Chemistry Notes The Properties of Gases
I. Gases
A. 11 elements are gases under normal conditions (100's of compounds)
B. Most gases are molecular (H2, O2, NH3, SO2)
1. Exception = Noble Gases - He, Ne, Ar, Kr, Xe, Rn
C. Goals of chapter:
1. To understand the properties of the gaseous state.
2. To explain how to use reaction stoichiometry to predict the volumes of gases that take part in
reactions
a) Both are interrelated b/c the volume of a gas changes sharply with changes in
temperature and pressure and thus any predictions made about reacting volumes
must account for these factors.
D. Characteristics of Gases
1. The GAS LAWS
a) Gas - the state of matter that fills any container it occupies.
- (molecular definition) a gas is a collection of widely separated molecules in chaotic,
random motion.
b) characteristics
1) large distances between molecules - thus highly compressible
2) expanding gases are heated (when pressure & moles remain constant)
3) compressed gases are cooled (when pressure & moles remain constant)
4) low densities with units of g/L which is different from the gases which have g/mL or g/cm3.
5) expand to fill container
6) weak IMFA (although the ideal gas laws would suggest that there are no IMFA b/w gas molecules)
c) Our Atmosphere - Read about it.
1) Layers of our atmosphere and the importance of the statosphere.
a) changes in the ozone
b) affects of CFC's on ozone
2) Global Warming:
a) Changes in [CO2] and its affect on the Global Warming
FORCE
d) Pressure (P) - the force exerted per unit area P = AREA
1) fast moving gas molecules exert a pressure on the object containing it as the molecules collide
with the surface.
a) > # gas molecules, > # collisions, > pressure
2) Units of pressure
Torr
atm
Pascal (Pa) = SI unit 1 kg/(m.s2) = 1 N/m2
kilopascal (kPa) 1x103 Pa
a) conversions
1 atm = 760 Torr (exactly) = 760 mmHg
1 Torr = 1 mmHg
1 atm = 101.3 kPa
1 atm = 1.013 bar
1 bar = 100 kPa
1 atm = 14.7 lb/in2 or psi
1 kPa = 1 x 103 Pa
3) barometer measures atmospheric pressure.
Normal atmospheric pressure at sea level = 760 mm Hg or = 760 Torr
a) Torr – unit named in honor of Evangelista Torricelli who invented the mercury
barometer in 1643.
4) Chemists have found it convenient to use the unit called the atmosphere (atm)
1 atm = 760 Torr
a) Problem 500 Torr = ? atm of pressure
P = 500 Torr x 1atm = 0.658 atm
760 Torr
E. Four Properties or Variables used to describe gas
1. Pressure (mainly use atm)
2. Volume (use L or mL)
o
3. Temperature (convert to K) C + 273 = K
4. Moles (symbol is “n”)
a. Avogadro's Law: different gases with equal volumes under the same conditions (temp. &
pressure) have the same number of molecules (particles, moles).
1) Standard molar volume Vm of a substance (any substance not only a gas) is the volume it
occupies per mole of molecules
Vm = volume occupied = V
moles of molecules n
2) There is a direct relationship b/w volume and quantity of a gas such that:
when quantity of gas (moles) increases, volume increases
3) at STP, for most gases:
a) standard molar volume = 22.4 L/mol of gas
b) (roughly the volume of a 1 foot sided cube)
F. To study the effects of changing temperature and pressure on a gas, scientists have a standard for
comparison.
1) Standard Temperature and Pressure (STP) – the conditions under which chemist report molar
Volume which are: 0oC or 273 K (really 273.15 K)
and 1 atm pressure.
