Download book problems c 10.

BOOK PROBLEMS C 10.
1. Pressure is force per unit area, F/m2 .
Force = mass x acceleration...Newton’s 2nd law. F = ma. “a” in this case is the acceleration due to
gravity on earth = 9.81 kg m/sec2 = 9.81 N (Newtons)
A
B
These two objects have the same mass (1 kg) and are, we presume, on the same planet in the
same gravitational field. A has less surface area than B (100 cm x 100 cm vs 1000cm x 100cm). Thus
you’d figure A would exert more pressure than B.
b. Getting some math into it:
length (m) width (m) area (m2) mass (kg)
A
B
1.00
10.00
1.00
1.00
1.00
10.0
1.000
1.000
acceleration
Force
m/sec2
kg m/sec2
9.810
9.810
9.810
9.810
Pressure
(kg/sec2 m, or Pascals)
9.810
0.981
2.
length (m) width (m)
A
B
C
10.00
10.00
1.00
1.00
0.10
0.10
↓Area and ↑pressure.
area (m2)
10.00
1.00
0.1
acceleration
Force
Pressure
mass (kg)
m/sec2
kg m/sec2 (kg/sec2 m, or Pascals)
1.000
1.000
1.000
9.80665
9.80665
9.80665
9.807
9.807
9.807
0.9807
9.807
98.07
Surface area A m2
Patm
h
Hg
USEFUL THINGS TO KNOW
1 atm = 101.325 kPa (kilopascals) = 760 mm Hg (h)
1 pascal = 1 N /m2 = 1 kg /m sec2
1 atm = 1.01325 x 105 kg/m sec2 = 760 mm Hg
Units of density = ρ = g/cm3
Acceleration due to gravity = a = 9.81 m/sec2
A = the cross-sectional area of the tube in m2.
We can figure that the force holding the Hg down in the tube = and that is equal to the force of the
atmosphere pushing it up:
Forcedown = Forceup.
Dividing both sides by the unit area to which the force is applied:
Forcedown/A = Force up/ A (A is the area of Hg the atm. is pushing up on).
Forcedown exerted by the Hg is (mass Hg in tube)( x acceleration due to gravity) = (mHg)(a):
1. mHg (a)/ A = Forcedown
Forceup /A is the atmospheric pressure.
2. Forceup = Patm
Forcedown/A = Forceup/A
3. mHg (a)/ A = Patm
OK, so what is the mass of Hg in the tube? No one tells us, but if we knew the volume of the tube
and the density of Hg (Ah Ha!) we could find out.
mHg = volume x density = (v)( ρ) = cm3 (g/cm3)
so, substituting into 3:
4. (v )( ρ)(a)/A = Patm
We can get the density if we know the volume. Can we get the volume? Sure. The tube is a cylinder,
and Vcylinder = h x A = 760 mm(1 m/1000 mm) x A
Substituting into eq’n 4:
5. (v )( ρ)(a) /A =
(h)(A)( ρ)(a) /A = Patm
The Area term cancels out, and we can rearrange 5 to:
6. ρ = Patm/(h)(a)
Plugging the numbers in:
5. ρ = 1.01325 x 105 (kg m/sec2)/( m2 ) /[760 mm(1 m/1000 mm) ( 9.81 m/sec2 )]
= 1.359 x 104 kg/m3
1.359 x 104 kg/m3 (1000 g/kg)[(1 m/100 cm)]3 = 13.59 g/cm3 = density of Hg.
A nice feature of this is that the area of the tube cancels. That means that you can make a barometer
with any diameter tubing, and the height of Hg will be the same.
