Chapter.10 Gases
Soo Jeon
10.1 Characteristics of Gases
Gases
 are composed entirely of nonmetal elements.
 expand spontaneously to fill its container( volume
of gas = volume of its container).
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are highly compressible.
form homogeneous mixtures with each other
regardless of the identities.
molecules are far apart.
different gases behave similarly.
10.2 Pressure
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Pressure conveys the idea of a force, a push
that tends to move something else in a given
direction.
Pressure = Force / Given area
Force(Newton) = Mass * Acceleration)
Pressure = 1*10^5 N/1m^2 = 1*10^5N/M^
= 1*10^5Pascal = 100 kpa = 1 bar
Blaise Pascal(1623-1662) : French Scientist,
discovered pascal(N/m^2)
Standard temperature and
pressure
STP : Typical pressure at sea level.
 At STP:
1) Pressure = 1atm = 760mmHg =760 torr
= 1.01325*10^5 pa =101.325 kpa.
2) 1 mole of gas occupies 22.4 liters.

Tools for measuring atmospheric
pressure
1)
Mercury barometer : height of mercury column
changes as the atmospheric pressure changes.
2)
Manometer : measures the pressure of enclosed
gases. Difference in the heights of mercury levels in
the two arms of the manometer relates the gas
pressure. If the pressure of enclosed gas is less than
atmospheric pressure, the mercury will be higher in
the arm exposed to the enclosed gas. Pgas = Patm +
P (difference in height of arms)
Problems 1
1.
Convert
a) 0.357atm to torr.
b) 6.6*10^-2 torr to atm.
c) 147.2kPa to mmHg.
Solution 1
a)
b)
c)
(0.357atm)(760torr/1atm) = 271 torr
(6.6*10^-2torr)(1atm/760torr) =
8.7*10^-5atm
(147.2kPa)(760mmHg/101.325kPa)
= 1104torr
10.3 The Gas Laws
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Boyle’s Law( the pressure-volume Relationship) :
when a volume of gas is compressed, the
pressure of gas increase. The volume of a
fixed quantity of gas maintained at constant
temperature is inversely proportional to the
pressure.
British Chemist Robert Boyle(1627-1691) :
first investigated the relationship between
pressure of gas and volume.
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V= constant * (1/p)
PV = Constant
Charles’s Law (temperature –
volume)
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Charles’s Law : the volume of a fixed
amount of gas maintained at constant
pressure is directly proportional to its
absolute temperature.
French Scientist Jacques Charles
(1746-1823): found that volume of
fixed quantity of gas at constant
pressure increases linearly with
temperature.
`
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In 1848 William Thomson(1824-1907)
proposed an absolute temperature
scale, Kelvin.
Absolute zero = -273.15°C.
Avogadro’s Law
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In 1808, Gay Lussac(1778-1823)
observed the law of combining volumes;
at a given pressure and temperature,
the volumes of gases that react with
one another are in the ratios of small
whole numbers.
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Avogadro interpreted Lussac’s
observation and concluded.
Avogadro’s law : The volume of a gas
maintained at constant temperature and
pressure is directly proportional to the
number of moles of the gas.
Volume = constant X n
STP; 22.4L, 0°C, 6.02*10^23 molecules,
1atm
10.4 The Ideal-Gas Equation
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Ideal-gas equation:
Pressure*Volume= #of moles*
R(0.0821L-atm/mol-K)* temperature
Ideal gas is STP.
The ideal-gas equation does not always
accurately describe real gas.
P1/T1=P2T2, P/T=nR / V,
P1V1/T1 = P2V2/T2
Problem 2

An ideal gas is contained in a 5.0L
chamber at a temperature of 37°C. If
the gas exerts a pressure of 2.0atm on
the walls the chamber, what expression
is equal to the number of moles of the
gas?
Solution 2
n = PV/RT
= (2.0atm)(5.0L)/(0.0821L-atm)(310K)
moles
Problem 3
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A 0.5 mol sample of oxygen gas is
confined at 0°C in a cylinder with a
movable piston. The gas has an initial
pressure of 1.0atm. The gas is then
compressed by the piston so that its
final volume is half the initial volume.
The final pressure of gas is 2.2atm.
What is the final temperature?