STP = 273.16 K and 1 atm
STP = 273.16 K and 760 torr
STP = 273.16 K and 760 mmHg
STP = 273.16 K and 101.325 kPa
2) Problem: Calc. the volume occupied by 10. g of CO2 at STP by using the typical molar volume of
22.4 L/mol.
10 g CO2 x 1_mol_CO2 x 22.4_L = 5.2 L CO2 at STP
44.01 g CO2 1 mol CO2
3) Problem: Calc. the vol. occupied by 1.0 kg of Hydrogen at STP. 1.1 E4 L
1.0 kg H2 x 1000g_H2 x _1_mol_H2 x _22.4L_H2 = 1.1 E4 L H2
1 kg H2 2.02 g H2 1 mol H2
G. IDEAL GAS
1. In studying gases, scientist have found that most gases follow all the laws that are about to be
studied at room temp and pressure (which is near that of 1 atm.)
2. However, as we learned more about the properties of gases at high pressures and extremely low
temperatures, scientists found that gases did NOT follow the gas laws.
3. SO a model was developed to address the "IDEAL GAS" that would follow all the gas laws under
all conditions.
IDEAL GAS ACTUAL
a) Individual gas particle is so small that its V = 0 Individual gas particles have a measurable
volume. We know b/c when they condense, the
liquid has volume.
b) Individual gas particles are in constant, Particles not necessarily moving in straight lines
rapid, random motion, moving in straight Speeds and pathways may vary due to
lines until they collide w/ other particles or attractions or repulsion
the walls of the container
c) COLLISIONS are ELASTIC Collisions not elastic
(no loss or gain of NRG)
IDEAL GAS ACTUAL
d) No attractive or repulsive forces b/w particles van der Waal forces do exist when gases are
are compressed & particles are close to each
other. (electrical attract)
e) Aver. KE is directly prop. to absolute temp. Not directly prop at extremely low temp. & high
in K pressure.
& the product of P x V is exactly proportional
to absolute Temp.
f) Does not condense even if compressed and condenses at high pressure & low Temp.
cooled
II. The Ideal Gas
Experiments have shown that the pressure, temperature and volume of a fixed gas all depend on one
another and a change in affects the others.
A. Boyle’s Law P 1
V
1. At constant temp. for a given amount of gas, the volume is inversely proportional to the pressure.
a) an increase in one results in a decrease of the other (always think of an inner tube)
b) Ex. volume 2x pressure ½ x
Ex. pressure 3x volume 1/3 x
c) To make the proportion into an equality we need a constant (lower case b/c it is specific to this
system). V = k x 1 where k is the constant
P
d) Graphing y = mx + b where b = y intercept at the origin
where x = 1/P
where m = constant or slope = k
V m=k or V
1/P P
e) Boyle’s Law can be tested by plotting V against 1/P for any gas resulting in a straight line in each
case. With temperature kept constant, each set of data points should result in the same value
for the constant “k”. (Conditional: at or below 1 atm pressure)
ex. P1V1 = k P2V2 = k P3V3 = k etc.
f) Since every PxVx = k, then we can say P1V1 = P2V2 in solving Boyle’s Law problems
-What pressure is needed to compress a gas occupying 100 L at 100 kPa to 10 L?
P1V1 = P2V2 100 L x 100 kPa = P2 x 10 L
1 E4 L.kPa = 1 E3 kPa
1 E1 L
-What pressure is needed to confine a sample of argon in a 300 mL container if it occupies 500 mL
at 750 Torr P. (ans: P1V1 = P2V2 500mL x 750 Torr = 300 mL x P 2 1250 Torr = P2)
B. Charles’ Law - At constant pressure, the volume of a sample of given sample of gas is proportional to
the absolute temperature.
1. If the volume increases, the temperature increases proportionally.
2. Double the temp = doubling of the volume
3. To make the proportion into an equality use a constant. V = constant (k) x T
4. Graphing y = mx + b where b = y intercept at the origin
where x = T
where m = constant or slope = k
m=k
V With pressure kept constant, all Vx/Tx values = k
T
a) Plotted points of V vs T show a straight line which run through the origin (0,0) at absolute zero
(-273.15oC)
5. Use degrees Kelvin!!!
a) to convert from degrees C to Kelvin simple add 273.
