we live at the
bottom of an ocean of gases, called the atmosphere, and we depend on those
gases for our survival. but, because most gases are invisible . we're usually
much less aware of them in our surroundings than we are of liquids and solids.
gases are
difficult for a chemist to work with : they leak out of any container that
isn't tightly sealed, most of them can't be seen, and many of them are
poisonous or explosive . special equipment is needed to measure quantities of
gases and to move them from one container to another. because gases are
difficult to work with, early experiments in chemistry preferred to work with
liquids and solids.
when chemists
carried aout the first systematic studies of gases, in the seventeenth and
eighteenth centuries, they found that the behavior of gases is simpler than
that of liquids and solids, and easier to understand. in a liquid or a solid,
the atoms,molecules,or ions are packed close together,and this crowding allows
the particles to interact with one another in complicated ways. in a gas, the
particles to intersct with one another only very slightly : as a result, gases
behave more simply than liquide or silids. it was the study of the simple
properties of gases that led jhon dalton to the creation of the atomic theory.
gases are
important in our lives, and they provide simple models of basic chemical
theories. for these reasons, the study of gases is an important part of any
introduction to the fundamentals of chemistry.
13-1 the quantity of gas in a
sample can be described by stating its mass or by stating its
volume,temperature, and pressure.
imagine that we have a sample of
oxygen gas contained in a cylinder by a piston, as shown in figure 13,1.
obe way to
describe that quantity of gas in the sample is to state its mass.for the
quantities id gases used in laboratory work, the units usually used are grams
or milligrams.let's assume that the mass of oxygen in our sample is 50.0 mg.
another way to
describe the quantity of gas in the sample is to state its volume. but, as you
know from everyday experience, the volume of a sample of gas depend on its
temperature and pessure. a gas expands if it's heated andc ontracts if it's
cooled. if the pressure on a gas is increased (for example, by pshing down on
the piston in the system shown in figure 13.1) the volume of the gas decreases,
and if the pressure is decreased (by pulling up on the piston) the volume of
the gas increases. to express the quantity of gas in a sample by specifying its
volume, we also have to specify the temperature and pressure.
the volume
units used for laboratory work are usually milliliters or liters,and the
temperature units are usually c or k : you;re familiar with these units from
chapter 3. we assume that our oxygen sample has a volume of 42,4 ml and a
temperature of 25c.
the units most
commonly used to express gas pressures in chemical work are the atmosphere, the
torr, and kilopascal.
Balance among three pressures : the pressure exerted by air
molecules colliding with the inside surface of the rubber membrane , which
pushes the membrane outward ; the pressure exerted by air molecules bombarding
the outside surface of the membrane, which pushes it inward and the pressure
exerted by the elasticity of the stretched rubber which pulls the membrane
inward.
If you force more air into the balloon, the increased number of air
molecules inside inside the balloon will cause a larger number of collisions of
air molecules with the inside surface of the membrane. The preasure on the
inside surface will increase and the balloon will expand. At some new and
larger volume, the three pressures the outward pressure of the air inside the
balloon, the inward pressure of the air outside the balloon and the inward
pressure from the stretched rubber will again be balance.
The principles of the kinetic theory can also explain why a balloon
expands if you warm it. The temperature of a gas is a measure of the average
speed of the motions of its particles. As the air in a balloon absorbs heat,
the air molecules move more rapidly. As the molecules move more rapidly, they
collide more frequently with the inside surface of the rubber membrane, and
each collision occurs with a stronger impact. As a result, the inside pressure
on the membrane increases and the balloon expands.
Exercise
13.2
Explain
each of the following events in terms of the kinetic theory :
(a)
A
tire is punctured and goes flat
(b)
A
tire expands as the car is driven to higher altitudes. ( At higher altitudes
the air is less dense, so
fewer air molecules bombard the outside surface of the tire each second.)
13-3
Three mathematical laws- Avogardo’s Law, Boyle’s Law and Charles’s Law describe
the relationships between the volume of a gas and its mass,preassure, and
temperature.
Avogadro’s
Law
Imagine that we have a sample of hydrogen gas contained in a
cylinder by a piston, as shown in the first drawing in Figure 13.5. The number
of moles (n) of hydrogen is 1.00 its volume
(V) 250ml and its temperature (T) is 300 K.
The pressure exerted downward on the piston is 98.5 atm and this
pressure is exactly balanced by the pressure (P) exerted upward on the piston
by the sample of hydrogen gas.
Now imagine that we suddenly double the amount of gas in the
sample, from 1.00 mol to 2.00 mol by injecting another mole of hydrogen into
the eylinder. The immediate effect as shown in the middle drawing must be to
double the pressure on the underside of the piston,since we now have twice as
many gas molecules bombarding the face of the piston every second. Because the
downward pressure on the piston has stayed the same while the upward pressure
has doubled, the piston wil rise. As the piston rises,the volume of the sample
will increase,the hydrogen molecules will have more room to move,and three will
be fewer collisions per second with the piston face. As a result,the pressure
of the gas will decrease as the piston rises. Eventually as shown in the drawing
on the right,the pressure exerted upward by the gas will again be the same as
the pressure exerted downward on the piston and the piston will come to rest at
a new position. The new volume will be twice the original volume.
If we double the number of moles of gas in a sample at constant
temperature and pressure, the volume of the sample will double . Similarly, if
we reduce the number of moles of gas in a sample by half, the volume of the
sample will decrease by half (imagine going from the system shown o the right
in Figure 13.5 to the system shown on the left). Avogadro’s Law states that,at constant temperature and pressure,
the volume of a gas sample is directly propotional to the number of moles of
gas in the sample. The law can also be stated mathematically :
V1 /
V2 = n1 / n2
V1
is the volume of a gas samle that contains n1 moles of gas , and V2 is the
volume of a sample that contains n2 moles at constant T and P.
In figure 13.5
V1= 250 mL
V2= 500 mL
n1= 1.00 mol
n2= 2.00 mol
we
can show that these values are consistent with Avogadro’s law by solving the
law for V2 and substitusing the values for V1, n1,
and n2 into the equation:
V2 =
500 mL
the equation for avogadro’s lau
contains symbols for four variables. If we have numerical values for any three of them, we can solve
for the fourth.
Exercise
13.3
A
sample of 0:350 mol of helium gas had a volume of 2.25 L. The temperature and
pressure were held constant, more helium
was added, and the new volume was 3.75 L. How many moles of helium were added?
Boyle’s
law
In the middle of the seventeenth century , the english chemist
robert boyle (1627-1691) carried out a series of experiments that revealed
the relationship between the volume of a gas sample and its pressure. To
understand boyle’s discovery, imagine that we begin again, as shown in the
drawing in figure 13.6, with 1.00 mol of hydrogen gas contained in a cylinder
by a piston. The volume of the sample is 250 mL, its temperature is 300 K, and
its pressure is 98.5 atm. Now imagine that the downward presure on the piston
is doubled. The piston will move downward. As it does so, the molecules of hydrogen wiil be crowded into a smaller
space, and the frequency of collisions of hydrogen molecules with the face of
the piston will increase; that is, the pressure of the hydrogen gas will
increase. At some new and lower position of the piston, the increaset downward
pressure will again be balanced
by the upward pressure of the gas, as shown in the right-hand drawing in figure
13.6.
Boyle’s law states that, at constant tenperature, the volume of a gas sample is
inversely proportional to its pressure. In mathematical form,
V1 = is the volume of gas
sample at pressure P1, and
V2 = is its volume at
pressure P2, at constant T
and n
In our example,
V1 = 250 mL
V2 = 125 mL
P1 = 98.5 atm
P2 = 197 atm
V2
=
=
Exercise
13.4
Imagine that you have the sample of hydrogen
gas shown onthe left in figure 13.6. Now imagine that the pressure exerted
downward by the piston is changed to 49.2 atm. Calculate the new volume of the sample.