Solution 3
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= 1atm* XL/ 273K
= 2.2atm * .5XL/temperature
X/273 = 1.1X/temperature
= 300.3K = 27°C
10.5 Further Applications of the
Ideal-gas equation
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Gas densities and molar mass(M)
nM/V = PM/RT
moles/Liter * grams/mole = grams/liter
= density(g/L)
M = dRT / P
Density = PM/RT
1) higher the molar mass and pressure, the
more dense the gas
2) higher temperature, less dense gas
Problem 4
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Molar mass is 28.6g/mol, temperature is
95K, and the pressure is 1.6atm.
Calculate the density
Solution 4
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D = PM/RT =
(28.6g/mol)(1.6atm)/(0.0821Latm/mol-k)(95K) = 5.9 g/L
Problem 5
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Find the molar mass of an unknown gas. First,
a large flask is evacuated and found to weigh
134.567g. It’s then filled with the gas to a
pressure of 735torr at 31° and reweighed. Its
mass is now 137.456g. Finally, the flask is
filled with water at 31°C and found to weigh
1067.9g.(the density of the water at this
temperature is 0.977g/ml.) calculate the
molar mass of unknown gas.
Solution 5
The mass of the gas : 137.456g – 134.567g = 2.889g
The mass of water : 1067.9g – 134.567g = 933.3g
Volume of flask = m/d = (933.3g)/(0.997g/ml) = 936 ml
So, the density of the gas = 2.889g/0.936L = 3.09g/L
Molar mass of the gas = dRT/ P
= (3.09g/L)(0.0821L-atm/mol-K)(304K)/(735/760)atm
= 79.7 g/mol
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Problem 6
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Volumes of Gases in Chemical Reactions
Ammonia reacts with oxygen gas at 850°C and
5.00atm in the presence of a suitable catalyst. The
following reaction occurs:
4NH3(g) + 5O2 → 4NO(g) + 6H20(g)
How many liters of NH3 at 850°C and 5.00atm are
required to react with 1.00mol of 02 in this reaction?
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Solution 6
850°C = 1123K, 5atm, 1 mol of O2
 V= nRT/P =
(1mol of O2)(0.0821L-atm/mol-)(1123K)/
(5atm) = 18.4 L
18.4L of O2 * (4 mol NH3)/(5 mol O2)
= 14.8L of NH3
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10.6 Gas Mixtures and Partial
Pressures
John Dalton(1766-1844), English
chemist, tells us that the total pressure
of mixture of gases is just the sum of all
the partial pressures of the individual
gases in the mixture.
 Dalton’s Law : Ptotal = Pa + Pb …..
Ptotal = # of toal moles(RT/V)
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Problem 7
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A gaseous mixture made from 6.00g O2
and 9.00g CH4 is placed 15L . What is
partial pressure of each gas and total
pressure?
Solution 7
nO2 = (6.00g O2)(1molO2/32.0g O2)=0.188mol O2
nCH4 =(9.00gCH4)(1molCH4/16g CH4)= 0.563mol
PO2= no2 RT/ V = (0.188mol)(0.0821)(273K)/(15L)
= 0.281atm
PCH4 = (0.563mol)(0.0821)(273K)/(15L)
= 0.841atm
Pt = 0.281atm + 0.841atm = 1.122atm
Partial pressures and mole
fraction
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P1/Pt = (nRt/V)/(ntRT/V)= n1/nt
P1 = (n1 / n total)Pt = X Pt
Thus, the partial pressure of a gas in a
mixture is its mole fraction times the
total pressure.
Problem 8
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The atmosphere is composed of 1.5mol
percent CO2, 18.0mol percent O2, and
80.5 mol percent Ar. a) calculate the
partial pressure of O2 in the mixture if
the total pressure is 745torr. b) if this
atmosphere is to be held in a 120L
space at 295K, how many moles of O2
are needed?
Solution 8
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a) PO2 = (0.180(745torr)= 134torr
b) PO2 =
(134torr)(1atm/760torr)=0.176atm
V = 120L, T = 295K
nO2 = PO2(V/RT)=
(0.176atm)(120L)/(0.0821L-atm/Kmol)(295K) = 0.872mol
Collecting gases over water
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The volume of gas collected is
measured until the inside water levels
and outside the bottle are same.
The total pressure inside is the sum of
the pressure of gas collected and
pressure of water vapor in equilibrium
with liquid water.