1) Ex. 25oC = 25 + 273 or = 298 K
6. Problem V1 = V2
T1 T2
a) Calc. the vol. that a sample of air would occupy at 40oC if it occupies 1.00 L at 20oC at
constant pressure.
V1 = V2 1.00_L = __x___
T1 T2 40 + 273 20 + 273
1.00_L_x_293_K = 1.07 L
313 K
b) What temp. should a sample of gas be cooled from 25oC to reduce its volume to half its
initial value at constant pressure? Answer = 149 K or -124oC
7. We can see that Volume and Temp. are DIRECTLY PROPORTIONAL one factor goes down, so
must the other!
C. NO NAME LAW PT
P1 = P2 At constant volume, a gas at 23oC and 0.989 atm has it temp. raised to 62oC.
T1 T2 What is the new pressure? 1.12 atm.
The graph looks the same as with Charles’ Law except swap the V with P for pressure and k = Px/Tx
D. The COMBINED GAS LAW - combining Charles’ and Boyle’s Law we get:
V = constant x T or PV = constant
P T
b. Problems use P1V1 = P2V2
T1 T2
Ex. a diver exhales a 15.0 mL bubble of air at a depth where pressure is 12.0 atm and the temp. is
8.0oC. What is the volume of the bubble at the surface where the atmospheric pressure is 770
Torr and the temp is 20oC.
P1V1 = P2V2 1 atm = 760 Torr 9120Torr_x_15mL = 770Torr_x_V2
T1 T2 12 atm = 9120 Torr 8 + 273 20 + 273
9120_x_15_x_293 = V2 = 185 mL
770 x 281
Ex. A sample of methane gas is stored in a 227 mL flask at 600 Torr and 10.0 C. The gas is allowed
to expand into another flask of 1.78 L at 27.0 C. What is the pressure of methane in the second
flask? Ans. 81.1 Torr
E. IDEAL GAS LAW
1. How do you calculate the volume when the conditions are NOT at STP? OR
2. How can you determine the properties of a gas when the conditions do not change?
You CANNOT use the 22.4 L per mol. You must solve for volume using the Ideal Gas Law which
states that PV = nRT.
a) P (pressure can be in atm or kPa ) is given or you may have to solve for this factor.
b) V (volume in L - must be L) is given or you may have to solve for this factor.
c) n = # moles of the gas. (may be calculated from mass) is given or you may have to solve for
this factor.
d) R is proportionality constant or GAS CONSTANT which is the same for all gases and equals
0.08206 atm . L if you are using atm OR 8.314 kPa . L if you are using kPa
mol . K mol . K
e) T (temperature in K – must be in Kelvin) is given or you may have to solve for this factor.
3. NOTICE that this law does not involve a change in the conditions of the characteristics of a gas the
as is the case with the Combined Gas Law.
4. The ideal gas law is approximately true for all gases under normal lab conditions which means
room temperature and 1 atm).
a) The physical explanation for why gases behave nearly ideally under normal conditions is that
their molecules move almost independently of each other due to the very large distances
between the molecules. The interaction between the molecules is very weak.
b) As the pressure is lowered, the gases interact even less significantly. More and more gases
obey the ideal gas law as the pressure is reduced.
c) In a completely ideal gas, the molecules move completely independently of one another
except during collisions.
5. A gas that obeys the law is called an "IDEAL GAS"
6. NO real gases are ideal at all temperatures & pressure but we usually assume that they are ideal for
the calculations that we will do.
