#28 notes
Unit 4: Gases
Ch. Gases
I. Pressure and Manometers
Gas particles move in straight line paths. As they collide, they create a force, pressure.
Pressure = Force / Area
Standard Atmospheric Pressure = 14.7 psi
(at sea level)
= 101325 Pa
= 1 atm
= 760 mm Hg*
= 760 torr*
(1b./ in2 = force / area)
(pascals)
{1013.25 millibars}
(atmospheres)
*equivalent
A. Closed Manometers:
Ex. 1) Find the pressure in Pa and atm, if
the height of the mercury is 45 mm.
Height of Mercury = the Pressure
45 mm Hg │ 101325 Pa = 6.0 X103 Pa
│ 760 mm Hg
45 mm Hg │ 1 atm = 5.9 X10-2 atm
│ 760 mm Hg
B. Open Manometers:
Ex. 2) The mercury is 45 mm higher in the atmospheric arm. Find the pressure in Pa and atm,
if the atmospheric pressure is 100000. Pa.
45 mm Hg │ 101325 Pa = 6.0 X103 Pa
│760 mm Hg
Patm + PHg = Pgas
←The gas pushes the hardest (strongest), so it is alone!
100000. Pa + 6.0 X103 Pa = Pgas
106000 Pa = Pgas
1.1 X 105 Pa = Pgas
1.1 X105 Pa │ 1 atm = 1.1 atm
│101325 Pa
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Ex. 3) Same as Ex. 2, except the mercury is higher in the gas arm.
(height of Hg = 45mm, Patm = 100000. Pa)
45 mm Hg │ 101325 Pa = 6.0 X103 Pa
│760 mm Hg
Patm = PHg + Pgas
←The air pushes the hardest (strongest), so it is alone!
3
100000. Pa = 6.0 X10 Pa + Pgas
94000. Pa = Pgas
9.4 X104 Pa
9.4 X104 Pa│ 1 atm = 0.93 atm
│101325 Pa
II. Ideal Gas Law:
In this ideal gas law, the gas molecules are spread out. There will not be any attractions, so
Pobserved will not need any adjustments. The molecules will be much smaller than the
distances separating them. They are treated as point masses. They have mass, but negligible
volume. (This works at low pressures and high temperatures.) Because of this the volume of
the container can be used without subtracting the volume of the gas particles.
PV=nRT where R= 0.08206 L·atm/(mol·K)
o
C + 273 = K
Ex. 1) What pressure is exerted by 0.622 mol of CO2 contained in a 922 ml vessel at 16 oC?
P = ?, n = 0.622 mol, V = 922 ml, T = 16 oC
922 ml │1 X10-3 L = 0.922 L
│ 1 ml
16 oC + 273 = 289 K
PV = nRT
P (0.922 L) = (0.622 mol) (0.08206 L·atm/(mol·K)) (289K)
P = 16.0 atm
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#29 Notes
III. More Gas Laws (Derived from the Ideal)
Ex. 1) A gas in a container at 25 oC, having a volume of 300 cm3 exerts a pressure of 520 torr.
What is its volume at standard atmospheric pressure?
T1 = 25 oC
V1 = 300 cm3
P1 = 520 torr
V2 = ?
P2 = 760 torr = 1atm etc.
PV = nRT
n, R, T are constant
PV = constant so, P1V1 = constant
and
P2V2 =constant
P1V1 = P2V2 Boyle’s Law
(520 torr) (300 cm3) = (760 torr) V2
205 cm3 = V2
Ex. 2) Find the volume at standard temperature for a sample of H2 that at 9.00 oC has a
volume of 613 cm3.
**Standard Temperature = 273 K
V1 = ?
T1 = 273 K
V2 = 613 cm3
T2 = 9.00 oC
PV = nRT
P, n, R are constant
9.00 oC + 273 = 282 K
*Temperatures must be in Kelvin!
(Adding conversions will not cancel.)
V = nRT
P
V = nR
T P
V = constant
T
so, V1 = V2
T1 T2
V1 = (613 cm3)
(273K) (282 K)
V1 = 593 cm3
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Charles’ Law
**Cross Multiply!!
Ex. 3) 45.3 mol of CH4 gas occupies 916 L of space. What volume would 0.214 mol occupy?
n1 = 45.3 mol
V1 = 916 L
PV = nRT
n2 = 0.214 mol
V2 = ?
P, R, T are constant
V = nRT
P
V = RT
n P
V = constant
n
V1 = V2
n1
n2
Avogadro’s Law
(916 L) = V2
(45.3 mol) (0.214 mol)
4.33 L = V2
Ex. 4) CO2 in a 45 ml vessel at 0.98 atm and 25 oC is transferred to a 0.067 L container at
35 oC. What is its new pressure?
