Homework Problems

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Homework Problems Chapter 10 Homework Problems
22, 24, 34, 40, 44, 48, 50, 56, 62, 66, 70, 81,
82, 88, 94, 106, 128
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CHAPTER 10 Gases
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Properties of Gases Gases were the first state
of matter extensively studied. This is because
small changes in temperature and pressure lead to
measurable changes in volume, and because gases,
to a good first approximation, behave in a simple
manner.
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Units For Pressure Pressure Force/Area The
MKS unit for pressure is the Pascal 1 Pascal
1 Pa 1 N/m2 1 Newton 1 N 1 kg.m/s2
Other common units include 1 bar 100,000
Pa 105 N/m2 (exact) 1 atmosphere 1 atm
101,325 Pa 1.01325 bar (exact) 1 atm
760. Torr 760. mm Hg (exact)
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Barometer
A barometer is a device for measuring
atmospheric pressure. In a barometer pressure
force/area hdg h height (m) d density
(kg/m3) g gravitational constant (9.807
m/s2) The pressure unit atmosphere (atm)
represent the approximate value for the pressure
of the Earths atmosphere at sea level. The unit
torr (mm Hg) comes from the most commonly used
barometer for exper-imentally measuring
pressure, the mercury barometer.
1 torr 1 mm Hg (exact) d(Hg) 13.53 g/cm3
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Manometer A manometer is a device for measuring
differences in pressure. In a mercury
manometer h (in mm Hg) pgas - patm
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Boyles Law Consider an experiment where volume
is measured as a function of pressure under
conditions of constant temperature and amount of
gas. Experimentally, the following is
observed pV constant This observation is
called Boyles law. pressure volume pV
(torr) (L) (L.torr)
160.0 100.0 16000. 320.0 50.0 16000.
640.0 25.0 16000. 800.0
20.0 16000. 1200.0 10.0 16000.
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Charles Law Now consider an experiment where
volume is measured as a function of temperature
under conditions of constant pressure and amount
of gas. Experimentally, a plot of V vs T is
found to be linear. When different amounts of
gas or pressures are used different lines are
obtained, but all lines appear to intersect at
approximately the same temperature, T ? - 273. ?C.
If we define a new tempera-ture scale, the
Kelvin scale, with T(K) T(?C) 273.15 then
V (constant) T This observation is called
Charles law.
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Relationships From Boyles Law and Charles
Law Useful relationships can be derived from
Boyles law and Charles law. Boyles law. If
n, T constant, then pV constant. Therefore
piVi pfVf Charles law. If n, p constant,
then V T (or V cT, where c is a constant).
From this, it follows that V/T constant, or
Vi/Ti Vf/Tf. Example A sample of gas at T
300. K and p 1.00 atm occupies a volume V
586. mL. What volume will the gas occupy if the
pressure is changed to 5.00 atm while keeping
temperature constant?
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Example A sample of gas at T 300. K and p
1.00 atm occupies a volume V 586. mL. What
volume will the gas occupy if the pressure is
changed to 5.00 atm while keeping temperature
constant? Since n, T are constant, it follows
that piVi pfVf Vf Vi (pi/pf) (586. mL)
(1.00 atm/5.00 atm) 117. mL
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Avogadros hypothesis Based on the observed
combining ratios for gas reactions, Amadeo
Avogadro (1811) proposed the following hypothesis
(Avo-gadros hypothesis) Equal volumes of any
two gases at the same pressure and temperature
contain the same number of molecules.
Avogadros hypothesis was generally ignored
until revived in the 1850s by Stanislao
Cannizzaro.