7. When solving problems, set the problem up first and then solve for the unknown value.
8. You can solve for P, V, T or n using P = nRT or V = nRT or T = PV or n = VP
V P nR RT
F. Density - Mass/ Volume (units for gases = g/L)
1. d = P M where M = molar mass
RT
2. Density is directly proportional to the molar mass of a substance
3. Use the ideal gas law to solve for the number of moles (n) and then divide that value into the
number of grams of a substance to calculate its molar mass
g gas = Molar mass
n (# moles of gas)
III. Dalton's Law of Partial Pressure - states the total pressure in a gas mixture is the sum of the partial
pressures of the individual components.
a. Partial Pressure - is the pressure of an individual gas in a gas mixture that contributes to the total
pressure of the mixture.
b. Formula: Ptotal = P1 + P2 + P3 + etc.
c. Most often used to calculate the pressure of a gas that was collected over water.
- First carry out a chemical reaction that will generate a gas in a gas Ptotal =
measuring tube (or eudiometer) which is inverted in a beaker of Pgas from the rxn
water as seen in the diagram to the right. + Pwater vapor
- Water is generally ideal for this type of reaction because many gases are G at XoC
NOT very soluble in water. a
- As the reaction takes place, gas rises upward into the tube and displaces s
the water which originally filled the tube.
- Depending on the temperature and the position of the gas in the gas
measuring tube in relation to the level of the water in the bucket, the
total pressure of the Dry gas and the water vapor can be determined.
- Pinside the column = Poutside the column
Pinside = Pgas from the rxn + PH2O vapor When the level of the gases in the Reaction producing
tube is at the same level as the water in the beaker, then … a gas in the water.
Pgases inside the column = Poutside the column
- If the column is above the water level in the beaker, then the pressure inside the column is
the combined total of the PDry gas + PH2O vapor + PH2O liq/13.6
- Pinside = Pcollected gas, + PH2O vapor + PH2O in column / 13.6
Read from chart Measured in mmH2O and converted to mmHg by
Since pressure outside below or on pg dividing the height by 13.6 (which is the density
equals the pressure inside 376 of text. difference between mercury and water)
use the atmospheric pressure
Then solve for P of the collected gas (also called the dry gas)
Temperature (oC) Pressure H2O (kPa) Temperature (oC) Pressure H2O (kPa)
0 0.61 55 15.75
5 0.87 60 19.93
10 1.23 65 25.02
15 1.71 70 31.18
20 2.34 75 38.56
25 3.17 80 47.37
30 4.25 85 57.82
35 5.63 90 70.12
40 7.38 95 84.53
45 9.59 100 101.32
50 12.34 105 120.79
d. Problem:
Hydrogen gas is collected over water at a total pressure of 175.93 kPa. The volume of hydrogen
gas collected is 28mL at 50oC and the gas is at the same level as the water it was collected over.
What is the partial pressure of the hydrogen gas?
Ptotal = Pwater vapor + Pgas or
Ptotal - Pwater vapor = Pgas
175.93 kPa - 12.34 kPa (at 50oC) = PH2 answer = 163.59 kPa
Problem: At 20oC and 745.0 mm Hg air pressure, 12.0 mL of H2(g) is collected by water
displacement. The liquid level inside the gas-measuring tube is 95 mm higher than that outside
PH2 + PH2O vapor + PH2O liq in column/13.6 = Poutside
PH2 + 2.34 kPa (or 0.0231 atm) + 95mmH2O/13.6 (or 0.00919 atm) = 745.0 mmHg (or 0.980 atm)
PH2 = 0.947 atm
Practice: What is the partial pressure of carbon dioxide gas (in kPa) that is collected over water at 45 C when the
o
total pressure of the system is 0.9790 atm with no water in the column? (Be careful with the pressure units)
(answer: PCO2 = 89.58 kPa)
Practice: At 40oC, 89.0 ml of fluorine gas is collected in a eudiometer in which the water level was 184 mm
above the outside level. The barometric pressure was 103.5 kPa. a) What is the PF2 in atm? b) How many moles of F2
formed?c) How many grams of F2 formed? [ answers: a) PF2 = 0.93 atm, b) 0.0032 mol F2, c) 0.12 g F2 ]
IV. Mole Fraction – is the number of moles of a particular gas compared to the total number of moles in a
mixture. It is very useful in determining the partial pressure of the gas in the mixture.
