Chapter 6
Gas Laws
The Gas Phase
• Gases have no distinct volume or shape.
• Gases expand to fill the volume of their
container.
• Gas particles are miscible with each other.
• Evidence for gas particles being far apart :
 We can see through gases
 We can walk through gases
 Gases are compressible
 Gases have low densities
Kinetic Theory of a Gas
• Gas particles are sizeless relative to their
volume.
• Gas particles are in constant random motion.
• Gas particles have elastic collisions.
• Gas particles do not have attractive forces
with each other.
• The Average kinetic energy of a gas is
directly proportional to absolute temperature.
Kinetic Theory of a Gas
• Sizeless particles relative to the volume means
that most of a gas is empty space.
 We can see through a gas
 We can walk through a gas
• Elastic collision means that no kinetic energy is
lost on impact.
• If gas particles had attractive forces then they
would not be mostly empty space, and be in the
solid or liquid state.
• Since gas particles are in constant rapid motion,
then they must possess kinetic energy
Atmospheric Composition
Composition of Earth’s Atmosphere
Compound
Nitrogen
Oxygen
%(Volume)
78.08
20.95
Mole Fractiona
0.7808
0.2095
Argon
Carbon dioxide
Methane
Hydrogen
0.934
0.033
2 x 10-4
5 x 10-5
0.00934
0.00033
2 x 10-6
5 x 10-7
a. mole fraction = mol component/total mol in mixture.
Parameters Affecting Gases
• Pressure (P)
• Volume (V)
• Temperature (T)
• Number of Moles (n)
Pressure
Pressure is equal to force/unit area (P =F/A)
 On earth the force related to gravity
 From physics the F=ma, where m is kg and
a=9.8m/s2
 The units kgm/s2 is called a Newton (N)
 Since P = F/A, then pressure unit is Kgm/s2m2, or
N/m2
 Pascal is the abbreviation of N/m2 called Pa
Pressure
The atomosphere is pushing with a force of 14.7
lbs/in2 on every surface in this room.
For example each square inch on the whiteboard
has 17.7 lbs of air pushing on each square inch,
which is probably several tons. Why is the white
board not pushed into the next room????
Pressure Units
•
•
SI units = Newton/meter2 = 1 Pascal (Pa)
1 standard atmosphere (atm) = 101,325 Pa
 1 atm =760 mm Hg
 1 atm = 760 torr (torr is abbreviation of mmHg)



1 atm = 14.7 lbs/in2
1 atm = 1.013 barr
Barr = 100 kPa
Measurement of Pressure
What is above mercury?
Measurement of Pressure
760 mm Hg
Measurement of Pressure
How many pounds of
mercury are in the tube
if its opening is 1.0 in2,
assuming 14.7 psi
pressure?
How about if the
opening is 2.0 in2?
760 mm Hg
Measurement of Pressure
How many pounds of
mercury are in the tube
if its opening is 1.0 in2,
assuming 14.7 psi
pressure?
How about if the
opening is 2.0 in2?
How high would the
column of Hg be in the
2.0in2 device?
760 mm Hg
Units for Expressing Pressure
Unit
Value
Atmosphere
1 atm
Pascal (Pa)
1 atm = 1.01325 x 105 Pa
Kilopascal (kPa) 1 atm = 101.325 kPa
mmHg
1 atm = 760 mmHg
Torr
1 atm = 760 torr
Bar
1 atm = 1.01325 bar
mbar
1 atm = 1013.25 mbar
psi
1 atm = 14.7 psi
Elevation and Atmospheric Pressure
Manometer
Pressure Measurement
Open Tube Manometer
= 15 mm
Is the atmosphere or the gas
in the canister pushing
harder?
Pressure Measurement
Open Tube Manometer
Is the atmosphere or the gas
in the canister pushing
harder? Gas in the canister
If the atmospheric pressure is
= 15 mm 766 mm, then what is the
pressure of the canister?