Exercise 13.5
A sample of 2.50 mol of nitrogen gas, confined
in a cylinder by a piston at 40
and a pressure
of 375 torr, had a volume of 330 L. The pressure was changed, and the
sample had a new volume of 52.5 L, at the same temperature . what was the new
pressure , in atmospheres?
Carles’s Law
Imagine again that we have. 1.00 mol of
hydrogen gas contained in a chylinder by a piston, as shown in figure 13.7; the
volume of the sample is 250 mL, its temperature is 300 K, and its pressure is 98.5 atm.
Nomw imagine that we double the temperature of our sample of hydrogen,
form 300 K to 600 K, while holding the number of moles of gas and the pressure
constant. As the temperature increases, the average speed at which the hydrogen
molecules are moving will increase. This increased speed will cause an
increased frequency of collisions of hydrogen molecules with the face of the
piston and make each collision more forceful. As aresult, the gas pressure will
increase and the piston will move upward. As the piston rises, the hydrogen
molecules will have more room to move, so the frequency of their collisions
with the face of the piston (the pressure of the gas) will decrease. The piston
will again come to rest when the upward
pressure of hydrogen on the piston equals the downward pressure on it. The new volume
will be twice the original volume.
charles's law establishes the
relationship between the volume of a gas sample and its temperature: At
constant pressure, the volume of a gas sample is directly proportional to its
absolute temperature. in mathematical form,
V1
is the volume of a gas sample at temperature T1, and
V2
is its volume at temperature T2, at constant P and n
the mathematical form of charles's law is
true only for temperatures expressed on the absolute scale, and not on the
Celsius or Fahrenheit scales.
an
illustration of charles's law: as air-filled balloons are immersed in liquid
nitrogen at 77 K, they shrink to much smaller volumes; when the balloons are
removed from the liquid nitrogen they warm up and expand to their original
volumes. (Charles D. Winters)
in
the exsample shown in figture 13.7:
V1
= 250 mL T1 = 300 K
V2
= 500 mL T2 = 600 K
V2
=
=
exercise
13.6
a
sample of fluorine gas had volume of 175 mL at a temperature of 325 K and a
pressure of 700 torr. to what temperature would the sample have to be cooled to
change its volume to 150 mL, at the same pressure?
graphs
based on charles's law show that absolute zero, the lowest limit of
temperature. occurs at about -273
. for example,
the graph figure 13.8 shows plots of volume versus temperature for samples of
oxygen gas and carbon dioxide gas, at constant pressure. if the line for each
gas is extended to a theoretical volume of zero, the corresponding temperature
is about -273
figure
13.8
graph
of charles's law for O2 and Co2. each blue line is a plot of actual
measurements of temperature and volume for a sample of gas, at constant
pressure. if these plot are extended, as shown by the red lines, they converge
on a theoretical volume of 0 mL as about
-273
.
13-4 Gas-Law
calculations can be passed on memorized
equations or on reasoning
Avogadro’s
law, Boyle’s law, and Charles’s law describe relationships among the number of
moles (n)
of gas in a sample ,its volume (V) , its pressure (P) , and its temperature (T).
Avogadro’s law :
at
constant T and P
Boyle’s law :
at
constant T and n
Charles’s law :
at
constant T in K , at constant P and n
You can use
these laws to make calculations in one of two ways; by substituting values into
the preceding equations or by reasoning from what you know about the behavior
of gases. The following example illustrates the difference between these two
approaches.
Example 13.1
A sample of helium gas has a volume of 75.0 ml at 750
torr and 23,6°C , What will its volume be at 750 torr and 41,4°C
Solution
The first step in solving this problem is to recognize that
it describes a gas sampel with a constant number of moles(n)
of gas and the constant pressure(P)
. We know that n is constant because there is no mention in the problem
that any helium is added to or taken from the sample during its temperature
change. We know that P is constant the
pressure is 750 torr before and after the temperature change.
At the next
step we have a choice of how the procced. One approach treats the example as a
problem in mathematics. We identify it as a Charles’s –law problem, a change in
V and T at constant n and P . Then we write the equations for Charles’s law ,
rearrange it to solve for V2 and substitute valus from the problem
into it, remembering to change the tempratures from celcius to kelvin
In the other approach we think about the problem , not in
terms of mathematics , but in term of what we know about gases behave .
According to the problem , the temperature of gas sample is increasing . Gases
expand when they’re heated, so the final volume of sample will be large than
its onital volume. To calculate the final volume , we’ll multiply the initial
volume by a fraction , made by putting one tempraturre over the other . there
are two possibilities:
We know that
the second calculation must be the correct one, because it shows a volume
increase.
Each of
these approaches to a gas-law calculation has its advantages . Working from a
memorized equation avoids the effort if reasoning. Working by reasoning avoids
memorization of equations, avoids the steps of rearranging the equation to
solve for a specific unknown, and minimize the possibility that a mistake in
mathematics will lead to wrong answer.
The two
approaches can also be combined, so that a calculation is set up with a
memorized equation and checked by reasoning. This method, illustrated in
example 13.2, is probably the one most commonly used the solve gas-law
problems.
Example 13.2
A sample of methane (natural
gas, CH4) had
a volume of 1.44L at 23,0°C and 2.62 atm. The pressure on the sample was
increased to 5.11 atm, at 22.0°C. Calculate the new of the sample.
Solution
V2
=
=
= 0.738 L
The volume (V) and pressure (P) of the sample are changing while its number of moles(n) and its temperature (T) are constant, so this is a
Boyle’s-Law problem.
V2 =
=
= 2.99 L
We can decide whether the answer is a reasonable, based on
what we know about the behavior of gases. The pressure on the sample was
increased, so its volume should have decreased. Our answer shows a final volume
smaller than the initial volume, so the answer is reasonable one. When the pressure on a gas sample
is increased, its volume should decrase, but our calculation shows a volume
increase. Reasoning points out the mistake and lets us correct it.
As you work the following Exercise, use your understanding of
the behavior of gases to verify that your calculations are reasonable.
Exercise 13.7
A sample of 0.549 mol of chlorine gas had a volume of 1.40 L
at 30.0°C and 1.25 atm. Calculate the volume of 0.862 mol of chlorine
at same temperature and pressure.
avogadro's
law, boyle law, and charles's law can be combined into one law.
avogadro's law, boyle law,
and charles's law describe the changes in the volume (V) of a gas sample that
occur when we change the number of moles (n) of gas, the pressure (T).
Avogadro’s law boyle
law charles’s
law
these equations can be
combined into one equation, called the combined gas law.
the combined gas law
can be used in the place of any of the three laws it contain. the following
exsample uses it instead of boyle's law.
example 13.3
a sample of nitrogen
gas has a volume of 3.75 L at 130°C and 1.10 atm. calculate its volume at 130°C and
0.900 atm.
solution
in this problem,the
temperature remains constan , so T1=T2 and the number of the moles of gas
remains constant, so n1=n2. because of these equalities, T1 and T2 cancel one
another and disappear from the equation, and the same is true for n1 and n2.
Solve for
:
use the combined gas
law to solve the following proble.
Exercise 13.10
a sample of oxygen gas
had a volume of 85.0 mL at 290 K and 1.00 atm. the gas was heated at constant pressure, and its
volume changed to 125 mL. what was its new temperature ?..
the combined gas law
contains eight variables, and can it be used to solve for any one of them, if
values for the other seven are known.
example 13.4
a sample of 0.650 mol
of hydrogen gas, contained in a cylinder by a piston, had a volume of 13.8 L.
the temperature of the gas was 25.0'c and the pressure on it was 1.15 atm. the
temperature was changed to 20.0'c, the pressure was changed1.50 atm, and 0.150
mol of hydrogen was edded to the sample. calculate its new volume,
solution...
reasoning
can show whether we've set this calulation up correctly. the problem describles
a pressure increase, so the effect of the pressure change should be a volume
decrease. in our calculation, we multiply the initial volume by 1.15 atm/1.50 atm, which will lower the
volume, so we've set the calculation up correctly. similarly the number of
moles of gas is being increased, causing a volume increase, so we multiply by
0.800 mol/0.650 mol. the temperature is being decreased, causing a volume
decrease, so we multiply by 293 K/ 298 K.
use the combined gas law to solve the
following problem, and use reasoning to check your calculation.