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The Ideal Gas Law The above observations
(Boyles law, Charles law, and Avo-gardos
hypothesis) can be summarized in a single
equation, called the ideal gas law. pV nRT p
pressure n moles of gas T
temperature V volume R gas constant All
of the previous observations can be recovered
from the ideal gas law. For example, if n and T
are held constant, then pV nRT constant
(Boyles law)
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Comments 1) The ideal gas law is
approximate. 2) The ideal gas law is an example
of an equation of state, that is, a relationship
among state variables (p, V, n, T). 3) The state
variables can be divided into two types a)
Extensive variables - Depend on the size of the
system b) Intensive variables - Independent of
the size of the system
pA pB VA VB TA TB nA nB p pA pB V
VA VB 2 VA 2 VB T TA TB n nA nB
2 nA 2 nB Vm molar volume V/n
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4) The ideal gas law is based on two underlying
assumptions a) The volume occupied by the gas
molecules is small b) Attractive forces
between gas molecules are small These
assumptions become true in the following
limits V ? ? (T, n held constant) p ? 0 (T, n
held constant) T ? ? (p, n held constant) 5)
The units for the gas constant, R, depend on the
units used for p, V, and n. R pV/nT can be
used to determine the units. R 8.314 J/mol.K
(MKS unit) (1 J 1 kg.m2/s2) R 0.08314
L.bar/mol.K R 0.08206 L.atm/mol.K

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Sample Problems 1) A sample of N2 gas with mass
m 8.238 g is confined in a container of volume
V 2.000 L at a temperature T 30.0 ?C. What
is the pressure exerted by the gas? Give your
answer in atm and torr. 2) A glass bulb has a
mass m 49.618 g when empty and m 51.203 g
when filled with an unknown pure gas. The volume
of the bulb is V 1.283 L. The measurements are
made at p 0.978 atm and T 21.5 ?C. What is
the molecular mass of the gas?
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1) A sample of N2 gas with mass m 8.238 g is
confined in a container of volume V 2.000 L at
a temperature T 30.0 ?C. What is the pressure
exerted by the gas? Give your answer in atm and
torr. pV nRT so p nRT M(N2) 28.01
g/mol
V n 8.238 g 1 mol 0.2941 mol T
30.0 ?C 303.2 K 28.01
g p (0.2941 mol) (0.08206 L.atm/mol.K)
(303.2 K) 3.659 atm
2.000 L In units of torr
p 3.659 atm 760 torr 2781 torr
1 atm
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2) A glass bulb has a mass m 49.618 g when
empty and m 51.203 g when filled with an
unknown pure gas. The volume of the bulb is V
1.283 L. The measurements are made at p 0.978
atm and T 21.5 ?C. What is the molecular mass
of the gas? M m/n mgas mfilled - mempty
51.203 g - 49.618 g 1.585 g pV nRT so n
pV T 21.5 ?C 294.6 K
RT n (0.978
atm) (1.283 L) 0.05190 mol
(0.08206 L.atm/mol.K) (294.6 K) So M
m 1.585 g 30.5 g/mol
n 0.05190 mol
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Density Density (d) is defined as d m/V For
solids and liquids density is approximately
independent of temperature. For gases, density
depends strongly on the conditions for which it
is measured. Gas density is often reported at
STP, standard temperature and pressure, which is
T 0 ?C 273.15 K p 1.000 atm Note that
the molar volume for an ideal gas at STP is V
RT (0.08206 L.atm/mol.K) (273.15 K)
22.41 L/mol n p
(1.000 atm)
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Density and Molecular Mass The density of a gas
can be related to its molecular mass. Since pV
nRT m nM n m/M So pV mRT
M M mRT m RT dRT
pV V p p So
by measuring the density of a gas for known
conditions of pressure and temperature, the
molecular mass of the gas can be found. We can
also rearrange the above equation to solve for
other variables such as density, temperature, or
pressure.
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Reactions Involving Gases Since pV nRT, then V
nRT/p For gases at a particular pressure and
temperature, n V. We can therefore interpret a
balanced chemical equation involving gases either
in terms of moles of gas or volume of gas. For
example 2 CO(g) O2(g) ? 2 CO2(g) can be
interpreted in two ways. 1) 2 moles of CO will
react with 1 mole of O2 to form 2 moles of
CO2. 2) 2 liters of CO will react with 1 liter
of O2 to form 2 liters of CO2 (assuming the gases
obey the ideal gas law, and are at the same
temperature and pressure).
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Daltons Law of Partial Pressures To this point
we have discussed pure gases. For mixtures of
ideal gases we can use Daltons law of partial
pressures. For N gases p ptotal p1 p2
p3 pN p1 partial pressure of gas 1 p2
partial pressure of gas 2 etc. Now since pV
nRT, p nRT/V, and so p1 n1RT/V p2
n2RT/V ptotal p1 p2 pN ptotal
n1RT n2RT nNRT
V V V If we factor
out (RT/V) then
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ptotal RT n1 n2 nN ntotalRT
V
V Also, p1 n1RT n1 ntotalRT
X1ntotalRT X1ptotal
V ntotal V V
where X1 n1 is the mole fraction of
gas 1 in the mixture
ntotal Note that X1 X2 X3 XN 1
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Sample Problem A gas mixture is prepared by
mixing together 10.00 g of nitrogen (N2, M
28.01 g/mol) and 10.00 g of argon (Ar, M 39.95
g/mol). The gas is confined in a container with
volume V 20.00 L. The total pressure of the
gas mixture is ptotal 0.8488 atm. Find the
following 1) Partial pressure of N2 and Ar in
the gas mixture. 2) Temperature of the gas
mixture.