Pressure Measurement
Open Tube Manometer
Is the atmosphere or the gas in
the canister pushing harder? Gas
in the canister
If the atmospheric pressure is 766
= 15 mm mm, then what is the pressure of
the canister?
P = 766 + 15 = 781 mm (torr)
gas
Pressure Measurement
Open Tube Manometer
gas
Is the atmosphere or the gas
in the canister pushing
harder?
Pressure Measurement
Open Tube Manometer
= 13 mm
gas
Is the atmosphere or the gas
in the canister pushing
harder? The atmosphere
What is the pressure of the
gas if the atmosphere is 766
mm?
Pressure Measurement
Open Tube Manometer
= 13 mm
gas
Is the atmosphere or the gas
in the canister pushing
harder? The atmosphere
What is the pressure of the
gas if the atmosphere is 766
mm? 753 mm
Pressure Measurement
Open Tube Manometer
gas
Now what is pushing harder,
the gas or the atomosphere?
Pressure Measurement
Open Tube Manometer
gas
Now what is pushing harder,
the gas or the atmosphere?
Neither, both the same.
Pressure Measurement
Open Tube Manometer
Now what is pushing harder,
the gas or the atmosphere?
Neither, both the same.
Is the gas canister empty?
gas
Pressure Measurement
Open Tube Manometer
Now what is pushing harder,
the gas or the atmosphere?
Neither, both the same.
Is the gas canister empty?
No, completely full of gas!
gas
Boyles Law
Atm
●
●
●
●
Consider a gas in a closed system
containing a movable plunger. If the
plunger is not moving up or down,
what can be said about the pressure
of the gas relative to the
atmospheric pressure?
Boyles Law
Atm
●●
●
●
●
●
●
Suppose we add some red gas to the
container, what would happen to
the collisions of gas particles with
container walls. Would they
increase, decrease or stay the same?
Boyles Law
Atm
●●
●
●
●
●
●
Suppose we add some red gas to the
container, what would happen to
the collisions of gas particles with
container walls. Would they
increase, decrease or stay the same?
More particles, more collisions, and
more pressure.
What happens to the plunger?
Boyles Law
Atm
●●
●
●
●
●
●
Suppose we add some red gas to the
container, what would happen to
the collisions of gas particles with
container walls. Would they
increase, decrease or stay the same?
More particles, more collisions, and
more pressure.
What happens to the plunger?
Boyles Law
●
●●
●
●
●
●
●
●
The number of
particles remain the
same, but the surface
area they have to strike
increases, thus the
number of collisions
per square inch
decrease as the
plunger goes up
exposing more surface
area causing a
decrease in pressure.
Boyle’s Law
Pressure and volume are
inversely proportional.
• P  1/V (T and n
fixed)
• P  V = Constant
• P1V1 = P2V2
Charles’s Law
The kinetic theory of gases states that absolute
temperature and kinetic energy are directly
proportional. As temperature increases then particle
velocity increases and the particles reach the
container wall sooner, thus increasing collisions per
second. Since Charles law is for constant pressure,
then the volume must expand to keep a constant
pressure. The mathematical statement of Charles
Law is below, showing pressure and volume directly
proportional.
V1
V2
T1 = T2
Charles’s Law
Avogadro’s Law
• For a gas at constant
temperature and
pressure, the volume is
directly proportional to
the number of moles of
gas (at low pressures).
V n
V1 = V2
n1
n2
Amonton’s Law
• P T
• P/T = Constant
• P1 = P2
T1
T2
Combined Gas Law
• Combining the gas laws the relationship
P T(n/V) can be obtained.
• If n (number of moles) is held constant,
then PV/T = constant.
P1V1
P2V2
=
T1
T2
Temperature, Kelvin, only
Pressure, atm, torr, mmHg,
Volume, L, mL, cm3, etc
Example
A balloon is filled with hydrogen to a pressure of 1.35
atm and has a volume of 2.54 L. If the temperature
remains constant, what will the volume be when the
pressure is increased to 2.50 atm?