V1 = 45 ml
P1 = 0.98 atm
T1 = 25 oC
PV = nRT
V2 = 0.067 L
P2 = ?
T2 = 35 oC
45 ml │1 X10-3 L = 0.045 L
│ 1 ml
o
25 C + 273 = 298 K
35 oC + 273 = 308 K
n, R are constant
PV = nR
T
PV = constant
T
P1V1 = P2V2
T1
T2
(0.98 atm) (0.045 L) = (P2) (0.067 L)
(298 K)
(308 K)
0.68 atm = P2
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#30 Notes
IV. Gas Density
PV = nRT
mols =
mass
molar mass
PV = (mass)
RT
(molar mass)
(molar mass) = (mass) RT
PV
(molar mass) = (mass) · RT
V
P
Molar mass = dRT
P
but density is in g/L !!!
Ex. 1) A compound has the empirical formula CH2. A 200 ml flask at 298 K and 755 torr
contains 0.57 g of the gaseous compound. Give the molecular formula.
200 ml │ 1 X10-3 L = 0.200 L
│ 1 ml
d = m/V
d = 0.57 g
0.200 L
755 torr │ 1 atm = 0.993 atm
│ 760 torr
d = 2.85 g/L
mm = dRT/P mm = (2.85 g/L) (0.08206 L·atm/(mol·K)) (298 K)
(0.993 atm)
mm = 70. g/mol
Empirical Formula = CH2 = 14.027 g/mol
X5
Molecular Formula = C5H10
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70. g/mol = 5
14.027 g/mol
#31 Notes V. Dalton’s Law of Partial Pressures
In a mixture of gases, each particular gas will exert the same pressure it would, if it were
alone. In this room PO2 = 0.20 atm, PN2 = 0.79 atm, What is Ptot ?
Ptot = P1 + P2 + P3 + …
The total pressure is the sum of the partial pressures.
Ex. 1) CO2 is collected over H2O at 10 oC in a 20.0 cm3 vessel. The manometer indicates a
pressure of 60.0 kPa. What is the mass of gas collected? ( PH2O at 10 oC = 1.2 kPa)
Ptot = Pgas + PH2O
60 kPa = Pgas + 1.2 kPa
58.8 kPa = Pgas
P = 58.8 kPa │1000 Pa │ 1 atm = 0.580 atm
│1kPa │101325 Pa
T = 10 oC = 283 K
V = 20.0 cm3 │ 1 ml │1 X10-3 L = 0.0200 L
│ 1 cm3│ 1 ml
n=?
PV = nRT
(0.580 atm) (0.0200 L) = n (0.08206 L·atm/(mol·K)) (283 K)
4.995 X10-4 mol = n
4.995 X10-4 mol CO2 │ 44.00915 g = 2.2 X10-2 g CO2
│ 1 mol
** Remember: All gases diatomic (N2, H2, O2) except noble gases (He, Ne, Ar).
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Ex. 2) 6.00 dm3 of H2 and 2.00 dm3 of O2 at 25 oC and 760. torr are pumped into a 0.250 L
container. What is the pressure in the container?
H2
V = 6.00 dm3
T = 25 oC
P = 760. torr
Constant T, so ignore
O2
V = 2.00 dm3
T = 25 oC
P = 760. torr
new container
V = 0.250 L
P=?
PV = nRT
n, R, T are constant
P1V1 = P2V2
H2
O2
3
P1V1 = P2V2
**1L =1dm
P1V1 = P2V2
3
(760 torr) ( 6.00 dm ) = P2 (0.250 L)
(760 torr) ( 2.00 dm3) = P2 (0.250 L)
18240 torr = P2 of H2
6080 torr = P2 of O2
Ptot = PH2 + PO2 = 18240 torr + 6080 torr = 24320 = 2.43 X104 torr
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#32 Notes
VI. Gas Stoichiometry
Derive a conversion factor between mols and volume at STP for 1 mol of any gas.
STP = standard Temperature (273 K) and standard Pressure (1atm)
P = 1 atm
V =?
n = 1 mol
T = 273 K
PV = nRT
(1 atm) V = (1 mol) (0.08206 L·atm/(mol·K)) (273 K)
V = 22.4 L
** 1 mol of any gas = 22.4 L at STP
Ex. 1) An excess of nitrogen gas reacts with 5.0 g of hydrogen gas. What volume of ammonia
is produced at STP?
Write the reaction:
N2(g) + 3 H2(g) → 2 NH3(g)
5.0 g H2 │1 mol H2 │ 2 mol NH3 │ 22.4 L NH3 = 37 L NH3
│2.016 g H2│ 3 mol H2 │ 1 mol NH3
How many grams of H2 are produced from 39 L NH3?
