2 V 2
T2
P1 V 1 T2 = P2 V 2 T1
T2
= P 2 V 2 T1
P1V1
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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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Data Table
Set up Data Table
P1 = 0.800 atm
V1 = 0.180 L
P2 = 3.20 atm
V2= 90.0 mL
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T1 = 302 K
T2 = ?? ??
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Solution
Solve for T2
Enter data
T2 = 302 K x
T2 =
atm x
atm
K - 273 =
mL =
mL
K
°C
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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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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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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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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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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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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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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
n1
initial
=
V2
n2
final
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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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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 =
P2 =
V1 =
T1 =
V2 = ?? T2 =
K
V2 = 15 L x
atm
atm
x
K
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K
K
= 6.8 L
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Solution C4
P1 = 1.0 atm
P2 = 2.0 atm
V1 = 15 L
V2 = ??
T1 = 273 K
T2 = 248 K
V2 = 15 L x 1.0 atm x 248 K
2.0 atm
273 K
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= 6.8 L
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Molar Volume
At STP
4.0 g He
1 mole
(STP)
V = 22.4 L
16.0 g CH4
1 mole
(STP)
V = 22.4 L
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44.0 g CO2
1mole
(STP)
V = 22.4 L
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Molar Volume Factor
1 mole of a gas at STP = 22.4 L
22.4 L
1 mole
and
1 mole
22.4 L
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Learning Check C5
A.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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Solution C5
A.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
22.4 He
4.00 g He = 1.43 g He
1 mole He
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Daltons’ Law of Partial Pressures
Partial 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.
P T = P1 +
P2 + P3 + .....
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Gases in the Air
The % 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
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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
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3) 0.109
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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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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
1 mole H2
0.5 mole O2
+ 0.3 mole He
+ 0.2 mole Ar
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Health Note
When 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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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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Solution C7
P = nRT
V
PT
=
PT = PO + PHe
2
1.47 mol x 0.0821 L-atm x 298 K
=
5.00 L (K mol)
7.19 atm
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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 =
mol
0.250 g = 35.6 g/mol
0.00703 mol
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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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Substitute
(1.00 atm ) mol-K
=
(0.0821 L-atm) (273 K)
0.0446 mol O2/L
Change moles/L to g/L
0.0446 mol O2
1L
x
32.0 g O2 =
1 mol O2
1.43 g/L
Therefore the density of O2 gas at STP is
1.43 grams per liter
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