Example
A sample of oxygen gas is at 0.500 atm
and occupies a volume of 11.2 L at 00C,
what volume will the gas occupy at 6.00
atm at room temperature (250C)?
Ideal Gas Law
PV = nRT
R = universal gas constant
= 0.08206 L atm K-1 mol-1
P = pressure in atm
V = volume in liters
n = moles
T = temperature in Kelvin
Example
Calculate the pressure of a 1.2 mol sample of
methane gas in a 3.3 L container at 25°C.
This problem is not a “changing condition”
problem, therefore use the ideal gas law.
Practice
A 1.22 g sample of a gas is contained in a 125 mL
flask at 25⁰C and 766 torr. Find the molar mass.
Practice
A 1.22 g sample of a gas is contained in a 125 mL
flask at 25⁰C and 766 torr. Find the molar mass.
What are the units of molar mass?
Practice
A 1.22 g sample of a gas is contained in a 125 mL
flask at 25⁰C and 766 torr. Find the molar mass.
What are the units of molar mass? g/mole
Practice
A 1.22 g sample of a gas is contained in a 125 mL
flask at 25⁰C and 766 torr. Find the molar mass.
What are the units of molar mass? g/mole
0.0821 L-atm
mole-K
Practice
A 1.22 g sample of a gas is contained in a 125 mL
flask at 25⁰C and 766 torr. Find the molar mass.
What are the units of molar mass? g/mole
0.0821 L-atm 298.15 K
mole-K
Practice
A 1.22 g sample of a gas is contained in a 125 mL
flask at 25⁰C and 766 torr. Find the molar mass.
What are the units of molar mass? g/mole
0.0821 L-atm 298.15 K 760 torr
mole-K
atm
Practice
A 1.22 g sample of a gas is contained in a 125 mL
flask at 25⁰C and 766 torr. Find the molar mass.
What are the units of molar mass? g/mole
0.0821 L-atm 298.15 K 760 torr
mole-K
atm
Practice
A 1.22 g sample of a gas is contained in a 125 mL
flask at 25⁰C and 766 torr. Find the molar mass.
What are the units of molar mass? g/mole
0.0821 L-atm 298.15 K 760 torr 1000 mL
mole-K
atm
L
Practice
A 1.22 g sample of a gas is contained in a 125 mL
flask at 25⁰C and 766 torr. Find the molar mass.
What are the units of molar mass? g/mole
0.0821 L-atm 298.15 K 760 torr 1000 mL
mole-K
atm
L
Practice
A 1.22 g sample of a gas is contained in a 125 mL
flask at 25⁰C and 766 torr. Find the molar mass.
What are the units of molar mass? g/mole
0.0821 L-atm 298.15 K 760 torr 1000 mL
mole-K
atm
L
766 torr
Practice
A 1.22 g sample of a gas is contained in a 125 mL
flask at 25⁰C and 766 torr. Find the molar mass.
What are the units of molar mass? g/mole
0.0821 L-atm 298.15 K 760 torr 1000 mL
1.22 g
mole-K
atm
L
766 torr
125 mL
Practice
A 1.22 g sample of a gas is contained in a 125 mL
flask at 25⁰C and 766 torr. Find the molar mass.
What are the units of molar mass? g/mole
0.0821 L-atm 298.15 K 760 torr 1000 mL
1.22 g
mole-K
atm
L
766 torr
125 mL
STP
•
“STP” means standard temperature and standard
pressure
 P = 1 atmosphere
 T = 0C
 The molar volume of any ideal gas is 22.42 liters
at STP (for 1 mole at STP
Dalton’s Law of Partial Pressures
• For a mixture of gases in a
container
• PTotal = P1 + P2 + P3 + . .
.
Mole Fraction
• Mole Fraction: the ratio of the number of
moles of a given component in a mixture to
the total number of moles in a mixture.