Problem. A) If there are 5 moles of hydrogen gas in a mixture that contains a total of 25 moles of gas,
the mole fraction of hydrogen gas is 20%. B) If the total pressure is 60.kPa, what is PH2?
A) 5 moles of H2(g) x 100 = 20 % B) 20% or 1/5 of 60. kPa = 12 kPa
25 moles of gas mixture
Problem A) If a mixture of a gas has 3 moles of ammonia gas and 6 moles of sulfur dioxide gas, then
what is the mole fraction of the ammonia gas? B) What is PNH3 if PTotal = 780. Torr?
A) 3 moles of NH3 x 100 = 33% B) 780. Torr x 0.33 = 260. Torr = PNH3
9 moles of gas in the mixture
V. Manometer – Lab equipment that is used to determine the pressure of a confined gas when compared to the
outside atmospheric pressure.
gas
The air pressure outside is greater than the confined gas
pressure. In fact, it is 20 mmHg greater than the confined gas
pressure. If the outside pressure is 1 atmosphere or 760 mmHg,
120 mm then the confined gas pressure = 740 mmHg or 0.974 atm.
100 mm
VI. Diffusion – the process by which particles disperse from regions of higher concentration to regions of lower
concentration.
a. Movement of gas molecules in the system continues until equilibrium is reached.
b. Equilibrium – state in which there are equal concentrations of a substance throughout the system.
VII. GRAHAM's LAW
a. Based on the idea that different sized gas molecules move at different rates.
b. Lighter molecules diffuse more quickly than heavier molecules.
c. EFFUSION - the motion of a gas or gases through an opening
d. Lighter gases effuse faster than heavier gases.
e. DEMO:
Cotton ball soaked Cotton ball soaked
Draw in where the ring forms?
in concentrated NH3 in HCl (conc)
f. Graham’s law can be derived from the formula: KE = ½ mv2
-If two gases (a & b) are at the same temperature, then KE is the same for both so,
½ mav2a = ½ mbv2b
va2 mb
-Thus the ½’s cancel and we rearrange the formula vb2 = ma take the square of both
va mb
vb = ma
g. Problems: (1)
Rate Xfaster/lighter = Molar Mass Y Ex. Smaller/Faster X is N2 = 28.02 g/mol
Rate Y slower/heavier Molar Mass X Larger/Slower Y is CO2 = 44.01 g/mol
X or N2 is 1.25 times faster than Y or CO2
(2) An unknown gas effuses 1.366 times slower than fluorine gas. Identify it.
(1.366)2 * 38.00 g/mol = Y Y = 70.91 g/mol = Cl2
(3) An unknown gas effuses 5.729 times faster than xenon gas. Identify it.
(5.729)2 * X = 131.29 g/mol X = 4.00 g/mol = H2
VIII. Reaction Stoichiometry using Gas Volumes
a. Gay Lussac's Law of Combining Gases - Volumes of gases combine in small whole # ratios
1. Ex. N2(g) + 3H2(g) = 2NH3(g)
1mole 3 moles 2 moles
2. Solving problems: Determining the volume of one gas based on the volume of another gas using
reaction stoichiometry and mole to mole conversions.
a) For the reaction above, how many liters of ammonia gas will be produced if 6.5 L of hydrogen gas
react with excess nitrogen gas at STP?
