Transcript Document

Gas Laws
Chapter 11
11.1 Pressure and Force
• Pressure = Force per unit area
• Measure force in Newtons (N)
– 1N = force that will increase the speed of a
1 kg mass by one meter per second each
second that the force is applied
– Force of gravity = 9.8m/s2
• A 51 kg person
(112 lb) exerts
a force due to
gravity of 500
N
• Pressure
exerted on the
floor depends
on the area of
contact
Pressure = Force
Area
Units of Pressure
1 pascal (Pa) = 1 N/m2
1 atm = 760 mmHg = 760 torr
1 atm = 101,300 Pa
1 atm = 101.3 kPa
1 atm = 14.7 lb/in2
Standard pressure is
pressure exerted by a
column of mercury 760
mm high
Barometer
10 miles
4 miles
Sea level
0.2 atm
0.5 atm
1 atm
Crushing a Coke Can
Click on the
picture
As P (h) increases
V decreases
Graham’s Law
• Suppose 2 containers at the same
temperature are opened across the
room from you. One contains HCl & the
other NH3.
• Which will you smell first?
• What moves faster, a large particle or a
small particle?
• The rate of diffusion will depend on both
the average velocity and the mass of
the particle
Graham’s Law
• T1 = T2  KE1 = KE2
• Cancel ½
• Move m1 to right
1
1
2
2
m1v1  m2 v2
2
2
m2v2
v 
m1
v m
v m
2
1
2
• Move v2 to left
1
2
2
• Take square root of both
sides
2
1
v1
m2