Ptotal = Pgas + Pwater
Problem 9
Ammonium nitrate, NH4NO2,
decomposes upon heating to form N2
gas.
NH4NO2→ N2 + 2H2O
When a sample of NH4NO2 is
decomposed in a test tube, 511ml of
N2 gas is collected over water at 26°C
and 745torr total pressure. How many
grams of NH4NO2 were decomposed?
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Solution 9
Pressure of water vapor = 25torr
 745torr – 25torr= 720torr
 Number of moles N2 = PV/RT
= (720torr*(1atm/760torr)(0.511L)
/(0.0821)(299K) =0.0197 mol N2

0.0197molN2(1molNH4NO2/1Mol N2) *(64.04g
NH4NO2/1molNH4NO2) = 1.26g NH4NO2
10.7 kinetic-Molecular Theory
Kinetic molecular theory developed
over 100 years, culminating in 1857
when Rudolf Clausius(1822-1888)
published a complete thm
1)The volume of an ideal gas particle is
insignificant when compared with the
volume of its container.
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2) Attractive and repulsive forces between
gases molecules are negligible.
3) Energy can be transferred during
collision, but the average kinetic energy
does not change.
4) The average kinetic energy of
molecules is proportional to the
absolute temperature.
Continued..
Root – mean – square(rms) speed, u,
: the speed of a molecule possessing K.E
 Average kinetic energy of gas mole(J)
=
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m= mass of mole(kg) u= speed of mole(m/sec)
★K.E increases with increasing temperature implies that
rms speed increases as temperature inreases.
Total Kinetic Energy of gas sample
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Total KE= (3/2)nRt
R= 0.0821L-atm/K-moles
t = temperature (K)
n = number of moles
Application to the gas law
1.
2.
At a constant temperature, if the
volume is increased , the molecules
must move a longer distance between
collision. So, the pressure decreases.
An increase in temperature means in
the average kinetic energy of the
molecules, and thus increase in
rms(speed).
Problem 10
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How is the rms speed of N2 molecules
in a gas sample changed by
a) increase in temperature
b) increase in volume of sample
c) mixing with a sample of Ar at same
temperature
Solution 10
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A) increase
B) no change
C) no change
10.8 molecular Effusion and Diffusion
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Graham’s law of Effusion : In 1846, Thomas
Graham discovered that the effusion rate of
gas is inversely proportional to the square
root of its molar mass.
Average speed of gas=
=
 Lighter molecules move faster than heavier
molecules.
M is molar mass
Problem 11
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An unknown gas composed of
homonuclear diatomic molecules
effuses at a rate that is only 0.355times
that of O2 at the same temperature
what is the gas?
Solution 11
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r1 = 0.355 * r2
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r1/r2= 0.355= square root of(32g/mol)/M1
(32g/mol)/M=(0.355)^2=0.126
M1 = (32g/mol)/(0.126)= 254g/mol
Only di iodine has this atomic weight.
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Diffusion and mean free path
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Diffusion: spread of one substance
throughout a space or a second substance.
Diffusion of gas is slower than molecular
speed, because of molecular collision.
Mean free path : the average distance
traveled by a molecule between collisions.
the mean free path for air molecules at sea
level is about 60nm.
10.9 Real Gases: deviation from
ideal behavior
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Real molecules do have finite volumes
and they do attract one another.
The difference that remains at high
temperature stems mainly from the
effect of the finite volume of the
molecuels.
The Van der Waals Equation
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Johannes van der Waals(1837-1923)
proposed the useful equation to predict
the behavior of real gases.
Ideal gas: P= nRT/V
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According to Van der Waals:
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Both a and b are given. A has units of
L^2-atm/mol^2.
Van der Waals equation:
Problem 12
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If 1.00mol of an ideal gas were
confined to 22.41L at 0℃,it would
exert a pressure of 1.00atm. Use Van
der Waals equation to estimate the
pressure exerted by 1.00mol of Cl2 in
22.4L at 0℃.
Solution 12
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n = 1mol, R= 0.0821, T=273K, V=22.4L
a= 6.49L^2atm/mol, b= 0.0562L^2atm/mol
P = (1mol)(0.0821)(273.2)/22.4-(1)(0.0562)
-(1^2)(6.49)/(22.4)^2
= 1.003 atm – 0.013atm = 0.990atm
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_Laws-Real_Gas.htm
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