UNIT VI REVIEW – PHASES & GAS LAWS
I. Good Stuff to Know
 definitions of solid, liquid, gas, fluid, condensed state
 Pressure - relationship between force and area
 barometer - atmospheric pressure, air pressure, barometric pressure
 standard temperature and pressure (STP): 1 atm = 76θ Torr = 76θ mmHg = 101.3 kPa and 273K
 manometer
 vaporization – boiling (how does it relate to vapor pressure?), evaporation
 amorphous solid- example
 crystal lattice – crystals
 density of solids, liquids, gases - water (most dense at 4oC)
 intermolecular forces: dipole-dipole interactions, London Dispersion Forces, hydrogen bonding, ionic crystals,
network solid. metallic crystals
 Phase Diagrams – melting, freezing, boiling, condensation, sublimation, deposition
 Phase Diagrams - triple point, critical temperature, unique properties of water
 substances that exist as gases
 Avogadro’s Law
 Kinetic Theory  use it to explain Boyle’s Law, Charles’ Law, Avogadro’s Law, Dalton’s Law,
compressibility of gases
 root-mean-square speed, mean free path
 diffusion vs. effusion
 deviations from ideal behavior, vanderWaal’s equation
II. Multiple Choice
For questions 1-4, select from the following answers
a. metallic bonding
b. network covalent bonding
d. ionic bonding
e. London Dispersion Forces
a
c
e
d
c. hydrogen bonding
1. Solids exhibiting this kind of bonding are excellent conductors of heat.
2. This kind of bonding is the reason that water is more dense than ice.
3. All of the inert gases are held together with this weaker type of bonding.
4. This type of bonding results in solids that are poor conductors of heat and electricity, but which,
when melted, are good conductors of electricity.
Questions 5-8 refer to the following phase diagram:
B
5. At this point the substance represented by the phase diagram will be solely in the solid phase at
equilibrium.
D
A
C
6. This point represents the boiling point of the substance.
a
7. At this point, the substance represented by the phase diagram could be undergoing sublimation.
8. At this point the substance represented by the phase diagram will be solely in the liquid phase
at equilibrium.
9. Which of the following lists of species is in order of increasing boiling point?
a. H2 , N2 , NH3 b. N2, H2, NH3 c. NH3, H2, N2 d. NH3, N2, H2 e. H2, NH3, N2
UNIT VI REVIEW – PHASES & GAS LAWS
III. Essay
1. Explain why the boiling point of argon is -186oC, but the boiling point of neon is -246oC.
Both are nonpolar, therefore both exhibit London Dispersion Forces, but because the
formula mass of argon is greater, it has more electrons, therefore the dispersion forces are
stronger, making it more difficult to separate the atoms, therefore causing the boiling point to
be greater.
2.
The phase diagram for a substance is shown above. Use the diagram and your knowledge of phase changes to
answer the following questions.
a. Describe the phase change that the substance undergoes as the temperature is increased from
points B to C to D at constant pressure.  MELTING
b. Could this phase diagram represent water? Why or why not?
No. as you increase the pressure from point D (liquid) the state will change
from liquid to solid. Because increasing the pressure makes the molecules become more
compressed that means that the density is increasing under greater pressure. The phase
diagram for water, in which the liquid state is more dense than the solid state has the border
between liquid and solid (line CD on this diagram) slanted the other direction.
IV. Problems
1) What is the pressure of an object which exerts 45.6 N of force over an area of 0.00245 m2?
P
2) 143.2 kPa = ? atm
143.2kPa 
1atm
 1.41atm
101.3kPa
F
45.6 N

 18,600 Pa
A 0.00245m 2
3) 629 Torr = ? kPa
629Torr 
101.3kPa
 83.8kPa
760Torr
4) 0.802 atm = ? Pa
0.802atm 
5) What is the pressure of the gas in the enclosed container?
P = 1.27 atm
243 mm
1.27 atm 
760mmHg
 965mmHg  965  243  1208mmHg
1atm
101.3kPa 103 Pa

 81,200 Pa
1atm
1kPa
UNIT VI REVIEW – PHASES & GAS LAWS
6) Calculate the root mean square velocity of the following molecules in a sample of the following gases at
273K and 546 K.
3RT
3(8.31)( 273K )