39 L NH3 │ 1 mol NH3 │ 3 mol H2 │2.016 g H2_ = 5.3 g H2
│22.4 L NH3 │ 2 mol NH3 │1 mol H2
If it is not at STP, do stoichiometry, but stop at mols (after the mol ratio)
and then use PV = nRT.
OR Change given amounts to STP by using gas equations (T = 273 K, P = 1atm),
then do stoichiometry.
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#33 Notes VII. Kinetic Molecular Theory of Gases
1) The gas particles are so small compared to the distances separating them that we can
assume their volume is negligible (zero). Ideal Gases
2) The gas particles are in constant motion and their collisions cause the pressure exerted
by the gas.
3) The particles are assumed to exert no attractive/ repulsive forces on each other.
Ideal Gases
4) The average Kinetic Energy (KE) is directly proportional to temperature.
KE = ½ m v2
↑↑
mass velocity
KE = 3/2 RT
*high temperature, high velocity of particles, high KE
where R = 8.31 J/(mol·K)
1 joule = 1 Newton · meter
1 Newton = 1 (kg · m) /s2
VIII. Root Mean Square Velocity
KE = 1/2 mv2 = 3/2 RT
mv2 = 3RT
v2 = 3RT
m
_______
vrms = √ (3RT/m)
** molar mass must be in kg/mol, velocity in m/s
IX. Effusion/ Diffusion
Effusion:
-filling of a gas into a chamber by a hole.
Diffusion:
-mixing of 2 gases.
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Graham’s Law of Effusion (works for diffusion too.)
_______
____
__
Rate of effusion for gas #1 = vrms #1 = √(3RT/m1) = √(1/m1) = √m2
Rate of effusion for gas #2 = vrms #2 = √(3RT/m2) = √(1/m2) = √m1
__
Rate gas #1 = √m2
Rate gas #2 √m1
**Rate must be in volume/time, but
units don’t matter if they agree.
Ex. 1) Find KE and vrms for F2 at 25 0C.
KE = 3/2 RT = 3/2 (8.31 J/(mol·K)) (298 K) = 3.71 X103 J/mol
______ ______________________
Vrms = √3RT/m = √ (3) (8.31 J/(mol·K)) (298 K) = 442 m/s
(0.03800 kg/mol)
↑↑↑
F2 = 38.00 g │ 1 kg = 0.03800 kg/mol F2
mol│1 X103 g
Ex. 2) A gas has an effusion rate of 15 ml/min. If CH4 has a rate of 45 ml/min, what is the
molar mass of the other gas?
__
R1 = √m2
CH4 = 16.04 g │ 1 kg = 0.01604 kg/mol
R2
√m1
mol │1 X103 g
_____
Rate ? gas = √ mCH4
Rate CH4 = √ m ?gas
________________
15 ml/min
= √0.01604 kg/mol CH4
45 ml/min CH4
√ m ?gas
0.333
= 0.1266
√ m ?gas
____
0.333 √ m ?gas = 0.1266
____
√ m ?gas = 0.380
m ?gas = 0.144 kg/mol
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#34 Notes
IX. Real Gases
PV = nRT is for Ideal Gases
The particle volume was assumed to be zero, but molecules really do take up space.
Vvolume around the gas = Vvolume of the container – nb volume of the gas particles
Where n = mols and
b = an experimentally found empirical constant
The attractive forces were also assumed to be zero, but attractive forces between molecules
will keep them from hitting the container as hard as predicted.
Pobserved pressure = P’the real pressure – a (n/V)2 decrease in pressure caused by the attractive forces
Where a = an experimentally found proportionality constant and
n = mols and V is volume
so P’real = Pobserved + a (n/V)2
PrealVaround the gas = nRT
[Pobserved + a (n/V)2] (V – nb) = nRT
a,b: see Table in textbook
Van der Waals Equation for Real Gases
Ex.1) Calculate the pressure exerted by 0.2500 mol of Krypton in a 1.000 L container
at 25.0 oC.
P=?
n = 0.2500 mol
Look up for Kr: a = 2.32 atm•L2/mol2
V = 1.000 L
b = 0.0398 L/mol
o
T = 25 C = 298 K
[Pobserved + a (n/V)2] (V – nb) = nRT
[P + (2.32 atm•L2/mol2) (0.2500 mol/ 1.000 L)2] [1.000 L – (0.2500 mol)( 0.0398 L/mol)] =
(0.2500 mol)(0.08206 L•atm/(mol•K))(298 K)
[P + 0.145 atm] [0.99005 L] = (6.11347 L•atm)
[P + 0.145 atm] = 6.1749 atm
P = 6.03 atm
*End of Notes* (Assignments #35-36 are Review Assignments. There are no notes for these assignments.)
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