C1 =
n1
nTOTAL
=
n1
n1 + n2 + n3 + •••
• Mole Fraction in terms of pressure (n = PV/RT)
C1 =
P1(V/RT)
P1(V/RT) + P2(V/RT) + P3(V/RT) + •••
Mole Fraction
C1 =
C1 =
P1
P1 + P2 + P3 + •••
n1
nTOTAL
=
=
P1
PTOTAL
P1
PTOTAL
Mole Fraction Example
At 250C, a 1.0 L flask contains 0.030 moles
of nitrogen, 150.0 mg of oxygen and
4 x 1021 molecules of ammonia.
A. What is the partial pressure of each gas?
B. What is the total pressure in the flask?
C. What is the mole fraction of each?
Collecting a Gas Over Water
Practice
A sample of KClO3 is heated and decomposes to
produce O2 gas. The gas is collected by water
displacement at 25°C. The total volume of the
collected gas is 229 mL at a pressure of 754 torr.
How many moles of oxygen formed?
Hint: The gas collected is a mixture so use
Dalton’s Law to calculate the pressure of oxygen
then the ideal gas law to find the number of
moles oxygen.
PT = PO + P
2
H2O
Vapor Pressure of Water
Kinetic Molecular Theory
• K.E. = 1/2mu2rms
 Urms - the root-meansquared speed of the
molecules
•
Urms =
3RT
M
At Constant Temp
Pressure Derivation
From the kinetic molecular theory
Since pressure = P/A, then the following are required to
understand this concept.
Translational kinetic energy of molecules Ek the greater
the Ek the greater the force Ek=1/2mu2. Frequency,
collisions per secondThe greater the frequence the greater
the force Collision frequence Cf  u(n/V)Impulse, the
transfer of momentum when particles collideI  um
Pressure Derivation
Pressure of a gas is the product of the impulse and
the collision frequency
P  (mu)[(u)(n/V)]  n/V(mu2)
Not all molecules are moving at the same speed, thus
average of the squares speed must be used
Why squares, because pressure is proportional to the
speed squared.
P  n/V(muave 2)
Because there are three dimensions, then P =
1/3n/V(muave 2)
Derivation Urms
Rearranging the equation P = 1/3n/V(muave 2), gives
PV = 1/3n(muave2) (multiply both sides by V)
Since PV = nRT, or PV=RT for one mole of gas
Then 3RT = 1/3Na(muave 2) since a mole = Na
Since (Nam)is the molar mass (M), then 3RT = Muave2)
Solve for uave = (3RT/M)1/2 since square root is
involved uave is now renamed as urms = (3RT/M)1/2
It should be noted that this equation, says lighter
molecules of gas travel faster than heavier ones
Velocity Comparision
um = modal (most)
uave = sum u/total
urms from previous slide
The mean Free Path
1.The small distance a gas travels before hitting
another gas particle 10-7 m
2.This is a small distance, but relative to gas particles
it is not!
3. One effect of many collisions, in a small distance
Diffusion and Effusion
•
Diffusion: describes the mixing of gases.
The rate of diffusion is the rate of gas mixing.
Effusion: describes the escape of gas
through a tiny hole into a space of lower
pressure.
•
Relative Rates of Effusion
u1 = (3RT/M1)1/2 average velocity particle #1
u2 = (3RT/M2)1/2 average velocity particle #2
U1
U2
=
(3RT/M2)1/2
(3RT/M1)1/2
=
M2 1/2
Graham’s Law
M1
Practice
List the following gases, which are at
the same temperature, in the order of
increasing rates of diffusion.
O2, He, & NO
Practice
Calculate the molar mass of a gas if equal volumes
of oxygen gas and the unknown gas take 3.25 and
8.41 min, respectively to effuse through a small hole
under conditions of constant pressure and
temperature.
Practice
Calculate the molar mass of a gas if equal volumes
of oxygen gas and the unknown gas take 3.25 and
8.41 min, respectively to effuse through a small hole
under conditions of constant pressure and
temperature.