v2
m1
2
v1
m2

v2
m1
or
R1
m2

R2
m1
11.2 The Gas Laws
• Measuring the
density of a gas is
harder than for
liquids & solids
– Volume can change
• Four quantities affect measuring gases:
– Volume: amount of space
– Pressure: # collisions with walls of
container
– Temperature: average Kinetic Energy
– Quantity: # of moles
• We can make general statements about
any 2 quantities if 2 are held constant
Boyle’s Law
P a 1/V
P x V = constant
P1 x V1 = P2 x V2
Constant temperature
Constant amount of gas
A sample of chlorine gas occupies a volume of 946 mL
at a pressure of 726 mmHg. What is the pressure of
the gas (in mmHg) if the volume is reduced at constant
temperature to 154 mL?
P1 x V1 = P2 x V2
P2 =
P1 = 726 mmHg
P2 = ?
V1 = 946 mL
V2 = 154 mL
P1 x V1
V2
726 mmHg x 946 mL
=
= 4460 mmHg
154 mL
Learning Check 1
A sample of nitrogen gas is 6.4 L at a pressure
of 0.70 atm. What will the new volume be if
the pressure is changed to 1.40 atm? (T
constant)
1) 3.2 L
2) 6.4 L
3) 12.8 L
4) I don’t understand
Take a few minutes to work the problem before
moving forward.
PV Calculation
Learning Check 2
A sample of helium gas has a volume of
12.0 L at 600. mm Hg. What new
pressure is needed to change the
volume to 36.0 L? (T constant) Explain.
1) 200. mmHg
2) 400. mmHg
3) 1200 mmHg
4) I don’t understand
PV Calculation
As T increases
V increases
Variation of gas volume with temperature
at constant pressure.
Charles’ Law
VaT
V = constant x T
Temperature must be
in Kelvin
V1/T1 = V2/T2
T (K) = t (0C) + 273
Learning Check 3
Use Charles’ Law to complete the statements
below:
1. If final T is higher than initial T, final V
is 1) greater, or 2) less than the initial V.
2. If final V is less than initial V, final T is
1) higher, or 2) lower than the initial T.
Solution GL3
A sample of carbon monoxide gas occupies 3.20 L at
125 0C. At what temperature will the gas occupy a
volume of 1.54 L if the pressure remains constant?
V1/T1 = V2/T2
V1 = 3.20 L
T1 = 398 K
T2 =
V2 x T1
V1
=
V2 = 1.54 L
T2 = ?
1.54 L x 398 K
3.20 L
= 192 K
Learning Check 4
A sample of oxygen gas has a volume of
420. mL at a temperature of 18.0°C.
What temperature (in °C) is needed to
change the volume to 640. mL?
1) 443.°C
2) 170.°C
3) - 82.0°C
Solution GL4
Variation of gas pressure with temperature
at constant volume.
Gay-Lussac’s
Law
PaT
P = constant x T
P1/T1 = P2/T2
Temperature must be
in Kelvin
T (K) = t (0C) + 273
Learning Check 5
Use Gay-Lussac’s law to complete the
statements below:
1. When temperature decreases, the
pressure of a gas 1) decreases or 2)
increases.
2. When temperature increases, the pressure
of a gas 1) decreases or 2) increases).
Argon is an inert gas used in lightbulbs to retard the
vaporization of the filament. A certain lightbulb
containing argon at 1.20 atm and 18 0C is heated to
85 0C at constant volume. What is the final pressure of
argon in the lightbulb (in atm)?
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K
P1
P2
=
T1
T2
P1T2
= 1.20 atm x 358 K = 1.48 atm
P2 =
291 K
T1
Learning Check 6
Complete with
1) Increases 2) Decreases
3) Does not change
A. Pressure _____, when V decreases
B. When T decreases, V _____.
C. Pressure _____ when V changes from 12.0 L to 24.0 L
(constant n and T)
D. Volume _____when T changes from 15.0 °C to 45.0°C
(constant P and n)
Homework
• Find the density of the following gases
at STP: He, N2, Cl2, CO2
• WS #1:18-23
• WS #2: 10-14
• #3: 5, 7
• #4: 26, 36
• #6: 2, 3
Combined Gas Law
If it is true that:
P x V = constant1
V1/T1 = constant2
P1/T1 = constant3
We can also say that:
PV
= constant
T
Which gives us:
P1V1 P2V2
=
T1
T2
Combined Gas Law
If you should only need one of the other gas
laws, you can cover up the item that is
constant and you will get that gas law!
P1 V1 =
T1
P2 V2
T2
Boyle’s Law
Charles’ Law
Gay-Lussac’s
Law
Learning Check 7
A 28.5 mL sample of gas was collected at
751 mm Hg and 25.6 C. Find the
volume at STP.
A). 31.1 mL
B). 25.7 mL
C) 31.5 mL
Avogadro's Law: Volume and
Moles
In Avogadro’s Law
• the volume of a gas is
directly related to the
number of moles (n) of
gas.
• T and P are constant.
V1 = V2
n1
n2
Think of a balloon
• Pressure in the
balloon stays equal
to air pressure
• As you add more air
molecules…
• the balloon expands
(increases in
volume)
Learning Check 8
If 0.75 mole helium gas occupies
a volume of 1.5 L, what volume
will 1.2 moles helium occupy at
the same temperature and
pressure?
1) 0.94 L
2) 1.8 L
3) 2.4 L
Solution
No Name Law
Relates number of moles to pressure
• The pressure of a gas is directly related to the
number of moles (n) of gas.
• T and V are constant.
P1 = P2
n1
n2
Density of Gases
m
• D=
v
• Mass depends on the number of
particles and identity of the gas
• Volume depends on # moles,
temperature and pressure
• Density changes as volume changes
Avogadro’s Hypothesis
• At the same temperature and
pressure, equal volumes of
gases contain equal numbers
of molecules (moles)
– Size of molecules very small in
relation to overall volume
• Molar volume of a gas: The
volume of one mole of any
gas at a particular
temperature and pressure.
Molar Volume of a Gas
• Each temp and pressure combination
has its own molar volume
– 1 mole occupies 30.6L @ 100 C and 1
atm
– 1 mole occupies 24.5L @ 25 C and 1 atm
• At STP, standard molar volume of
any gas is 22.4 L (memorize)
A chemical reaction produces 0.35 moles
of a gas. What volume will it occupy at
STP?
Use molar volume as conversion factor:
1 mole = 22.4 L
0.35 mol 22.4L
=7.8 L
1 mol
65.0 mL of hydrogen sulfide gas is produced at
STP. How many moles are produced?
65.0 mL 1.00 L 1 mol
1000 mL 22.4L
=2.90 x 10-3 mol
Finding Density
m
• D=
v
Molar mass
or D=
Molar volume
• At STP: D= Molar mass
22.4 L
Finding Density
• Given the volume & mass of any
sample
– 0.519 g in 200.mL
m 0.519 g
D 
 2.60 g
L
v 0.200L
• From formula at STP
– F2 at STP
molar mass
38.0 g
D

 1.70 g
L
molar volume 22.4 L
Finding Molar Mass from
Density
• Can use density formula as a
proportion:
mass
molar mass
D

volume molar volume
Sample problem
• What is the molar mass of a gas with a
density of 1.28 g/L at STP?
mass
molar mass
D

volume molar volume
1.28 g
x
D

1L
22.4 L
x  (1.28 g )(22.4 L)  1L
x  28.7 g
mol
Sample problem
• Find the mass of 2.50L of ammonia at
STP.
mass
molar mass
D