Answer: 6.5 L H2 x 1 mol H2 x 2 mol NH3 x 22.4 L NH3 = 4.3 L NH3
22.4 L H2 3 mol H2 1 mol NH3
Remember you can determine the Volume when the conditions are at STP by using
22.4 L / 1 mole of a gas.
b. More Stoichiometry and Gases
a) How many grams of NaN3 , sodium azide, the component used in air bags, is needed to fill an air bag
with of 45.5 L of nitrogen gas at a pressure of 828 mm Hg and a temp. of 22.8 oC based on the
reaction:
2NaN3(s) 2Na(s) + 3N2 (g)
Answer: n = PV P = 828 mm Hg x 1 atm = 1.89 atm
RT 760 mmHg
V = 45.5 L
R = 0.08206 atm L / mol K
T = 22.8 oC + 273.2 = 295 K
n = 2.05 mol N2
2.05 mol N2 x 2 mol NaN3 x 65.02 g NaN3 = 88.9 g NaN3
3 mol N2 1 mol NaN3
c) Find the Empirical Formula for Crx(CO)y based on the following reaction and information
Crx(CO)y x Cr + y CO
And knowing that 0.112 g of Crx(CO)y decomposed to form the 155 mL CO gas formed under the
conditions of 369 mmHg and 27.0oC.
Answer: n = PV P = 369 mm Hg x 1 atm = 0.486 atm
RT 760 mmHg
V = 155 mL x 1L = 0.155 L
1000 mL
R = 0.08206 atm L / mol K
T = 27.0 oC + 273.2 = 300.2 K
n = 3.06 E-3 mol CO
3.06 mol CO x 28.01 g CO = 0.0857 g CO
1 mol CO 1 mol NaN3
0.112 g Crx(CO)y
- 0.0857 g CO
0.0263 g Cr
0.0263 g Cr x 1 mol Cr = 0.000506 mol Cr / 0.000506 mol Cr = 1
52.00 g Cr
0.0857 g CO x 1 mol CO = 0.00306 mol CO / 0.000506 mol Cr = 6
28.01 g CO
Thus, the empirical formula = Cr(CO)6
IX. Conditions at Which a GAS CONDENSES.
A. Forces of Attraction
1. Normal gases do exert forces of attraction/repulsion when they are near to each other under
conditions of:
a. high pressure
b. low temp.
2. At low temp., gas molecules slow down.
3. At high pressure, gas molecules are close enough to each other that the IMFA affect the
direction/motion of the gas molecules.
4. If the pressure is high enough & the temp. low enough, the attractive forces b/w the gas
molecules/atoms will cause the gas to condense.
B. Vapor Pressure
1. Common gas that condenses is water vapor. (on cool surfaces)
2. Equilibrium Vapor Pressure - is the measure of the tendency of the particles of a liquid substance
to enter the gas phase at a given temp.
3. As the Temp. of water increases, its maximum vapor pressure increases.
4. When the vapor pressure = atmospheric pressure, liq boils.
5. Boiling Point - a) the temp. at which a liq. boils.
b) the temp. at which vapor pressure = atm pressure (or external pressure)
C. Volatile Substances
1. Volatile - a term used to describe a substance that is readily vaporized at low temp.
a. gasoline, turpentine, purfume, moth balls, acetone.
2. These substances that vaporize easily have weak IMFA thus, they evaporate quickly.
3. Non volatile substances have stronger IMFA which keep them as liquids. (thus low vapor pressures.)
D. Phase Diagram relate Temp and Pressure to the physical state of matter.
1. Notice that if the pressure is greater than 1 atm, water boils at a higher temp.
2. Notice that if the pressure is less than 1 atm, water boils at a lower temp. (in the mountains.)
3. Notice that water can be a solid with temperatures above 0oC if pressure is very low.
4. Notice that water can be a liquid with temperatures below 0oC if pressure is very high
5. Notice that water can be a liquid with temperatures above 100oC if pressure is very high
See Diagram on Next Page
Phase Diagram or Water
218
Liquid
Solid
Pressure in atm
Gas
1.00
0.0060 -
273 373 647
Temperature in K
On the diagram above, label the following terms and explain what each means.
A. Solid Phase
B. Liquid Phase
C. Gaseous Phase
D. Melting Point Curve
E. Vapor Pressure Curve
F. Sublimation Curve
G. Triple point
H. Normal melting point
I. Normal boiling point
J. Critical Point
K. Critical Temperature