Exercise 13.11
a sample of 0.500 mol
of helium gas had a volume of 12.5 L at 280 K and 700 torr. the pressure and
temperature were changed. the new temperature was 310 K and the new volume was
12.0 L. what was the new pressure ?
13-6 the most
general and most useful gas law is the ideal equation, PV=nRT.
the left and right
sides of the combined gas law refer to a sample of gas under two different sets
of conditions of P, V, n, and T
The
two sides of this equation are equa, meaning that, if we have any values of P.
V.n, and T for a gas sample, the value of PV/nT must always be the same :
Eachof
these fractionshas the same value, and the symbol R represents their common
value:
One
equation summarizes all of these relationships:
This
equation, called the ideal- gas equation, is usually written
in this form :
PV=nRT
An
ideal gas that behaves exactly as the kinetic theory and the gas law predict.
In the problems in this book, we’ll assume that all gases behave as if they
were ideal gases.
We
can find the numerical value of R and its unit from the information in figure
13.9. which shows that 1.00 mol of any ideal gas confined in a cylinder by a
piston at a pressure of 1.00
atm and a
temperature of
(273 K) will have a volume of 22.4 L. By
substituting these values of P, V, n, and T in the ideal aquation, we can
calculate the value of R:
PV = Nrt
R =
The
value of R is abbreviated 0.0821 atm-L/mol-K. The units, atm-L/mol-K, are
necerssary for R to function in the ideal-gas aquation is solved for one of its
variables, as shows in the following examples.
Example 13.5
A
sample of 0.500 mol of hydrogen gas had a volume of 8.00 L at a pressure of
2.63 atm. What was its temperature ?
Solution
PV
= nRT
T
=
= 512 K
Example
13.6
A
sample of neon gas had a volume of 8.37 X
mL at a pressure of 587 torr and a temperature
of 42.0
. what was the
mass of the sample in grams ?
Solution
P
= 587 torr = 0.772 atm
V
= 8.37 X
mL = 8.37 L
T
= 42.0
= 315 K
PV
= nRT
N
=
Mol
Ne ®
g Ne
(
)= 5.05 g Ne
Use
the ideal-gas equation to solve the following Exercises.
Exercise 13.12
A
sample of 2.50 mol of oxygen gas had a volume of 40.0 L at 290 K. Calculate its
pressure.
Exercise 13.13
Calculate
the volume of 5.00g of chlorine gas at 125
and 600 torr.
In
discussing the behavior of samples of gases it’s often useful to compare them
under the same conditions of temperature and pressure. The temperature 0
and the pressure 1.00 atm are used as standard
conditions for gases and are referred to as standard temperature and pressure, abbreviated STP. As shown in
figure 13.9 on page 281, 1.00 mol of an ideal gas occupies 22.4L at STP; for
this reason, 22.4 L is called the molar
volume of an ideal gas at STP.
13-7 GAS
VOLUMES CAN BE USED IN STOICHIOMETRY
Avogadro’s law shows that, at constant
temperature and pressure, the volume of a gas sample will vary directly with
the number of moles of gas in the sample:
In other words, as constant temperature and
pressure, the volume of a gas sample is a direct measure of the number of moles
of gas in the sample.
As we saw in chapter 12, a balanced chemical
aquation can be read in moles, and the corresponding molar relationships can be
used to provide factors for stoichiometry. For example, the equation
Can be read 3 H2(g) + N2(g)
2 NH3(g)
3
mol H2 1 mol N2 2 mol NH3
Because the volumes of gas samples, at constan temperature and
pressure, are measures of the numbers of moles of gases in the samples,
quantities of reactans and products that are gases can also be expressed in
volume units. For example, the equation.
Or
in any other volume units. The relationship of the volume unit shown in the
balanced chemical equation can be used in stoichimetry, just as molar
relationship are used.
Example 13.7
Assumming
that the temperature and pressure are constant, how many liters of ammonia can
be produced from 4.25 L of hydrogen, according to the following equation?
Solution
This
equation can be read in moles, and, because at constant T and P te volume of a
gas sample is a measure of the number of moles of gas in the sample, te
equation can also be read in volume units:
These
volume relationsips are used in stoichiometry just as molar relationships are
used:
4.25
L=? L NH3
Example:
Assuming
that the temperature and pressure are constant, how many milliliters of
chlorine will be needed to produce 50.0mL of hydrogen chloride, according to
this equation:
Solution
At
constant T and P, the quantities of gases shown in a balanced chemical equation
can be read in any volume units:
Solve the following Exercise by a
stoichiometric using gas volumes.
Exercise 13.14
The reaction
Was carried out at constant temperature
and pressure, using 53.6 mL of hydrogen. How many millilitersof oxygen were
required for the reaction?
The gas law can be combined with stoichiometry to make calculations
in which quantities of reactants or products that are gases are expressed in volume
units, in moles, or in mass units, as shown in the following Examples and
Exercises.
Example
13.9
At
STP, how many grams of oxygen will react with 135 mL of sulfur dioxide,
according to the following equation?
Solution
At
STP, 1.00 mol of an ideal gas has a volume of 22.4 L (Section 13-6), so, at
STP,
1.0
mol
of O2=? & O2
135
mL SO2=? & O2
The
problem can also be solved this way:
Inside
chemistery : why is it important to prevent the loss of ozone from the
atmosphere ?.
Light from the sun
includes ultraviolet radiation (section 3-8), which is potentially harmful to
animals because it can penetrale and damage delicate animal tissues. In human
beings ultraviolet rays can cause blindness and skin cancer.
Two moleculer
forms of okygen in the otmosphere absorb much of the harmful ultraviolet
radiation from the sun before it an reach the surface of the earth. Diatomic
oxygen
, which makes
up about 20% of the atmosphere, absorbs radiation with shorter wavelengths in
the ultraviolet range. Tritomic oxygen,
, which occurs
in small amounts in the upper atmosphere, bsorbs longerwavelenggth ultraviolet
radiation. The triatomic form of oxygen is called ozone.
Evidence
gathered during the past ten years shows that the ammount of ozone in the upper
atmosphere is decreasing. Because a decrease in the ozone layer increases our
risk of harm from ultraviolet radiation, there has been an urgent research
effort to find the cause of ozone depletion.
This
research has shown that the loss of ozone in the upper atmosphere is probably
caused by gases called chlorofluorocarbons, often referred to as CFCs. Two
example are
and
:
F
F
Throughout the past there decades, great
quantities of CFCs have been used for many purpose. There are probably CFCs in
your home now, as the circulating fluid i the refrigenerator and as the
propenlant gas in spray cans of such products as deodorant, shaving cream, and
spray paint. When CFCs escape from their containers, they move upward though
the atmosphere and eventually reach the ozone layer.
In
the upper atmosphere CFCs may destroy ozone in several ways. In the one
example, the process occurs in two steps. The first steps is the reaction of a
CFC molecule with light to form a chlorine atom.
The
second steps is the reaction of the chlorine atom with an ozone molecule.
The two-step process shown in the equation
above is especially effective destroying ozone because the molecule of chlorine
monoxide that is produced in the second step can be converted back into a
chlorine atom. In the upper atmosphere, there are oxygen atom. A molecule of
chlorine monoxide will react with an oxygen atom.
