modul gas b.ing dan translateannya

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 V_1, n₁, and n₂ into the equation:

V_{2 =}  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 P_2,  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

            V₂  =  =    = 79.3 mL   

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: 

  = 70.9 mL                                or                       = 79.3 mL   

            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.

  or, in a form that is easier to remember,

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:

    at constant T and P

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

Cl         C          Cl                     F          C          Cl

            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 65^oC. At 215^oC 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 78.5^oC. At 300^oC 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 35^oC. Calculate its volume at 1.25 atm and 35^oC.

28. A sample of 0.488 mol of neon gas had a volume of 17.4 L at 0.921 atm and 128^oC. 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 25^oC  and 85.0 kPa. Calculate its volume in mL at 25^oC 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.

Bagian	Subyek	Ringkasan	Memeriksa ketika belajar
13-1 13-1 13-2 13-3 13-3 13-3 13-5 13-6 13-6 13-6 13-6 13-7	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	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.

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.