What are the units of molar mass?
Practice
Calculate the molar mass of a gas if equal volumes
of oxygen gas and the unknown gas take 3.25 and
8.41 min, respectively to effuse through a small hole
under conditions of constant pressure and
temperature.
What are the units of molar mass? g/mole
Practice
Calculate the molar mass of a gas if equal volumes
of oxygen gas and the unknown gas take 3.25 and
8.41 min, respectively to effuse through a small hole
under conditions of constant pressure and
temperature.
What are the units of molar mass? g/mole
U1
=
U2
M1
M2
1/2
Practice
Calculate the molar mass of a gas if equal volumes
of oxygen gas and the unknown gas take 3.25 and
8.41 min, respectively to effuse through a small hole
under conditions of constant pressure and
temperature.
What are the units of molar mass? g/mole
U1
=
U2
1
3.25
1
8.41
M1
M2
M1
= 32.0
1/2
1/2
Practice
Calculate the molar mass of a gas if equal volumes
of oxygen gas and the unknown gas take 3.25 and
8.41 min, respectively to effuse through a small hole
under conditions of constant pressure and
temperature.
What are the units of molar mass? g/mole
U1
=
U2
1
3.25
1
8.41
M1
M2
M1
= 32.0
1/2
1/2
8.41
3.25
1/2
M1
= 32.0
Practice
Calculate the molar mass of a gas if equal volumes
of oxygen gas and the unknown gas take 3.25 and
8.41 min, respectively to effuse through a small hole
under conditions of constant pressure and
temperature.
What are the units of molar mass? g/mole
U1
=
U2
1
3.25
1
8.41
M1
M2
M1
= 32.0
1/2
1/2
8.41
3.25
8.41
3.25
1/2
M1
= 32.0
1/2
32.0 = M1
Practice
Calculate the molar mass of a gas if equal volumes
of oxygen gas and the unknown gas take 3.25 and
8.41 min, respectively to effuse through a small hole
under conditions of constant pressure and
temperature.
What are the units of molar mass? g/mole
U1
=
U2
1
3.25
1
8.41
M1
M2
M1
= 32.0
1/2
1/2
8.41
3.25
8.41
3.25
1/2
=
M1
32.0
1/2
32.0 = M1
M1 = 51.5 g/mole
Real Gases
Must correct ideal gas behavior when at high
pressure (smaller volume) and low temperature
(attractive forces become important).
Deviations from Ideal Behavior
Real Gases
• van der Waals Equation
2
[ Pobs + a (n / V ) ] x (V - nb ) = nRT
corrected pressure
Pideal
corrected volume
Videal
ChemTour: Ideal Gas Law
Click to launch animation
PC | Mac
In this ChemTour, students manipulate the variables of the
ideal gas law to explore the relationship between the
pressure, volume, and temperature of a gas. Includes
worked examples and interactive Practice Exercises.
ChemTour: Dalton’s Law
Click to launch animation
PC | Mac
This ChemTour uses animation to explore Dalton’s law of
partial pressures.
ChemTour: Molecular Speed
Click to launch animation
PC | Mac
This ChemTour explores kinetic molecular theory.
Interactive graphs illustrate the concepts of kinetic energy
and root-mean- square speed.
The hydrostatic pressure on a diver’s
lungs is 1 atm at the surface and
increases by 1 atm for each additional
10 m below the surface.
Which of the following ascents poses the gravest
danger to a diver holding his or her breath?
A) 10 m → 0 m
Diving and Gas Pressure
B) 40 m → 20 m
C) 60 m → 40 m
Please consider the following arguments for each
answer and vote again:
A. The ascent from 10 m to 0 m is the most dangerous
because the twofold decrease in pressure causes the
greatest increase in lung volume.
B. The ascent from 40 m to 20 m is the most dangerous
because it involves a larger change in pressure.
C. The ascent from 60 m to 40 m is the most dangerous
because it occurs at the lowest depth.