The
chlorine atom that’s released in this reaction can react with another ozone
molecule. Because a chlorine atom can recycle in this way, one chlorine atom
from one CFC molecule an destroy many
ozone moleules.
The
evidence presented by environmental chemists to show that CFCs are probably
depleting the ozone layer has led to important steps toward eliminating this
potential hazard. In 1990, one hundred nations agreed, throught the united
nations environment program, that a ban on the use of CFCs will begin in the year
2000.
Second step can be
converted back into a chlorine atom.in the upper atmosphere, there are oxygen
atoms. A molecule of chorine monoxide
will react with an oxygen atom :
Chlorine monoxide ,molecule ,oxygen atom, chlorine atom,oxygen
molecule,
The chlorine atom thats released in this reaction can react with
another ozone molecule. Because a chlorine atom can recycle in this way, one
chlorine atom from one CFC molecule can destroy many ozone molecules.
The evidence presented
by environmetal chemists to show that CFCs are probably depleting the ozone
layer has led to important steps toward eliminating this potential hazard. In
1990,one hundred nations agreed, throught the United Nations Environment
Program,that a ban on the use of CFCs will begin in the year 2000.
|
Section
|
Subject
|
Summary
|
Check when learned
|
|
13-1
13-1
13-2
13-3
13-3
13-3
13-5
13-6
13-6
13-6
13-6
13-7
|
Ways to describe the quantity of gas in a sample
Pressure units :
Atmosphere (atm)
torr mm of Hg Pascal (pa)
Description of gases according to kinetic theory
Avogadro’s Law
Boyle’s Law
Charles Law
Combined gas Law
Ideal gas
Ideal gas equation,including meaning and units for
each term
STP
Molar volume of an ideal gas
Basis for using gas volume in stoichiometry
|
State the mass of the sample or state its volume, temperature,and pressure.
Presure exerted by the earth”s atmosphere at sea
level .
Defined by the 1 atm = 760 torr.
Same as torr .defined by 1 atm = 101 k Pa.
(1)A gas consists of particles (atoms or molecules).
The total volume of the particles is negligible compared with the total
volume of the gas.(2) The particle are in rapid,random motion, and thy
constantly collide with one another and with the walls of theirs container .
The collisions are assumed to be perfectly elastic . (3) The number of
particles and their motion are responsible for the volume, temperature, and
pressure of the sample.
V1/V2 = n1/n2 at constant T and P , V1 is the volume
of a gas sample that contains n1 mol of gas,and V2 is the volume of a sample
that contains n2 mol.
V1/V2 = P2/P1 at constant n and T .V1 is the volume
of gas sample at pressure P1 and V2 is the volume of the sample pressure P2.
V1/V2 = T2/T1 at constant n and T.V1 is the volume of
a gas sample at pressure T1,and V2 is the volume of of the sample at
temperature T2. Temperatures must be in K.
P1V1/n1T1=P2 Temperaturs must be in K.
Gas that behaves exactlyas the kinetic theory and the
gas laws predict.
PV=nRT .P is pressure in atm ; V is.27 volume in L .n is quantity of gas
in mol ; R is gas constant, 0.0821 atm-L/mol-K ;T is temperature in K.
Standard temperature and pressure ; 273 K and 1 atm.
The volume,22,4 L,occupied by 1,00 mol of an ideal
gas at STP.
At constant T and P the volume of a gas sample is
directly proportional to the number of moles of gas in the sample,so the
quantities of gases in a balanced chemical equation can be read either in
moles or in any volume units.
|
|
Problems
Assume you can
use the periodic table at the front of the book
unless you’re directed otherwise.Answer to odd-numbered problems are in
Appendix 1
Units for
Mass,volume,Temperature,and Pressure ( Section 13-1)
1.
Make these conversions : (a)
345 mg of O2 to g of O2 .(b) 6.88 g of 02 to mol of O2 . (c) 763 mg of O2 to
mol of 02.
2.
Make these conversions : (a)
394 mL to L (b) 1.44 L to mL (265 cm2) to L.
3.
Make these conversions :
(a) 122⁰F to ⁰C (b) 34⁰C to K (c)-137⁰C to K.
4.
Make these conversions : (a)
0.494 atm to torr (b) 855 torr to atm (c) 337 kPa to atm.
5.
A sample of chlorine gas had a
mass of 2.00 g and a volume of 561 mL at a pressure of 950 torr and and a
temperature of 33⁰C. Convert these
units to mol,L,atm,and K.
6.
A sample of hydrogen gas had a
mass of 0,777 kg and a volume of 13,7 dL at
pressure of 1213 mm of Hg and a temperature of -15⁰C.convert
these units to mol,L,atm,and K.
7.
A sample of ammonia gas had a
mass of 933 mg and a volume of 825 torr and a temperature of 135⁰C.Convert these units to mol,L,atm,and K.
8.
A sample of methane gas had a
mass of 4.29 lb and a volume of 921 in at a temperature of 242⁰F and a pressure of 953 mm of Hg. Convert these units to
mol,L,atm,and K.
9.
Imagine that a mercury
barometer is carried from the top off a mountain to its base.Would you expect
the height of the column of mercury in the glass tube to increase or
decrease?Explain.
10.
Imagine that a mercury
barometer is placed in a sealed chamber and that the air in the chamber is then
removed by a pump. As the air is pumped out,would you expect the column of mercury
in the glass tube to rise or fall ? Explain.
11. Imagine that a sample of gas is contained in a
cylinder by a piston. If you push do on
the piston, you’ll feel increasing
resistance the farther down you push it. Use the kinetic theory to explain this
fact.
12. Using
the kinetic theory, explain why a tire expands as it’s inflated.
13. Using
the kinetic theory, explain why a ballon collapses when it’s puncutered.
14. Using
the kinetic theory, explain why cooling the air in a ballon will cause the
ballon to shrink.
15. Using the kinetic theory, explain why
a ballon expands as it rises through the earth’s atmosphere.
16. In
a common experiment in basic physics, the air is pumped out of a one-gallon
metal can, and the can collapses. Use the kinetic theory to explain why the can
collapses.
17. In
theory, all of the oxygen in a room could move to one corner, so that anyone in
the room would suffocate. Use the kinetic theory to explain why this danger is
not spignificant.
18. If
you inflate two identical ballons to the same size, one with helium and the
other with air, both ballons will eventually deflate as gas leaks out of them
through the rubber membrane, but the helium-filled ballon will deflate more
rapidly than the air-filled ballon. Can you sugest a reason?
19. Methanol (wood alcohol) is a colorless liquid that has
a density of 0.792 g/mL and boils at 65oC. At 215oC and
1.25 atm, 355 mg of methanol gas has a volume of 356 mL. Assuming that, in the
liquid state, the molecules are packed close to one another, so the volume of
the molecules, calcute the percentage of the methanol gas that is empty space.
20. Ethanol
(beverage alcohol) is a colorless liquid that has a spesific gravity of 0.798
and boils at 78.5oC. At 300oC and 0.800 atm, 500 mg of
ethanol gas has a volume of 635 mL. Assuming that, in the liquid state, the
molecules are packed close to one another, so the volume of the molecules,
calcute the percentage of the methanol gas that is empty space.
11. Avogadro’s
Law, Boyle’s, and Charles’s Law (Section 13-3 and 13-4
21. Use the kinetic theory to explain Avogadro’s
Law.
22. Use the kinetic theory to explain Boyle’s Law.
23. Use the kinetic theory to explain Charles’s
Law.
24. Write a mathematical statement of this Law: At
constant volume, the pressure of a sample of gas is directly proportional to
its absolute temperature.
25. Use the kinetic theory to explain the Law
stated in problem 24.
26. If the preassure on a sample of gas is doubled
and its absolute temperature is doubled, what will be the change in its volume?