volume molar volume
x
17.0 g
D

2.50 L 22.4 L
x  (2.50 L)(17.0 g )  22.4 L
x  1.90 g
Wording
• Molecular weight = molar mass (WS1)
• Review finding the empirical and
molecular formula
• Find the mass of exactly 1 L means find
the Density (WS2)
• Quiz on Friday: Gas Laws and Density
with Empirical/molecular Formula
Homework
• Find the density of the following gases
at STP: He, N2, Cl2, CO2
• WS #1:18-27
• WS #2: 10-14
• #3: 5, 7, 8
• #4: 26, 36
• #5: 3
• #6: 2, 3
• #8: 12
Homework
• Eudiometer Problems
– WS#2: 15-20; 25
• Dalton’s Law Problems
– WS#3: 9
– WS#8: 1, 3, 7, 9
Dalton’s Law of Partial
Pressures
He
1.00
atm
CO2
1.00
atm
H2O
1.00
atm
3.00
atm
The total
pressure of a
mixture of gases is
equal to the sum
of the partial
pressures of the
individual gases
Dalton’s Law of Partial Pressures
• Based on Avogadro’s hypothesis: equal
# particles occupy equal volumes
• Total pressure of a gas mixture is the
sum of the partial pressures of the
component gases
• PT = P1 + P2 + P3 + …..
Ex 1: A container holds three gases: oxygen,
carbon dioxide, and helium. The partial pressures
of the three gases are 2.00 atm, 3.00 atm, and
4.00 atm, respectively. What is the total pressure
inside the container?
Ptotal = P1 + P2 + P3 + …
Ptotal = PO2 + + PCO2 + PHe
Ptotal = 2.00 atm + 3.00 atm + 4.00 atm
Ptotal = 9.00 atm
A sample of natural gas contains 8.24 moles of CH4,
0.421 moles of C2H6, and 0.116 moles of C3H8. If the
total pressure of the gases is 1.37 atm, what is the
partial pressure of propane (C3H8)?
Eudiometer Tubes
(mercury or
water displacement)
• Eudiometer – a tube used to collect a gas. It
is closed at one end and is graduated.
• Reaction = 2HCl (aq) + Ca (s)  H2 (g) +
CaCl2 (aq)
Eudiometer with mercury
Situation #1
• If there is just enough gas to make the
mercury level inside the tube the same
as the level of the mercury in the bowl
then the pressure of the hydrogen gas is
the same as that of the atmosphere!
• Pgas = Pair
Eudiometer, Situation #2
• Suppose enough hydrogen gas is added to
make the level inside the tube lower than
the level of the mercury in the bowl. The P
of the hydrogen gas is greater than the
atmospheric pressure. (To determine the
hydrogen gas pressure one must add the
level difference to the barometric reading.)
d
• Pgas = Pair + d
Eudiometer with mercury,
situation #3
Suppose there was not enough gas to make the
mercury level the same. Then, the pressure of
d
hydrogen gas would not be the same as the
pressure of the air outside the tube. (So, we must
subtract the level difference from the barometric
reading.)
Pgas = Pair - d
Problem 1
• The volume of oxygen in a eudiometer
is 37.0 mL. The mercury level inside
the tube is 25.00mm lower than the
outside. The barometric reading is
742.0 mm Hg. What is the pressure of
the gas?
A. 717.0 mm
B. 767.0 mm
C. I don’t know
Solving Problem 1
Problem 2
What is the P of the gas in an eudiometer, when the
mercury level in the tube is 14mm higher, than that
outside? That barometer reads 735mm Hg.
A. 749 mm
B. 721 mm
Answer to Problem 2
Water in Place of Mercury
• Calculations are done the same way but
the difference in water levels must first
be divided by 13.6 to convert it to its
equivalent height in terms of a column
of mercury since water is 1/13.6 as
dense as mercury.
• Also, water evaporates much more readily
than mercury and so is in with the collected
gas.
• So, you need to determine the partial
pressure of the dry gas (unmixed with water
vapor).
• The vapor pressure of water, at the given
temperature, must be subtracted from the
total pressure of the gas within the tube.
Water in Place of Mercury
Bottle full of oxygen
gas and water vapor
2KClO3 (s)
2KCl (s) + 3O2 (g)
PT = PO2 + PH2 O
d
• Problem : Oxygen is collected using the
water displacement method. The water
level inside the tube is 27.2mm higher
than that outside. The temp. is 25.0 C.
The barometric pressure is 741.0mm.
What is the partial pressure of the dry
oxygen in the eudiometer?
Step 1: 27.2mm/13.6 = 2.00mm
Step 2: 741.0 mm - 2.00mm = 739.0 mm
So: 739.0 mm = P total = Poxygen + Pwater
Step 3: To correct for water vapor pressure, you need to
know the pressure of water vapor at 25.0C and it is
23.8mm (always given - see handout). Subtract this
water partial pressure from the pressure total found in