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A gas mixture is prepared by mixing together
10.00 g of nitrogen (N2, M 28.01 g/mol) and
10.00 g of argon (Ar, M 39.95 g/mol). The gas
is confined in a container with volume V 20.00
L. The total pressure of the gas mixture is
ptotal 0.8488 atm. Find the following 1)
Partial pressure of N2 and Ar in the gas
mixture. 2) Temperature of the gas
mixture. moles N2 10.00 g N2 1 mol N2
0.3570 mol N2
28.01 g N2 moles Ar 10.00 g Ar 1 mol
Ar 0.2503 mol Ar
39.95 g Ar Total number of moles of
gas 0.3570 mol 0.2503 mol 0.6073 mol X(N2)
0.3570 mol 0.5878 X(Ar) 0.2503 mol
0.4122 0.6073 mol
0.6073 mol
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From Daltons law, pi Xi ptotal, so p(N2)
(0.5878)(0.8488 atm) 0.4989 atm p(Ar)
(0.4122)(0.8488 atm) 0.3499 atm Finally, since
pV nRT, it follows that T pV/nR so T
(0.8488 atm)(20.00 L) 340.6
K (0.6073 mol)(0.08206 L.atm/mol.K)
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Daltons Law in Gas Evolution One common use of
Daltons law is in analyzing data from
experiments where a gas is produced as the result
of a chemical reaction. For example, consider
heating KClO3 in the presence of MnO2, in the
apparatus below. 2 KClO3(s) MnO2 2 KCl(s) 3
O2(g) The total pressure in the gas
collection vessel is ptotal pO2 pH2O where
pH2O vapor pressure of water
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Kinetic Theory The above description of gas
behavior is based on experimental observation.
However, by use of kinetic theory, a description
of the behavior of gases on a molecular level,
the above equations can be derived. Assumptions
1) Gases are composed of molecules in random
motion. 2) The volume occupied by the molecules
is small. 3) The interaction forces between
molecules are weak. 4) Collisions between
molecules, or between a molecule and the walls of
the container, are elastic (total kinetic energy
remains constant).
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Average Speed and Kinetic Energy Based on the
above assumptions the following equations can be
derived for urms, the root mean square average
speed and (KE)ave, the average kinetic energy of
a gas molecule urms 3RT/M1/2 (KE)ave
3RT/2 (per mole of gas) Note the following 1)
For a particular value of temperature all gases
have the same value for average kinetic
energy. 2) urms T1/2 urms (1/M)1/2 So gas
speeds increase as temperature increases, and at
a particular temperature light gases move faster
(on average) than heavy gases.
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Distribution of Molecular Speeds In a gas
mixture there will be some molecules that are
moving faster or slower than the average speed.
The distribution of molecular speeds in a gas can
be derived. This distribution, called the
Maxwell-Boltzmann distribution, has the following
appearance.
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Temperature Dependence of the Maxwell-Boltzmann
Distribution As temperature increases the peak
in the Maxwell-Boltzmann distribution shifts to
higher speeds, and the height of the distribution
decreases.
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Example What is the average speed of an N2 (MW
28.01 g/mol) molecule at T 300. K and at T
1000. K?
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Example What is the average speed of an N2 (MW
28.01 g/mol) molecule at T 300. K and at T
1000. K? urms 3RT/M1/2 At T 300.
K urms 3 (8.314 J/mol.K) (300. K)/ (28.01 x
10-3 kg/mol)1/2 517. m/s (1160 mph) At
T 1000. K urms 3 (8.314 J/mol.K) (1000. K)/
(28.01 x 10-3 kg/mol)1/2 944. m/s Notice
we use R in MKS units (J/mol.K), and M in units
of kg/mol.
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Diffusion and Effusion Diffusion refers to the
process of mixing of gases due to their random
motion. Effusion refers to the escape of a gas
to vacuum through a small hole. For both
processes the rate of the process is proportional
to u, the average speed of the gas
molecules Therefore rate (1/M)1/2
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Comparison of Rates of Diffusion and
Effusion Since for both effusion and diffusion
rate (1/M)1/2 it follows that the relative
rate of either diffusion or effusion for two
gases at the same conditions of temperature and
pressure is (rate2)/(rate1) (M1/M2)1/2
where M1 molecular mass of gas 1 M2
molecular mass of gas 2 Example At 300. K,
which gas diffuses more quickly, N2 or Ar? How
much more quickly does the gas diffuse?