27. A sample of 0.232 mol of fluorine gas had a
volumeof 8.49 L at 525 torr and 35oC. Calculate its volume at 1.25
atm and 35oC.
28. A sample of 0.488 mol of neon gas had a volume
of 17.4 L at 0.921 atm and 128oC. The pemperature of the sample was
changed while its pressure was held constant, and its new volume was 844 mL.
What was its new temperature, in oC?
29. A sample of 0.232 mol of helium gas had a
volume of 2.80 L at 355 K and 2.41 atm. Calculate its volume at 255 K and 2.41
atm.
30. A sample of
0.741 molof oxygen gas had a volume of 12.7 L at 315 K and 1.51 atm. The
temperature of the sample was increased to 415 K, at constant pressure, and the
sample expanded to a new volume. To what value perature, to change the volume
of the sample from its new value back to its original value?
31. A sample of 0.0500 mol of ammonia gas had a
volume of 235 mL. The temperature and pressure were held constant, and some
ammonia was removed from the sample; its new volume was 175 mL. How many grams
of ammonia were removed?
32. A sample of oxygen gas had a volume of 0.755 ft3
at 25oC and 85.0 kPa.
Calculate its volume in mL at 25oC and 794 torr.
BAHASA INDONESIA
GAS IDEAL
Kata
pengantar
Kita hidup pada dasar lautan gas yang disebut dengan atmosper dan
kita sangat bergantung pada gas – gas tersebut agar dapat bertahan hidup. Akan
tetapi kita sering kali tidak menyadarimya karena gas tidak kelihatan dan lebih
nyata dibandingkan dengan materi yang ada disekitar kita seperti cairan dan
benda – banda padat.
Gas adalah materi yang sulit untuk diamati oleh ahli kimia, karena
beberapa hal diantaranya: beracun, (gas beracun dan bersifat explosiv) atau
dapat meledak. Peralatan – peralatan khusus sangat dibutuhkan untuk mengukur
beberapa kuantitas gas dan memindahkan gas dari suatu tempat ketempat lain,
maka dari itu beberapa ilmuwan para ahli kimia lebih memilih mengamati dan
meneliti benda cair dan padat dibandingkan gas.
Pada abad 17 dan 18 peneliti memulai penelitian pertama tentang gas
secara sitematik dan mereka menemukan bahwa sifat gas lebih simpel dibandingkan
cairan dan benda padat dan dalam hal ini gas lebih mudah dimengerti. Dalam cairan dan benda padat, atom , molekul
–molekul, ion – ion, terpaketkan secara berdekatan dan dalam kumpulan ini
partikel satu berinteraksi dengan partikel lainnya dengan cara yang sangat
kompleks. Namun pada gas, atom, molekul, ion – ion, terpisahkan dengan jarak
tertentu dan cara partikel tersebut berinteraksi dengan cara yang sederhana
hasilnya gas bersifat lebih sederhana dibandingkan cairan dan benda padat. Hal
inilah yang menjadi dasar dan pendorong Jhon Dalton untuk menemukan teori atom
13-1 kuantitas gas dapat digambarkan dengan volume, suhu, massa,
tekanannya.
Cara termudah dalam menggambarkan kuantitas gas adalah dengan cara
menyabutkan volumenya, volume gas disimbolkan dengan mg. Cara lain yag dapat
digunakan adalah dengan menyebutkan volume dengan satuan ml ataupun liter,
suhunya dalam
atau K, tekanannya. Atmosmfer, torr dan kili
pascal (kpa). Hal yang perlu diperhatikan adalah volume gas berbeda pada suhu dan tekanan tang berbeda.
Atmosfer, sebagai unit atau satuan dari tekanan. Harus diperhatikan
penggunannya kareana tekanan sangat dipengaruhi oleh elevasi. Sebagai contoh
atmosfer lebih kecil dari puncak sebuah bukit dibandingkan dibagian dasarnya.
Satu torr= 1 ml mercuri
Satu atmosfer (atm)=760 torr
Satu atm=101 kilo pascal (kpa)
13-2
gas dapat digambarkan sebagai kumpulan benda sangat kecil, partikel
dengan perpindahan tercepat diareal yang kosong diantara partikel-partikel
tersebut.
Bayangkan jika parfum diletakkan dalam sebuah ruangan tertutup dengan
sifat gas yang mampu bergerak cepat kesegala arah maka hanya dalam beberapa
menit aroma parfum tersebut akau tercium didetiap sudut dan sisi ruangan.
Partikel-pertikelgas juga dapat bertubrukan dengan berbagai objek dalam ruangan
seperti dinding,lantai dan ruangan itu semdiri. Kesimpulan dari parfum tersebut
dapat menggambarkan sifat gas pada umumnya yang dapat disebut sebagai “teori
kinetik gas”
1. Sebuah gas mengandung
pertikel-partikel dalam jumlah banyak
2. Partikel-partikel gas
3. Jumlah partikel dalam
jumlah dan gerakannya ditentukan oleh volume suhu dan tekanan dari gas itu
sendiri.
Sebagai ilutrasi dari teori kinetik gas kita
dapat menyelesaikan contoh dibawah ini diketahui massa dari oksigen sama dengan
50 gram volume 42,4 ml suhu 25
tekanan 684 torr. Kita dapat menghitung jumlah
partikel dalam khasus ini adalah moleul oksigen yang terkandung didalam sampel
Mg O2 → g O2 → mol O2 → O2
(50mgO2)(
)
= 9.41 x
O2
Didalam contoh soal diketahui bahwa
volume sampel adalah 42,4 ml seberapa banyak kah dari volume tersebut yang diambil oleh molekul oksigen
dan seberapa besar ruang kosong antara molekul tersebut.
Ø Massa jenis dari oksigen dalam bentuk cair
adalah 1,14 mg/L sehingga untuk 50 mg
darioksigen cair akan menjadi (rumus 2 hal 270)
Persentase dari volume gas total yang diambil
untuk mengisi molekul oksigen adalah
Mg O2 → g O2 →ml O2
(5 mg O2)(
=0.0439 ml O2
Didalam contoh sampel hanya mengandung
oksigen sebanyak 0,04 % sedangkan sisanya adalah ruang kosong antar
molekul-molekul itu. Persentase dari ruang kosong adalah 100% -1.04 % - 9,99%
contoh lain yang dapat digunakan untuk
menjelaskan untuk menggambarkan sifat kinetik gas adalah ketika kita meniuop
balon ataumemompa sebuah ban. Semakin banyak molekul-molekul gas yang
dimasukkan kedalam balon tersebut maka tekanannya akan semakin meningkat
disebabkan karena jumlah partikel gas yang bertubrukan dengan dinding wadah ,
semakin meningkat. Jika kamu menambah udara lebih banyak kedalam balon, maka
akan meningkatkan jumlah molekul kedalam balon itu semdiri dan ini akan
menyebabkan banyaknya molekul udara yang bertubrukan dengan dinding bagian
dalam membran (dalam balon). Tekanan udara didalam balon akan semakin tinggi
dan balon akam meledak.prinsip dari teori kinetik gas dapat dijelaskan saat
sebuah balon meledak ketika kamu memanaskannya.ketika udara didalam balon
mengabsorbsi panas maka molekul udara di dalamnya akan bergerak sangat cepat,
dan semakin cepat gerakan itu terjadi dan semakin cepat gerakan itu terjadi
membuat frekuensi bertubruknya molekul tersebut tinggi pada akhirnya menbuat
tekanan didalam balon meningkat dan balon meledak.
Tiga hukum matematika (avogadro), hukum boyle, hukum
charles, didalam menggambarkan volume gas dengan massa tekanan dan suhu.
hukum avogadro
Bayangkan kita memiliki sebuah sampel
yaitu gas hidrogen yang terkandung dalam sebuah silinder berpiston gambar(13.5)
sekarang bayangkan jika tiba-tiba kita menaikkan jumlah gas didalam
sampelsebanyak dua kali dalam jumlah awal yaitu dari satu mol menjadi dua mol.