Diving and Gas Pressure
Suppose 1 mole of an ideal gas is held
at a constant temperature (T). Which
of the following plots shows the correct
relationship between the product of
pressure and volume (PV) and the
pressure (P)?
A)
Ideal Gas Law: PV versus P
B)
C)
Please consider the following arguments for each
answer and vote again:
A. As the pressure increases, the product of the pressure
and the volume also should increase.
B. As the pressure increases at a constant temperature,
the volume should decrease so that the product PV
remains constant.
C. Pressure and volume are inversely proportional, so as
the pressure increases, the product PV should
decrease.
Ideal Gas Law: PV versus P
Suppose 1 mole of an ideal gas is held
at a constant temperature (T). Which
of the following plots shows the correct
relationship between the product of
pressure and volume (PV) and the
pressure (P)?
From the graph above: When P=10, V=1, PV 10: P=5, V=2,PV =10
A)
Ideal Gas Law: PV versus P
B)
C)
Suppose 1 mole of an ideal gas is held
at a constant temperature (T). Which
of the following plots shows the correct
relationship between the product of
pressure and volume (PV) and the
pressure (P)?
From the graph above: When P=10, V=1, PV 10: P=5, V=2,PV =10; Choice B
A)
Ideal Gas Law: PV versus P
B)
C)
An ideal gas at 10 atm and 50 ºC is
placed in a rigid cylinder fitted with a
pop-off valve set to open when the
pressure reaches 20 atm. How high
must the temperature be raised to open
the pop-off valve?
A) < 100 ºC
Ideal Gas Law: Pressure Valve
B) 100 ºC
C) > 100 ºC
Please consider the following arguments for each
answer and vote again:
A. Any ideal gas that is heated will expand and open the
valve.
B. If the temperature is increased by a factor of 2, the
pressure also will increase by a factor of 2 to 20 atm.
C. A 50 ºC increase in the temperature will only
produce a 15% increase in the pressure, which is not
enough to open the valve.
Ideal Gas Law: Pressure Valve
A reaction of 0.50 atm of H2 and 0.50
atm of O2 occurs in a sealed vessel at a
constant temperature to form H2O gas.
If the reaction goes to completion, what
will be the final pressure?
A) 0.5 atm
B) 0.75 atm
Pressure of Water Vapor Product
C) 1.0 atm
Please consider the following arguments for each
answer and vote again:
A. Because hydrogen is the limiting reactant, only 0.5
atm of H2O will be formed. So the final pressure
will be 0.5 atm.
B. Although all 0.5 atm of H2 will be converted into 0.5
atm of H2O, only 0.25 atm of O2 will be consumed.
Therefore, the final pressure will be 0.75 atm.
C. Because the total number of moles must be
conserved, the total pressure also must be conserved.
So the total pressure will remain at 1.0 atm.
Pressure of Water Vapor Product
Consider the reaction of calcium
carbonate, CaCO3(s), with perchloric
acid, HClO4(λ):
CaCO3(s) + 2 HClO4(λ) → Ca(ClO4)2(s) + CO2(g) + H2O(λ).
If there is initially 100 grams (1 mole)
of CaCO3, how many grams of HClO4
must be added to produce ~10 L of CO2
at STP (273 K and 1 atm)?
A) 50 grams
B) 100 grams
Production of CO From CaCO
C) 200 grams
Please consider the following arguments for each
answer and vote again:
A. 100 grams (1 mole) of CaCO3 is capable of producing
~20 L (~1 mole) of CO2, so only 50 grams (0.5 mole)
of HClO4 should be added.
B. 100 grams of HClO4 will react completely with 50
grams of CaCO3, producing ~10 L of CO2 at STP.
C. Two HClO4 molecules are required for every one
CaCO3 molecule, so 200 grams (2 moles) of HClO4 is
needed.
Production of CO From CaCO
The plot to the left shows the rootmean-velocity (vrms) of Br2 as a
function of the square root of the
temperature, assuming ideal behavior.