9. We can use the eq’n we derived in Q 8:
ρ = Patm/(h)(a), where ρ = density of liquid = 1 g/cm3(1 kg/g)[(100 cm/m)]3 = 1.00 x 103 kg/m3
h = height of liquid in barometer
a = acceleration due to gravity, 9.81 m/sec2
Patm = atmospheric pressure, 1.01325 x 105 (kg m/sec2)/( m2 )
h = Patm/ ρ a = 1.01325 x 105 (kg m/sec2) (1 m3
) (1 sec2 ) = 10.3 m
( m2 )
(1.00 x 103 kg ) ( 9.81 m)
10.3 m (100 cm) (1 in
) ( 1 ft )
( m ) ( 2.54 cm) ( 12 in)
= 33.9 ft. You can see why we use mercury in barometers!
11. ↑ P when ↓V , according to your own experience & Boyle’s Law. Arrange the ratio of pressures
to reflect this.
a. 1 L (1 atm ) = 0.1 L
( 10 atm)
b. 1 L (1 atm ) = 0.01 L
( 100 atm)
c. 1 L (1 atm ) = 10 L
( 0.1 atm)↓
d. 1 L (1 atm ) = 100 L
( 0.01 atm)
12
10
8
Expanded
view
6
4
2
0
0
2
4
6
8
10
12
13. V1 = 435.1 mL
V2 = 172.9 mL
V↓ when P↑ Set up a ratio to reflect what you know: Volume goes down when pressure goes up.
Another way to say this is Boyle’s Law: P1V1 = P2V2
435.1 mL(P1/P2) = 172.9 mL
P1/P2 = 172.9 mL/435.1 mL = 0.3974/1
Or, P2/P1 = 2.516/1
! Read a short biography of Robert Boyle: http://www.bbk.ac.uk/boyle/biog.html
17. Charle’s Law states what we know from our own experience: V↓
when T↓ . Set up ratios to reflect this, remembering to convert Celsius
degrees to Kelvin degrees.
1.00 L (-50.0 oC + 273.15 K) = (223 K ) = 0.817 L
( 0o C + 273.15 K )
(273 K)
19. Charle’s Law states what we know from our own experience: V↓ when T↓ . Set up ratios to
reflect this, remembering to convert Celsius degrees to Kelvin degrees.
XL
(500 oC + 273.15 K ) =
( 400 o C + 273.15 K )
(X L) (773 K ) = 2.38 L
(673 K)
X L = 2.38 L(673 K ) = 2.07 L
(773 K)
21. At STP (Standard Temperature & Pressure, 273.15 K & 1 atm), 1 mole of an ideal gas occupies
a volume of 22.4 L.
All these gases (CO2, Nitrogen, Oxygen, air) are at STP, and all have a volume of 22.4 L, so they all
have 1 mole of gas, or 6.022 x 1023 particles.
22.4 L (1 mole gas )(6.022 x 1023 particles) = 6.022 x 1023 particles
( 22.4 L)STP
( 1 mole
)
23. This question combines all the gas laws; Boyles, Charles, and Avogadro’s.
V = Kb/P
V = K cT
V = K an
Combining them all:
V = KbKcKa nT/P
All those constants are a nuisance, so call KbKcKa “R”. Move P over to the left side:
PV = nRT, the Ideal Gas Law, the equation of state for a gas.
R = 0.08205783 L atm/K mole
When In Doubt, Convert To Moles!
a. n = PV/RT = (1.00 atm
)( 22.4 L
) = 1.00 mole
( 0.08206 L atm/K mole)(273 K)
1.00 moles (6.022 x 1023 molecules of carbon monoxide) = 6.022 x 1023 molecules
(
Mole
)
R
a
b
c
P (atm)
V (L)
T (K)
(L atm/K
mol)
1.00
1.00
1.00
22.4
44.8
11.2
273
273
273
0.08206
0.08206
0.08206
n
(moles)
N
Particles
0.999894 6.02E+23
1.999788 1.2E+24
0.499947 3.01E+23
Amedeo Avogadro
Avogadro - the man
Lorenzo Romano Amedeo Carlo Avogadro, conte di Quarequa e di Cerreto (1776 - 1856), was born
in Turin, Italy, on 9th August, 1776. He was the son of Count Filippo Avogadro and Anna Maria
Vercellone. His father was a distinguished lawyer and civil servant, becoming a senator of Piedmont
in 1768, and was appointed advocate general to the senate of Vittorio Amedeo III in 1777. Under the
French rule of 1799 he was made president of the senate.
