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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
CHAPTER 10
Gases
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.
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)
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 Earth’s
atmosphere at sea level. The unit torr
(mm Hg) comes from the most
commonly used barometer for experimentally measuring pressure, the
mercury barometer.
1 torr = 1 mm Hg (exact)
d(Hg) = 13.53 g/cm3
Manometer
A manometer is a device for measuring differences in pressure.
In a mercury manometer
h (in mm Hg) = pgas - patm
Boyle’s 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 Boyle’s 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.
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 temperature scale, the Kelvin scale, with
T(K) = T(C) + 273.15
then
V = (constant) T
This observation is called Charles’
law.
Relationships From Boyle’s Law
and Charles’ Law
Useful relationships can be derived from Boyle’s law and
Charles’ law.
Boyle’s 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?
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
Avogadro’s hypothesis
Based on the observed combining ratios for gas reactions,
Amadeo Avogadro (1811) proposed the following hypothesis (Avogadro’s hypothesis)
Equal volumes of any two gases at the same pressure and
temperature contain the same number of molecules.
Avogadro’s hypothesis was generally ignored until revived in the
1850s by Stanislao Cannizzaro.
The Ideal Gas Law
The above observations (Boyle’s law, Charles’ law, and Avogardo’s hypothesis) can be summarized in a single equation, called the
ideal gas law.
pV = nRT
p = pressure
n = moles of gas
V = volume
R = gas constant
T = temperature
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 (Boyle’s law)
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
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)
R = 0.08314 L.bar/mol.K
R = 0.08206 L.atm/mol.K
(1 J = 1 kg.m2/s2)
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?
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
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 = pVT = 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
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)
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.
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).
Dalton’s Law of Partial Pressures
To this point we have discussed pure gases. For mixtures of ideal
gases we can use Dalton’s 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
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
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.
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
0.6073 mol
X(Ar) = 0.2503 mol = 0.4122
0.6073 mol
From Dalton’s 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)
(0.6073 mol)(0.08206 L.atm/mol.K)
= 340.6 K
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).
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/M]1/2
(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 speds increase as temperature increases, and at a particular
temperature light gases move faster (on average) than heavy gases.
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 MaxwellBoltzmann distribution, has the following appearance.
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.
Example: What is the average speed of an N2 (MW = 28.01
g/mol) molecule at T = 300. K and at T = 1000. K?
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/M]1/2
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.
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
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
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.
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.
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”
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.
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
(V - nb)
V2
a, b are constants
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.
Note that in general b increases as the size of the molecule
increases, and a increases as the strength of intermolecular forces increases.
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
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)
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.
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 Boyle’s work were immediately published, one
by Thomas Hobbes…and the other by…Franciscus 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