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Example At 300. K, which gas diffuses more
quickly, N2 or Ar? How much more quickly does
the gas diffuse? M(N2) 28.02 g/mol M(Ar)
39.95 g/mol Since rate (1/M)1/2 , the ligher
gas (N2) diffuses more quickly. How much more
quickly? rate(N2)/rate(Ar) M(Ar)/M(N2)1/2
39.95/28.021/2 1.19 So nitrogen
diffuses about 20 faster than Ar.
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Real Gases As previously discussed, the ideal
gas law is an approximate description of the
behavior of real gases. One way of showing this
is by measuring the volume occupied by one mole
of a real gas at STP (standard temperature and
pressure, taken as T 0 ?C, p 1.000 atm). For
an ideal gas at STP the molar volume is 22.414
L/mol.
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Compressibility Factor (Z) The compressibility
factor (Z) is defined as Z pV
nRT For an ideal gas Z 1 for all pressures and
temperatures. Deviations from this value for
real gases can be taken as a measure of nonideal
behavior.
T 300.
Gas N2
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Van der Waals Equation Various nonideal gas laws
have been proposed to account for deviations from
ideal behavior. A simple and useful equation is
the van der Waals equation. We can obtain the
equation as follows From the ideal gas law
p nRT V We can
correct for excluded volume (volume occupied by
other gas molecules) by modifying this
equation p nRT (V nb)
where nb excluded volume
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Now consider the effect of attractive forces on
the pressure
For molecules in the middle of the container the
attractive forces appear in all directions and
tend to cancel. However, when a molecule
approaches the walls of the container the
attractive forces tend to slow the molecule down
and thus lower the pressure.
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We correct for this decrease in pressure by
subtracting a term from the pressure whose value
depends on the strength of the attractive forces
and the molar density (n/V) of the gas. p
nRT - an2 a, b are constants
(V - nb) V2 the van der Waals
equation. a coefficient - depends of strength of
intermolecular attractive forces b coefficient -
depends on size of molecules Values for a and b
are different for different gases, and are found
by fitting to experimental data.
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Note that in general b increases as the size of
the molecule increases, and a increases as the
strength of intermolecular forces in-creases.
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Example What is the pressure of 1.000 mole of
ammonia (NH3) at T 300.0 K and V 10.000 L
according to the ideal gas law and according to
the van der Waals equation (a 4.17 L2.atm/mol2
b 0.0371 L/mol)? Ideal gas law p nRT
V van der Waals
equation p nRT - an2
(V - nb) V2

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Example What is the pressure of 1.000 mole of
ammonia (NH3) at T 300.0 K and V 10.000 L
according to the ideal gas law and the van der
Waals equation (a 4.169 L2.atm/mol2 b 0.0371
L/mol)? Ideal p nRT (1.000 mol) (0.08206
L.atm/mol.K) (300.0 K) V
10.000 L
2.462 atm van der Waals p (1.000) (0.08206
L.atm/mol.K) (300.0 K) 10.000 L -
(1.000 mol) (0.0371 L/mol) - (4.17
L2.atm/mol2) (1.000 mol)2
(10.000 L)2 2.4710 atm - 0.0417 atm
2.429 atm (or 1.3 lower)
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So V(ideal) 2.462 atm V(van der Waals)
2.429 atm Which volume is correct? The result
from the van der Waals equation is likely to be
closer to the correct (experimental) value
because the van der Waals equation usually does a
better job of predicting pressure, volume, etc.
than the ideal gas law. However, since all gas
laws are approximate, it is incorrect to say that
one result is wrong and the other result is
right. It is better to say that the van der
Waals equation does a better job of predicting
these quantities than the ideal gas law.
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End of Chapter 10 ...equal volumes of gases,
at the same temperature and pressure, contain the
same number of molecules. - Amadeo Avogadro
Two attacks on Boyles work were immediately
published, one by Thomas Hobbesand the other
byFranciscus Linus. Hobbs based his criticism
on the physical impossibility of a vacuum (A
vacuum is nothing, and what is nothing cannot
exist). Linus claimed that the mercury column
(in the barometer) was held up by an invisible
thread, which fastened itself to the upper end of
the tube. The theory seemed quite reasonable, he
said, for anyone could easily feel the pull of
the thread by covering the end of the barometer
tube with his finger.
- John Moore, Physical Chemistry