Amedeo Avogadro went to school in Turin. Coming from a family of well established ecclesiastical
lawyers, Avogadro was guided toward a legal career, and became a bachelor of jurisprudence in
1792, at the ripe old age of just 16 years. Four years later he gained his doctorate in ecclesiastical
law and began to practice. In 1801 he was appointed secretary to the prefecture of the department of
Eridano.
In spite of his successful legal career, Avogadro also showed an interest in natural philosophy, and in
1800 he began private studies of mathematics and physics. His first scientific research in 1803,
undertaken jointly with his brother Felice, was on electricity.
In 1806, Avogadro was appointed demonstrator at the Academy of Turin, and in 1809 became
professor of natural philosophy at the college of Vercelli. In 1820, when the very first chair of
mathematical physics in Italy was established at the University of Turin, Avogadro was appointed.
Unfortunately, his post was short lived, since political changes suppressed the chair and Avogadro
was out of a job by July, 1822. The chair was eventually reestablished in 1832, and Avogadro was
reappointed to the position in 1834. Here he remained until his retirement in 1850.
Avogadro had succeeded to his father's title in 1787. He married Felicita Mazzé, and they had a total
of six children. Avogadro led an industrious life, and was a modest man, working in isolation. This
probably contributed to his relative obscurity, particularly outside Italy. Avogadro died on the 9th July,
1856. He was described as religious, but not a bigot.
Avogadro - his contribution to chemistry
In order to understand the contribution that Avogadro made, we must consider some of the ideas
being developed at this time. Chemistry was just beginning to become an exact science. The Law of
Definite Proportions and the Law of Multiple Proportions were well accepted by 1808, at which time
John Dalton published his New System of Chemical Philosophy.
Dalton proposed that the atoms of each element had a characteristic atomic weight, and that it was
atoms that were the combining units in chemical reactions. Dalton had no method of measuring
atomic weights unambiguously, so made the incorrect assumption that in the most common
compound between two elements, there was one atom of each.
At around this time, Gay-Lussac was studying the chemical reactions of gases, and found that the
ratios of volumes of the reacting gases were small integer numbers. This provided a more logical
method of assigning atomic weights. Gay-Lussac did not carry through the full implications of his
work. However, Dalton realised that a simple integral relation between volumes of reacting gases
implied an equally simple relation between reacting particles. Dalton still equated particles with
atoms, and could not accept how one particle of oxygen could yield two particles of water. This was a
direct threat to the relatively new atomic theory, and therefore Dalton tried to discredit the work of
Gay-Lussac.
In 1811, Avogadro published an article in Journal de physique that clearly drew the distinction
between the molecule and the atom. He pointed out that Dalton had confused the concepts of atoms
and molecules. The "atoms" of nitrogen and oxygen are in reality "molecules" containing two atoms
each. Thus two molecules of hydrogen can combine with one molecule of oxygen to produce two
molecules of water.
Avogadro suggested that
equal volumes of all gases at the same temperature and pressure contain the same
number of molecules
which is now known as Avogadro's Principle.
The work of Avogadro was almost completely neglected until it was forcefully presented by Stanislao
Cannizarro at the Karlsruhe Conference in 1860. He showed that Avogadro's Principle could be used
to determine not only molar masses, but also, indirectly, atomic masses. The reason for the earlier
neglect of Avogadro's work was probably the deeply rooted conviction that chemical combination
occurred by virtue of an affinity between unlike elements. After the electrical discoveries of Galvani
and Volta, this affinity was generally ascribed to the attraction between unlike charges. The idea that
two identical atoms of hydrogen might combine into the compound molecular hydrogen was
abhorrent to the chemical philosophy of the early nineteenth century.