step 2.
739.0mm - 23.8mm = 715.2 mm = partial pressure
of
the dry oxygen.
dHO
2
-P
Pgas = Pair ±
HO
13.6
2
• A eudiometer contains 38.4 mL of air
collected by water displacement at a
temperature of 20.0C. The water level
inside the eudiometer is 140. mm higher
than that outside. The barometric
reading is 740.0mm. Water vapor
pressure at 20.0C is 17.5mm. Calculate
the volume of dry air at STP.
Problem 3:
The pressure of the dry gas is
A. 732.8 mm
B. 712.2 mm
C. 747.2 mm
D. I have no idea
Problem 4: The volume at STP is
A. 33.5 mL
B. 38.6 mL
C. I forgot the combined gas law
Solution Problem 4
Ideal Gas Equation
Boyle’s law: V a 1 (at constant n and T)
P
Charles’ law: V a T (at constant n and P)
Avogadro’s law: V a n (at constant P and T)
Va
nT
P
V = constant x
nT
P
=R
nT
P
R is the gas constant
PV = nRT
The conditions 0 0C and 1 atm are called standard
temperature and pressure (STP).
Experiments show that at STP, 1 mole of an ideal
gas occupies 22.4 L.
PV = nRT
PV
R=
=
nT
(1 atm)(22.4L)
(1 mol)(273 K)
R = 0.0821 L • atm / (mol • K)
What is the volume (in liters) occupied by 49.8 g of HCl
at STP?
T = 0 0C = 273 K
P = 1 atm
PV = nRT
nRT
V=
P
V=
1 mol HCl
n = 49.8 g x
= 1.37 mol
36.45 g HCl
L•atm
1.37 mol x 0.0821mol•K
V = 30.6 L
1 atm
x 273 K
Argon is an inert gas used in lightbulbs to retard the
vaporization of the filament. A certain lightbulb
containing argon at 1.20 atm and 18 0C is heated to
85 0C at constant volume. What is the final pressure of
argon in the lightbulb (in atm)?
PV = nRT
n, V and R are constant
nR
P
=
= constant
T
V
P1
P2
=
T1
T2
P1 = 1.20 atm
T1 = 291 K
P2 = ?
T2 = 358 K
T2
= 1.20 atm x 358 K = 1.48 atm
P2 = P1 x
291 K
T1
Density (d) Calculations
PM
m
d=
=
V
RT
m is the mass of the gas in g
M is the molar mass of the gas
Molar Mass (M ) of a Gaseous Substance
dRT
M=
P
d is the density of the gas in g/L
Gas Stoichiometry
What is the volume of CO2 produced at 370 C and 1.00
atm when 5.60 g of glucose are used up in the reaction:
C6H12O6 (s) + 6O2 (g)
6CO2 (g) + 6H2O (l)
g C6H12O6
mol C6H12O6
5.60 g C6H12O6 x
6 mol CO2
1 mol C6H12O6
x
= 0.187 mol CO2
180 g C6H12O6
1 mol C6H12O6
V=
nRT
=
P
mol CO2
V CO2
L•atm
x 310.15 K
mol•K
1.00 atm
0.187 mol x 0.0821
= 4.76 L
Consider a case in which two gases, A and B, are in a
container of volume V.
nART
PA =
V
nA is the number of moles of A
nBRT
PB =
V
nB is the number of moles of B
PT = PA + PB
PA = XA PT
nA
XA =
nA + nB
PB = XB PT
Pi = Xi PT
nB
XB =
nA + nB
Kinetic Molecular Theory of Gases
1. A gas is composed of molecules that are separated from
each other by distances far greater than their own
dimensions. The molecules can be considered to be points;
that is, they possess mass but have negligible volume.
2. Gas molecules are in constant motion in random directions.
Collisions among molecules are perfectly elastic.
3. Gas molecules exert neither attractive nor repulsive forces
on one another.
4. The average kinetic energy of the molecules is proportional
to the temperature of the gas in kelvins. Any two gases at
the same temperature will have the same average kinetic
energy
Kinetic theory of gases and …
• Compressibility of Gases
• Boyle’s Law
P a collision rate with wall
Collision rate a number density
Number density a 1/V
P a 1/V
• Charles’ Law
P a collision rate with wall
Collision rate a average kinetic energy of gas molecules
Average kinetic energy a T
PaT
Kinetic theory of gases and …
• Avogadro’s Law
P a collision rate with wall
Collision rate a number density
Number density a n
Pan
• Dalton’s Law of Partial Pressures
Molecules do not attract or repel one another
P exerted by one type of molecule is unaffected by the
presence of another gas
Ptotal = SPi
Apparatus for studying molecular speed distribution
The distribution of speeds
of three different gases
at the same temperature
The distribution of speeds
for nitrogen gas molecules
at three different temperatures
urms =
M
3RT
Gas diffusion is the gradual mixing of molecules of one gas
with molecules of another by virtue of their kinetic properties.
NH4Cl
NH3
17 g/mol
HCl
36 g/mol
Deviations from Ideal Behavior
1 mole of ideal gas
PV = nRT
PV = 1.0
n=
RT
Repulsive Forces
Attractive Forces
Effect of intermolecular forces on the pressure exerted by a gas.