Which of the following plots shows the
correct relationship between vrms and T
for N2?
A)
B)
Temperature and Molecular Speeds of Ideal Gases
C)
Please consider the following arguments for each
answer and vote again:
It takes less energy (and hence a lower temperature) to
increase the velocity of Br2 molecules, so the slope of
the Br2 line should be steeper.
Since both Br2 and N2 are assumed to be ideal gases,
their slopes should be the same. Only when you
consider nonideal behavior will the slopes diverge.
N2 is a lighter molecule than Br2, and so at any given
temperature its root-mean-square speed is always higher
than that of Br2.
Temperature and Molecular Speeds of Ideal Gases
Shown to the left is a plot of the speed
distributions for two gases, X and Y, at
temperatures of 400 K and 800 K,
respectively. Which of the following
gases could be X and Y, respectively?
A) Br2 and Ar
B) 2H2 and 1H2
Distribution of Molecular Speeds
C) N2 and C2H4
Please consider the following arguments for each
answer and vote again:
A. Br2 is four times heavier than Ar, so the root-meansquare speed (vrms) for Br2 should be half that of
argon.
B. Deuterium gas (2H2) has twice the molecular mass of
1H and is at a temperature half that of 2H . As a
2
2
result, the vrms for 2H2 is a factor of 2 slower.
C. N2 and C2H4 have the same molecular mass.
However, because the temperature of C2H4 is twice
that of N2, the vrms is twice as fast.
Distribution of Molecular Speeds
One mole of metal, when added to 2
atm of chlorine gas (Cl2) in a 25-L
vessel at 25 ºC, reacts to form a metal
chloride, thereby decreasing the
pressure to ~1 atm. Which of the
following could be the unknown metal?
A) Na
Reaction of Metal with Cl Gas
B) Ca
C) Al
Please consider the following arguments for each
answer and vote again:
A. In a volume of 25 L at 25 ºC, 1 atm of a gas
corresponds to ~1 mole. The only metal that can
react in an equal molar quantity (1:1) with chlorine is
Na, which can form NaCl.
B. At 25 ºC, 1 atm of Cl2 in a 25-L vessel is ~1 mole.
Since 1 mole of Ca can react with 1 mole of Cl2 to
form 1 mole of CaCl2, the unknown metal must be
Ca.
C. Al will form AlCl3, using up three-fourths of the Cl2
and leaving an atmosphere of Cl atoms.
Reaction of Metal with Cl Gas
An unopened soda can expands when
left outside on a hot day. Based on this
observation, what can be concluded
about the dissolution of CO2(g) in
water?
A) It is exothermic.
B) It is endothermic.
Expansion of Soda Can upon Heating
C) It is isothermal.
Consider the following arguments for each answer and
vote again:
A. The expansion is due to the decreased solubility of
CO2(g) in water at higher temperatures, so the
dissolution of CO2(g) is exothermic.
B. Energy is always required to dissolve a solute molecule
in water, because to do so requires the breaking of
hydrogen bonds within the water.
C. The dissolution of a gas into a liquid corresponds to an
isothermal compression of the gas.
Expansion of Soda Can upon Heating
A single drop of water is injected into
a balloon at 25 °C that contains
ammonia, NH3(g), at 1.0 atm. What
will happen to the volume of the
balloon? Assume that the Henry’s
constant for NH3(g) is ~0.02 atm/M.
A) It stays the same. B) It increases. C) It decreases.
Injection of H O into a Balloon of NH
Consider the following arguments for each answer
and vote again:
A. A single drop of water takes up very little space and
can hardly affect the volume of a 1-L balloon.
B. Once added, the drop of water will immediately
vaporize to mix with the NH3(g), thus increasing the
pressure and hence the volume.
C. The pressure inside the balloon will drop as part of
the NH3(g) dissolves in the drop of water, thus
decreasing the volume of the balloon.
Injection of H O into a Balloon of NH