Avogadro - his number
It was long after Avogadro that the idea of a mole was introduced. Since a molecular weight in grams
(mole) of any substance contains the same number of molecules, then according to Avogadro's
Principle, the molar volumes of all gases should be the same. The number of molecules in one mole
is now called Avogadro's number. It must be emphasised that Avogadro, of course, had no
knowledge of moles, or of the number that was to bear his name. Thus the number was never
actually determined by Avogadro himself.
As we all know today, Avogadro's number is very large, the presently accepted value being
6.0221367 x 1023. The size of such a number is extremely difficult to comprehend. There are many
awe-inspiring illustrations to help visualize the enormous size of this number. For example:

An Avogadro's number of standard soft drink cans would cover the surface of the earth to a
depth of over 200 miles.

If you had Avogadro's number of unpopped popcorn kernels, and spread them across the
United States of America, the country would be covered in popcorn to a depth of over 9 miles.

If we were able to count atoms at the rate of 10 million per second, it would take about 2
billion years to count the atoms in one mole.
Determination of the number
Cannizarro, around 1860, used Avogadro's ideas to obtain a set of atomic weights, based upon
oxygen having an atomic weight of 16. In 1865, Loschmidt used a combination of liquid density,
gaseous viscosity, and the kinetic theory of gases, to establish roughly the size of molecules, and
hence the number of molecules in 1 cm3 of gas.
During the latter part of the nineteenth century, it was possible to obtain reasonable estimates for
Avogadro's number from sedimentation measurements of colloidal particles. Into the twentieth
century, then Mullikens oil drop experiment gave much better values, and was used for many years.
A more modern method is to calculate the Avogadro number from the density of a crystal, the relative
atomic mass, and the unit cell length, determined from x-ray methods. To be useful for this purpose,
the crystal must be free of defects. Very accurate values of these quantities for silicon have been
measured at the National Institute for Standards and Technology (NIST).
To use this approach, it is necessary to have accurate values of atomic weights, often obtained by
measuring the mass of atomic ions. For example, an ion trap, employing extremely uniform and
stable magnetic and electric fields should allow such measurements to be made to better than 1 part
in 1010. The relative atomic mass of silicon is particularly important, since silicon crystals are used in
the x-ray methods mentioned above.
As a continuation of this approach, one of the 1999 NIST Precision Measurement Grants has been
awarded to David Pritchard, physics professor at the Massachusetts Institute of Technology. He will
conduct cyclotron frequency measurements on ions that could achieve a 100-fold improvement in the
accuracy of atomic mass measurements. MIT has developed the world's most accurate mass
spectrometer capable of measuring the atomic mass of atoms to one part in 10 billion. Pritchard
proposes to simultaneously measure the cyclotron frequencies of two different ions in order to
improve the values of several fundamental constants, including Avogadro's number.
At the present time, information on Avogadro's number from many different experiments is pooled
with other observations on other physical constants. A most probable and self-consistent set of
physical constants that best fits all reliable data is then found by statistical methods.
The size of Avogadro's number is determined by our definition of the mole. What it does demonstrate
is how small an atom or molecule is compared to the amounts of material we are familiar with in
everyday life, since the definition of the mole involves amounts of material we are completely familiar
with.
All comments and queries to Chris Johnson, the originator of these pages.
This page last modified 19th March 1999
E-mail: [email protected]
http://avogadro.che.hw.ac.uk/avoga.html
26 a. PV = nRT is the equation to remember.
R = 0.08205783 L atm/K mole
We will need to convert pressure from Torr to Atmospheres.