Dengan menambahkan satu mol hidrogen kedalam silinder tersebut.
Dampak langsung dari perlakuan
tersebut digambarkan dengan gambar ditengah jika kita melipatgandakan jumlah mol
didalam satu sampel pada saat suhu dan tekanan konstan maka volume darigas
tersebut bertambah dua kali lipat dari semula sebaliknya jika molekul dikurangi
hingga setengahnya maka volumenya akan
berkurang hingga dari volume awal.
Hk avogadro :
Pada suhu dan tekanan konstan volume
dalam sebuah gas akan proporsional dengan jumlah molekul yang ada dalam sampel
tersebut. Secara matematika dapat rumuskan :
Keterangan :
VI : volume awal
V2 : volume akhir
N1:
jumlah molekul atom awal
N2: jumlah molekul atom akhir
Pada gambar 1.5 V1 = 250 ml. V2
=500ml.n1 = 100 mol, n2 = 200 mol. Untuk membuktikan apakah hukum avogadro
sesuai dengan masalah tersebut maka silakah tentukan V2 dengan menggunakan
hukum avogadro ?
V2 =
V2 =
V2 = 500ml
1.13 sebanyak 0.350 mol gas
Helium memiliki volume 2,25 L, suhu dan tekanan dari gas tersebut di
pertahankan dan beberapa jumlah mol ditambahkan kedalam sampel sehingga volume
akhirnya menjadi 3,75 L. Berapa molkah
gas Helium yang telah ditambahkan ?
Diiketahui :
V1 : 2,25 L
V2 : 3,75 L
n1 : 0,350 mol
ditanya n2 ?
jawab :
maka : n2 =
n2 =
n2 = 0,583 mol
Hukum BOyle
Hukum Boyle dapat dipahami dengan menyelesaikan contoh: diketahui
n : 1 mol
V: 205 ml
T : 300 K
P : 98,5 atm
Jika volume diperkecil maka tekanan
dalam tabung tersebut akan meningkat. Hk Boyle “: pada suhu yang konstan maka
volume dari gas tersebut berbanding terbalik secara psoporsional dengana
tekanannya atau volume suatu gas berbanding terbalik dengan tekananya. Dengan
rumus matematika:
Hk Charles
Charles menemukan hubungan antara
volume gas tersebut dan suhunya. Hk Charles “ pada tekanan konstan volume suatu
gas berbanding lurus secara proporsional sedangkan suhunya secara matematika
dapat dituliskan
Kesimpulan
1. hk Avogadro “ jumlah mol berbanding lurus dengan volume
2. Hk Boyle “ volume suatu gas berbanding terbalik dengan
tekanannya.
3. hk Charles “ pada tekanan konstan volume suatu gas berbanding lurus”
Hukum Gas Ideal
Sebuah contoh yaitu gas Helium
memiliki volume 75 ml pada tekanan 750 torr dan pada suhu 23,6
. Berapakah
volume gas tersebut ! pada tekanan 750 torr dan suhu 41.4
.
Jawab :
Dik:
V2 =
=
= 79,3 ml
Contoh 13.2
Sebuah gas metal (CH4) memiliki volume 1,44 liter pada suhu 22
dan tekanan2,2 atm. Tekanan dalamsampel
dinaikkan menjadi 5,11 atm pada suhu tetap. Hitunglah volume akhir dari
sampeltersebut.
Jawab : sesuai dengan hukum boyle
V2 =
V2 =
=2,99 liter
Secara matematika
membuktikan bahwa volume 2 meningkat dari volume 1.
Latihan
13.7
Sebanyak 0,549 mol klorin memiliki 1 volume 1,40 liter pada suhu
30
dan P = 1,25 atm. Hitunglah volume dari 0,862
mol klorin pada temperatur dan tekanan yang sama.
Latihan 13.8
Sebanyak 2,125 mol gas asetelin memiliki volume 78,4liter pada
suhu 310 K pada tekanan 675 torr.berapakah volumepada tekananyang sama dalam
suhu 410 K.
Latihan 13.9
Sebanyak 1,25 gas karbon monoksral memiliki volume 12 liter pada
suhu 350 K dan P 540 torr. Hitunglah volume gad pada 350 K dan P = 730 torr.
13.5
Hukum Gabungan/hukum gas ideal
Contoh: sebuah gas nitrogen memiliki volume 3,75 liter pada suhu
130
dan tekanan 3,11 atm. Hitunglah volume gas
tersebut pada suhu 130
dan tekanan 0,900 atm.
Maka :
V2
=
=
4,58 L
Latihan
13.1
Sebuah sampel gas memiliki volume 85
ml pada suhu 24 k dan p= 10 atm gas
tersebut dipanaskan, tekanan tetap maka volume berubah menjadi 125 ml.
Berapakah suhu akhir dari gas tersebut!
Contoh
13.4
Sebuah
sampel gas hidrogen dengan 0,65 mol yang disimpan dalam tabung slinder
berpiston memiliki volume 13,8 liter, suhu dari gas tersebut 25°c dan
tekanannya 1,15 atm. Suhunya diturunkan menjadi 20°c dan p dinaikan menjadi
1,50 atm dan sebanyak 0,15 mol H2 dinyatakan ke dalam sampel tersebut berapakah
volum akhir dari sampel
n2 = n1 + 0,150 mol
= 0,650 mol + 0,150 mol
= 0,800 mol
V2 =
= 12.8 L
Latihan
13.11
Sebuah
sampel yaitu gas helium dengan 0,5 mol gas helium memiliki volume 12,5 liter
pada suhu 25 k dan tekanan 700 torr tekanan dan suhunya telah diubah. Suhu
akhir = 310k, volume = 12 liter berapakah tekanan dari gas helium tersebut
13 – 6
Hukum persamaan gas PV= nRT
gabungan dari hukum boyle , charles, avogadro dapat disimpulkan menjadi R= pv/
n. T
Persamaan ini di sebut sebagai persamaan gas ideal namun
dituliskan dalam bentuk rumus berikut
PV= nRT
n = 1 mol
v= 22,4 liter
T= 273 k
P= 1 atm
Dengan
mensubstupstitusikan P, n, T didalam persamaan gas ideal maka kita dapat
menentukan nilai R.
Pv= nRT
R =
Contoh
13.5
Sebanyak 0,5 mol H2 memiliki volume 8 liter pada tekanan 2,63 atm
berapakah temperaturnya?
Pv=nRT
P =
= 512 K
Sebuah contoh gas neon memiliki volume 8,37 x
ml pada tekanan 857 torr, dan suhu
42
. Berapakah massa dari gas tersebut didalam gram ?
Dik :
P = 587 torr =0,772 atm
V = 8,37 x
ml =8,37 LT = 42.0
= 315 K
Jawab :
PV = nRT
n :
mol Ne → g Ne
(
= 5,05 g Ne
Gunakanlah persamaan gas ideal untuk menyelesaikan permasalan-permasalahan
latihan dibawah ini.
Latihan 13.12
Sebuah sampel 2,50 mol O2 memiliki volume 40 liter pada suhu 290 k
Ditanya tekanan
Latihan 13.13
Ditanya volume dari 500 gram gas klorin pada T= 125°c dan 600 torr.
13-7
Volume gas dapat digunakan dalam stoikiometri.
Hukum avogadro menyatakan bahwa pada suhu dan tekanan yang konstan volume
suatu gas akan berbeda bergantung dari jumlah mol yang dimiliki gas tersebut.