 652 m s  922 m s @ 546 K
3 kg
M
16.0 x10
mol
a. CH4 u 
b. N2
3RT

M
3(8.31)( 273K )
 493 m s  697 m s @ 546 K
3 kg
28.0 x10
mol
* Note the N2 is slower because it is heavier
u
7) Consider a 1-mol sample of CH4 gas at 273K. If the temperature is increased, what happens to the average
kinetic energy and the average velocity of the CH4 molecules? Explain.
Both values will increase. Temperature is directly proportional to AKE, so the AKE
will increase with increased temperature. KE= ½mv2, so since mass (m) is the same,
as KE increases, the velocity of the particles will increase.
8) Consider three identical flasks filled with different gases:
Flask A: CO at 760 Torr and 0oC
Flask B: N2 at 250 Torr and 0oC
Flask C: H2 at 100 Torr and 0oC
all the same a.
C
In which flask will the molecules have the greatest average kinetic energy?
b. In which flask will the molecules have the greatest average velocity?
* Since T is the same, the lightest molecules will be the fastest
9) A gas occupies a volume of 34.2 mL at a temperature of 15.0oC and a pressure of 800.0 Torr. What will the
volume of this gas be at STP?
15.0 + 273 = 288K
P1V1 P2V2
(800.0Torr )(34.2mL) (760mmHg )V2



 V2  34.1mL
T1
T2
288K
273K
10) A dry gas has a volume of 100.0 cm3 at a pressure of 1600 Torr. At what pressure would this volume be
reduced to 50.0 cm3? Assume temperature is constant.
P1V1  P2V2  (1600Torr )(100.0mL)  P2 (50.0mL)  P2  3200Torr
11) What is the relative rate of diffusion of oxygen (O2) gas vs. xenon gas (Xe)?
rr 
FM Xe
131.3

 2.03  Oxygen diffuses faster
FM O2
32
12) Find the formula mass of a gas which diffuses at a rate 1.16 times faster than that of sulfur dioxide gas.
rr 
FM SO2
FM x
 1.16 
64.1
64.1
 1.3456 
 x  47.6 g mol
x
x
UNIT VI REVIEW – PHASES & GAS LAWS
13) The volume of a gas originally at STP was recorded at 488.8 mL. How many moles of the gas are present?
488.8mL  0.4888L 
1mol
 0.0218mol
22.4 L
14) If you had a 3.33 L container with 0.400 mol of a gas at a temperature of -2θoC, what pressure is the gas
exerting?
PV  nRT  P 
nRT (0.400mol )(0.0821)( 253K )

 2.50atm  253kPa
V
3.33L
15) At conditions of 785 kPa of pressure and 15.0oC, a gas occupies a volume of 20θ mL. Assuming the container
is rigid, what pressure will the gas be under if you drop it in the fireplace (650oC - hot!!)?
P1 P2
785kPa
P



 P  2520kPa
T1 T2
288 K
923K
16) 40.0 mL of helium gas is collected over water at 20.0oC. If this gas exerts a pressure of 790.0 Torr, what would
the volume of the dry gas be at STP? (Remember this is a mixture of two gases: water vapor and helium)
[Pwater = 17.54 mmHg at 20.0oC
PHe  790.0  17.5  772.5mmHg
P1V1 P2V2
(772.5mmHg )( 40.0mL) (760Torr )V



 V  37.9mL
T1
T2
293K
273K
17) What is the density of oxygen gas collected at 21.0oC and 103.5 kPa?
FM O2  32.0 g / mol  assume you have 1 mol (32.0g) of oxygen :
V 
nRT (1mol )(8.31)( 294 K )
m 32.0 g

 23.6 L  D  
 1.36 g / L
P
103.5kPa
V 23.6 L
18) If you collected 0.0355 g of a gas which occupied a volume of 45.0 mL at a temperature of 2θoC, and a pressure of 0.926 atm, what would the formula mass of the gas be?
FM 
mRT (0.0355 g )(0.0821)( 293K )

 20.5 g mol
PV
(0.926atm)(0.0450 L)
19) A chemist weighed out 5.14g of a mixture containing unknown amounts of BaO (s) and CaO(s) and placed
the sample in a 1.50 L flask containing CO2(g) at 30.0oC and 75θ Torr. After the reaction to form
BaCO3(s) and CaCO3(s) was completed, the pressure of CO2(g) remaining was 23θ Torr. Calculate
the mass percentages of CaO(s) and BaO(s) in the mixture.
Still unsure. Try it one more time. See notes…