760 Torr = 1 Atm = 760 mm Hg
589.1 Torr( 1 atm ) = 0.775 atm
(760 torr)
We will also need to convert Temp from oC to K
312.4 oC + 273.15 K = 585.6 K
V = nRT/P = 3.28 moles(0.08206 L atm/K mol)(585.6 K)/(0.775 atm) = 203 L.
b. If the volume and temperature decreases by ½, you’d predict that pressure would increase do
to the volume dropping, but decrease due to the temperature dropping. Let’s plug in the
numbers & see how it plays out:
V = 203 L/2 = 101.5 L T = 585.6 K/2 = 292.8 K
P = nR(T/2)/(V/2) = 3.28 moles(0.08206 L atm/K mol)(292.8 K)/(101.5 L) = 0.775 atm
No change, the one change cancels the other here.
BONUS QUESTION, AND A GOOD ONE TO SWOT UP FOR EXAM PURPOSES!
Air bags are an important safety feature in your car. How do they work? When you hit a tree, a
spring shoots a steel ball bearing into a detonator cap, which then blows up, igniting a small canister
containing 100.0 g of sodium azide. The sodium azide reacts quickly (40 ms), producing nitrogen gas
and sodium metal (of all things).
2NaN3(s)  2Na(s) + 3N2(g)
Calculate the volume of N2 gas produced at 758 Torr and 17.0 0C from 100.0 g of sodium azide.
When in doubt, convert to moles
I mole of sodium azide weighs 65.011 g/mole
100.0 g NaN3(1 mole NaN3) =1.538 moles NaN3
( 65.011 g
)
How many moles of N2 will be made?
1.538 moles NaN3 ( 3 moles N2 ) = 2.388 moles N2 produced.
( 2 moles NaN3)
Bonus question, continued.
PV = nRT
V = nRT/P
n = 2.388 moles N2
P = 758 torr( 1 atm/760 torr) = 0.997 atm
T = 17.0 oC + 273.15 K = 290.2 K
V = 2.388 moles(0.08206 L atm/K mole)(290.2 K)/(0.997 atm ) = 57.0 L
30. The reaction is:
CH4 + 2O2  CO2 + 2H2O Methane + oxygen gas  carbon dioxide and water
MW CH4 = 16.043 g/mol
MW O2 = 31.9988 g/mol
When in doubt, convert to moles!
160 g CH4( 1 mol/16.043 g) = 9.97 moles CH4
480 g O2 ( 1 mol/31.9988 g) = 15.0 mol O2 (We are only entitled to 2 sig fig here, but let’s round off at
the end.)
OK, which is the limiting reagent?
9.97 moles CH4 (2 moles O2/1 mole CH4) = 19.9 moles O2 needed to react completely with all the
CH4. We’ve only got 15.0 moles of O2. O2 is the LR.
a. After combustion, 0 moles of O2 are left (it’s the LR).
How much CH4 gets used up?
15.0 moles O2 (1 mole CH4/2 moles O2) = 7.50 moles CH4 used up.
9.97 moles CH4 – 7.50 moles CH4 = 2.5 moles CH4 left over.
b. 15.0 moles O2 (1 mole CO2/2moles O2 ) = 7.5 moles Carbon dioxide made.
15.0 moles O2 (2 mole H2O/2moles O2 ) = 15 moles water made.
c. Let’s see... when things stop blowing up we have 2.5 moles of CH4 left over
7.5 moles CO2 made
15 moles of H2O made
25 moles of gas total
in a volume of 1.00 L, and T = 1250C + 273.15K
= 398 K
PV = nRT
P = nRT/V = 25 moles(0.08206 L atm/K mol)(398 K)/(1 L) = 820 atm.
31. Oh, this looks weird. When in doubt, convert to moles, and see what falls out.
MW carbon dioxide = 44.01 g/mol
MW water = 18.015 g/mol
MW methane = 16.043 g/mole
2.00 g CO2 ( 1 mol CO2/44.01 g) = 0.0454 moles CO2.