Pada T dan P itu konstan
Dengan kata lain pada suhu dan tekanan konstan, volume gas tersebut
dapatdiukur secara langsung hanya dengan mengetahui jumlah molekul yang
terkandung dalam gas tersebut
3H2(g) + N2(g)
→ 2NH3(g)
3 mol H2 1mol N2 2mol NH3
Karena sifat gas tersebut diatas maka reaksi tersebut dapat dituliskan
sebagai berikut
3H2(g) + N2(g)
→ 2NH3(g)
3L H2 1L N2 2L NH3
3mL H2 1mL N2 2mL NH3
3cm H2
Hubungan antara volume dengan satuan lainnya dalam kesetimbangan kimia
dapat digunakan dalam stoikiometri unit tersebut menyatakan molar
Contoh 13.7
Asumsikan bahwa suhu dan tekanan konstan berapa liter kah amonia yang dapat
diproduksi 4,25 liter H2 berdasarkan persamaan kimia dibawah
3H2(g) + N2(g)
→ 2NH3(g)
Jawab:
Reaksu kimia tersebut dapat dibaca dalam mol dan karena kesetimbangan
reaksi terjadi pada suhu dan tekanan konstan maka secara langsung volume gas
dapat dihitung dengan persamaan reaksi dibawah ini
3H2(g) + N2(g)
→ 2NH3(g)
3L H2 1L N2 2L NH3
Volume 1 mol digunakan dalam stoikiometri yaitu hubungan antara molar dan
volume
4.25 L H2= ? LNH3
L H2→ L NH3
(
= 2,83 L NH3
Contoh 13.8
Pada T dan P konstan berapamili
liter kah klorin yang ditentukan untuk menghasilkan 50 ml gas hidrogen klorin
berdasarkan persamaan reaksi kimia dibawah ini
H2(g) + CL2(g) → 2HCL(g)
Jawab:
H2(g) + CL2(g)
→ 2HCL(g)
1ml H2 1ml CL2 2ML HCL
(50.0 ml HCl)
)
= 25.0 ml Cl2
Jadi jumlah klorida yang dibutuhkan untukmenghasilkan 50 ml HCL adalah 2,5
ml.
Masalah tersebut dapat juga terselesaikan jumlah di bawah ini
ml SO2 → L SO2 → mol SO2 → mol O2 →
g O2
(
(
)(
(
= 0,0964 g O2
Latihan 13.15
Berapa gramkah klorin yang akan dibutuhkan 575 ml carbon tetrafluorin pada
keadaan STP berdasarkan reaksi kimia berikut :
CH4 (g) + 4F2 (g) → CF4 (g) + 4HF (g)
Contoh 13.10
Ketika potasium klorad dipanaskan menjadi potasium klorida dan
O2 : 2KClO3(s) →2 KClS + 3O2 (g) pada suhu 28
dan P 375 torr. Berapa literkah O2
yang akan diproduksi dengan memecah 5,24 g potasium klorat.
Jawab : gunakanlah perhitungan pers reaksi kimia untuk menghitung berapa
mol oksigen yang dihasilkan.
5,24 g KCL O3 ? mol O2
PV = nRT untuk mengkomsumsi 0,0639 g mol O2dari
K = C + 273
= 28 + 273 = 301 K
(
= 0,967 atm
V=
= 1,63 L
Mengapa kita penting untuk mencegah hilangnya
ozon dari atmosfer
Cahaya dari matahari mengandung dari radiasi ultraviolet yang sangat
berbahaya pada binatang karena cahaya tersebut dapat terpenetrasl dan merusak
jaringan tubuh hewan. Pada manusia gelombang ultraviolet dapat menyebabkan
kebutaan dan kanker kulit. 2 molekul O2 di dalam atmosfer mengabsorbsi sangat
banyak radiasi ultraviolet yang sangat
berbahaya dari matahari sebelum cahaya matahari mencapai permukaan bumi. O2 sebagai
bahan pembentukan atmosfir sehingga 20% mengabsorsi radiasi panjang gelombang
pendek. Triatomik O2 ozon terbentuk pada jumlah yang sedikit di atmosfer bagian
atas, mengabsorsi panjang gelombang ultraviolet yang lebih panjang dari absorsi
O2.
Triatomik tersebut disebut sebagai ozon. Berdasarkan data yang telah
dikumpulkan selama sepuluh tahun lalu menujukkan bahwa jumlah ozon diatmosfer
menurun. Karena penurunan jumlah ozon tersebut menyebabkan cahaya dari radiasi
ultraviolet penelitian inimenunjukkan bahwa hilangnya ozon disebabkan oleh gas
CFCS.
Berikut ini adalah dua contoh CFCS yaitu CCL2 F3
Gambar:
Pada tiga dkd
belakangan ini banyak jumlah CFCS telah ditemukan pada banyak tujuan
mereka contohnya temukan saja dirumah
anda misalnya pada kulkas, cat semprot, dan produk – produk semprot seperti
parfum lainnya. Ketika CFCS keluar dari tempatnya mereka akan bergerak hingga
atmosfer dan mencapai lapisan ozon. Diisini dilapisan ozon ini CFCS
menghancurkan ozon dengan dua cara:
1.
moleku CFCS berinteraksi dengan cahaya membentuk atom klorin
2.
atomklorin bereaksi dengan molekul ozon
Langkah
kedua dari reaksi tersebut dapat kembali membentuk atom klorin, monoatom menjadi atom klorin tunggal diatas akan bereaksi dengan atom O2 dengan
reaksi :
Atom
klorin yang dilepaskan dari reaksi diatas dapat bereaksi dengan molekul ozon
lain karena klorin dapat mengulangi reaksinya sendiri maka satu atom klorin dan
CFCS dapat menghancurkan banyak molekul ozon.masalah ini ditunjukan oleh ahli
kimia lingkungan yang menujukan bahwa CFCS memungkinkan terjadinya eliminasi
potensial hazar (bahaya). Pada tahun 1990,100 bangsa menyetujui melalui 4 N
environtmen program bahwa penggunaan CFCS
akan dimulai pada tahun 2000.
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Bagian
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Subyek
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Ringkasan
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Memeriksa ketika
belajar
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13-1
13-1
13-2
13-3
13-3
13-3
13-5
13-6
13-6
13-6
13-6
13-7
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Cara untuk
menggambarkan jumlah gas dalam sampel
Unit tekanan:
Suasana (atm)
mm Hg torr Pascal (pa)
Deskripsi gas menurut
teori kinetik
Hukum Avogadro’s
Hukum Boyle
Hukum Charles
Hukum gas
dikombinasikan
Gas ideal
Persamaan gas ideal,
termasuk makna dan unit untuk setiap jangka
STP
Volume molar gas ideal
Dasar untuk menggunakan
Volume gas di stoikiometri
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Menyatakan massa sampel
atau menyatakan volume, suhu, dan tekanan.
Tekanan yang diberikan
oleh bumi "s atmosfer di permukaan laut.
Ditetapkan oleh 1 atm =
760 torr.
Sama seperti torr .defined
oleh 1 atm = 101 k Pa.
(1) gas terdiri dari
partikel (atom atau molekul). Total volume partikel diabaikan dibandingkan
dengan total volume gas. (2) partikel yang berada di cepat, gerak acak, dan
Mu terus bertabrakan dengan satu sama lain dan dengan dinding wadah mereka.
Tabrakan diasumsikan elastis sempurna. (3) Jumlah partikel dan gerakan mereka
bertanggung jawab untuk volume, suhu, dan tekanan dari sampel.
V1 / V2 = n1 / n2 pada
T konstan dan P, V1 adalah volume sampel gas yang mengandung mol n1 gas, dan
V2 adalah volume sampel yang mengandung mol n2.
V1 / V2 = P2 / P1
konstan n dan T .V1 adalah volume sampel gas pada tekanan P1 dan V2 adalah
volume P2 tekanan sampel.
V1 / V2 = T2 / T1
konstan n dan T.V1 adalah volume sampel gas pada tekanan T1, dan V2 adalah
volume sampel pada suhu T2. Suhu harus dalam K.
P1V1 / n1T1 = P2
Temperaturs harus dalam K.
Gas yang berperilaku
exactlyas teori kinetik dan hukum gas memprediksi.
PV = nRT .P tekanan di
atm; V Volume is.27 di L .n adalah kuantitas gas di mol; R adalah konstanta
gas, 0,0821 atm-L / mol-K; T adalah suhu dalam K.
Suhu dan tekanan
standar; 273 K dan 1 atm.
Volume, 22,4 L,
ditempati oleh 1,00 mol gas ideal di STP.
Pada T konstan dan P
volume sampel gas berbanding lurus dengan jumlah mol gas dalam sampel,
sehingga jumlah gas dalam persamaan kimia yang seimbang dapat dibaca baik
dalam mol atau di unit volume.
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Masalah
Asumsikan Anda
dapat menggunakan tabel periodik di depan buku kecuali Anda diarahkan
otherwise.Answer untuk masalah ganjil dalam Lampiran 1
Unit untuk Misa,
volume, suhu, dan tekanan (Bagian 13-1)
1. Membuat
konversi ini: (a) 345 mg O2 untuk g O2 (b) 6,88 g dari 02 ke mol O2.. (c) 763
mg O2 ke mol dari 02.
2. Membuat
konversi ini: (a) 394 mL ke L (b) 1,44 L untuk mL (265 cm2) untuk L.
3. Membuat
konversi ini: (a) 122⁰F ke ⁰C (b) 34⁰C ke K (c) -137⁰C ke K.
4. Membuat
konversi ini: (a) 0,494 atm ke torr (b) 855 torr ke atm (c) 337 kPa ke atm.
5. Sebuah sampel
gas klorin memiliki massa 2,00 g dan volume 561 mL pada tekanan 950 torr dan
dan suhu 33⁰C. Mengkonversi unit-unit ini ke mol, L, atm, dan K.
6. Sebuah sampel
gas hidrogen memiliki massa 0.777 kg dan volume 13,7 dL pada tekanan dari 1.213
mm Hg dan suhu -15⁰C.convert unit-unit ini ke mol, L, atm, dan K.
7. Sebuah sampel
gas amonia memiliki massa 933 mg dan volume 825 torr dan suhu 135⁰C.Convert
unit-unit ini ke mol, L, atm, dan K.
8. Sebuah sampel
gas metana memiliki massa £ 4,29 dan volume 921 di pada suhu 242⁰F dan tekanan
953 mm Hg. Mengkonversi unit-unit ini ke mol, L, atm, dan K.
9. Bayangkan
bahwa barometer merkuri dilakukan dari atas dari sebuah gunung untuk base.Would
yang Anda harapkan ketinggian kolom air raksa di tabung kaca untuk menambah
atau mengurangi? Jelaskan.
10. Bayangkan
bahwa barometer merkuri ditempatkan dalam ruang tertutup dan udara di ruang ini
kemudian dihapus oleh pompa. Seperti udara dipompa keluar, yang Anda harapkan
kolom merkuri dalam tabung kaca untuk naik atau turun? Menjelaskan.
11.
Bayangkan bahwa sampel gas yang terkandung dalam silinder dengan piston. Jika
Anda mendorong lakukan pada piston, Anda akan merasa meningkatkan daya tahan
lebih bawah Anda mendorongnya. Menggunakan teori kinetik untuk menjelaskan
fakta ini.
12.
Menggunakan teori kinetik, menjelaskan mengapa ban mengembang seperti itu
meningkat.
13.
Menggunakan teori kinetik, menjelaskan mengapa ballon runtuh ketika yang
tertusuk.
14.
Menggunakan teori kinetik, menjelaskan mengapa pendingin udara di ballon akan
menyebabkan ballon yang menyusut.
15.
Menggunakan teori kinetik, menjelaskan mengapa ballon mengembang seperti naik
melalui atmosfer bumi.
16.
Dalam sebuah percobaan dalam fisika dasar, udara dipompa keluar dari logam satu
galon bisa, dan bisa runtuh. Menggunakan teori kinetik untuk menjelaskan
mengapa bisa runtuh.
17.
Dalam teori, semua oksigen di sebuah ruangan bisa pindah ke salah satu sudut,
sehingga siapa pun di dalam ruangan akan mati lemas. Menggunakan teori kinetik
untuk menjelaskan mengapa bahaya ini tidak signifikan.
18.
Jika Anda mengembang dua ballons identik dengan ukuran yang sama, satu dengan
helium dan lainnya dengan udara, baik ballons akhirnya akan mengempis seperti
kebocoran gas dari mereka melalui membran karet, tetapi ballon helium akan
mengempis lebih cepat daripada berisi udara ballon. Dapatkah Anda sugest
alasan?
19.
Metanol (alkohol kayu) adalah cairan tidak berwarna yang memiliki kepadatan
0,792 g / mL dan mendidih pada 65oC. Pada 215oC dan 1,25 atm, 355 mg gas
metanol memiliki volume 356 mL. Dengan asumsi bahwa, dalam keadaan cair,
molekul yang dikemas dekat satu sama lain, sehingga volume molekul, calcute
persentase gas metanol yang ruang kosong.
20.
Etanol (minuman alkohol) adalah cairan berwarna yang memiliki berat jenis 0,798
dan mendidih pada 78.5oC. Pada 300oC dan 0.800 atm, 500 mg gas etanol memiliki
volume 635 mL. Dengan asumsi bahwa, dalam keadaan cair, molekul yang dikemas
dekat satu sama lain, sehingga volume molekul, calcute persentase gas metanol
yang ruang kosong.
Avogadro
Hukum, Boyle, dan Hukum Charles (Bagian 13-3 dan 13-4
21.
Gunakan teori kinetik untuk menjelaskan Hukum Avogadro.
22.
Gunakan teori kinetik untuk menjelaskan Hukum Boyle.
23.
Gunakan teori kinetik untuk menjelaskan Hukum Charles.
24.
Tulis pernyataan matematis dari Hukum ini: Pada volume konstan, tekanan dari
sampel gas berbanding lurus dengan suhu mutlak.
25.
Gunakan teori kinetik untuk menjelaskan Hukum tercantum dalam masalah 24.
26.
Jika tekanan pada sampel gas dua kali lipat dan suhu mutlak adalah dua kali
lipat, apa yang akan menjadi perubahan volumenya?
27.
Contoh 0,232 mol gas fluor memiliki volume 8,49 L pada 525 torr dan 35oC.
Hitung volume sebesar 1,25 atm dan 35oC.
28.
Contoh 0,488 mol gas neon memiliki volume 17,4 L pada 0,921 atm dan 128oC. The
pemperature sampel berubah sementara tekanannya diadakan konstan, dan volume
baru adalah 844 mL. Apa suhu baru, di oC?
29.
Contoh 0,232 mol gas helium memiliki volume 2,80 L pada 355 K dan 2,41 atm.
Hitung volume pada 255 K dan 2,41 atm.
30.
Contoh 0,741 gas oksigen molof memiliki volume 12,7 L pada 315 K dan 1,51 atm.
Suhu sampel meningkat menjadi 415 K, pada tekanan konstan, dan sampel diperluas
untuk volume baru. Untuk apa perature nilai, untuk mengubah volume sampel dari
nilai baru kembali ke nilai aslinya?
31.
Contoh 0,0500 mol gas amonia memiliki volume 235 mL. Suhu dan tekanan yang
konstan, dan beberapa amonia telah dihapus dari sampel; Volume baru adalah 175
mL. Berapa gram amoniak telah dihapus?
32.
Sebuah sampel gas oksigen memiliki volume 0,755 ft3 di 25oC dan 85.0 kPa.
Hitung volume di mL pada 25oC dan 794 torr.
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