MW (g/mol) Mass (g)
Moles
CO2
44.01
2.00
0.0454
H2O
18.015
2.00
0.111
CH4
16.043
2.00
0.125
Total moles
0.281
V = 1.00 L
P = 10.0 atm
n = 0.281 moles
The equation that relates pressure, volume & temperature of a gas is:
PV = nRT
T = PV/nR = 10.0 atm (1.00 L)/0.281 moles)(0.08206 L atm/K mole) = 434 K
The take-home from this, besides the simple mechanics of being able to do the problem, is that heat is
related to the number of particles whizzing about.
32. A chance to use Dalton’s Law of Partial Pressures, which says that if you have a mixture of gases,
each gas contributes to the total pressure, depending on the volume, temperature & # moles of that
gas. Ptotal = P1 + P2 + P3 +...
Data: Ptotal = 1.17 atm
PN2 = 1.00 atm.
T
= 285
V
= 1.00 L
a. Ptotal = 1.17 atm = PHe + PN2 = PHe + 1.00 atm.
PHe = 1.17 atm = 1.00 atm = 0.17 atm.
Mole fraction He = Moles He/(moles He + Moles N2)
We COULD use PV = nRT to convert each gas pressure to moles:
eg nHe = PHeV/RT = 0.17 atm (1.00 L)/(0.08206 L atm/K mole)(285 K) etc.
HOWEVER, Since pressure of a gas is directly proportional to moles of a gas, you can use pressure
directly. Watch:
Mole Fraction He =
=
nHe
nHe + nN2
.
=
PHeV/RT
. =
(V/RT) PHe . =
PHe
.
(PHeV/RT + PN2V/RT)
(V/RT) PHe + PN2
PHe + PN2
0.17 atm/(0.17 atm + 1 .00 atm) = 0.14
Mole Fraction N2 = 1 – 0.14 = 0.86
33. The problem calls for moles N2/L , which is n/V.
Data: P = 2.50 atm
T = -10oC = 263.15 K
PV = nRT, and we want n/V.
n/V = P/RT = 2.50 atm/(0.08206 L atm/K mole)(263.15 K) = 0.116 mol/L
35. A “Homonuclear diatomic gas” . What is THIS 50¢ word?
“Homo” the same.
“Nuclear” the nucleus of an atom
“Diatomic” 2 atoms.
A gas molecule made up of 2 identical nuclei. H2 , N2, O2 , F2 , Cl2 are homonuclear diatomic gases.
CO is a heteronuclear diatomic gas.
O3 , ozone, is a homonuclear triatomic gas...
What will help us identify this gas is ist molecular weight, g/mole. So if we knew how many
grams corresponds to how many moles...
DATA: mass of gas: 2.14 g
V = 1.50 L
T = 273.15 K (the problem states that we are at STP)
P = 1.00 atm (STP)
PV = nRT
n=
PV/RT = (1.00 atm)(1.50 L)/(0.08206 L atm/K mole)(273.15 K) = 0.0669 moles = 2.14 g
2.14 g/0.0669 moles = 32.0 g/mole. Looks like O2
H2
MW
(g/mol)
2.016
N2
28.014
O2
31.998
F2
37.996
Cl2
70.906
37.
Ek = 3/2nRT for any Ideal Gas.
We will use the value 8.3145 x 10-3 kJ/K mole as our units of choice for R (L atm is an energy term!)
R
MW (g/mol)
moles
Kj/K mole
150 K
Ek (KJ)
300K
2.016
1.00
0.008315
1.87
3.74
5.61
4.003
1.00
0.008315
1.87
3.74
5.61
20.18
1.00
0.008315
1.87
3.74
5.61
39.948
1.00
0.008315
1.87
3.74
5.61
H2
He
Ne
Ar
450K
46. You get to use Graham’s Law here, which says that the rate a gas particle moves is inversely
proportion to the square root of its MW.
If you have 2 particles, of different mass
Rate A
Rate B
=
√Mass B
√Mass A
MW
H2
He
g/mol
rate H2/rate He
2.016
4.003
1.4091
rateN2/rateHCN
N2
HCN
28.014
27.025
0.9822