Download presentation

Presentation is loading. Please wait.

1
Chapter 5 Gases

2
**The Nobel Gases Exist as Monatomic Gases He, Ne, Ar, Kr, Xe, Rn**

Diatomic Gases are H2, N2, O2, F2, Cl2. Oxygen also exists as the gas Ozone = O3

3
**HF, HCl, HBr, HI, CO, CO2, NH3, NO, NO2, N2O, SO2, H2S, HCN**

Compound Gases Ionic Compounds do not exist as gases at standard temperature and pressure (STP = 0 oC and 1 atm). This is because the cations and anions are held together by very strong electrostatic forces. To overcome these forces, we would have to raise the temperature well above 1000 oC. Molecular Compounds vary but their boiling points are much lower. Most still exist as solids or liquids at STP. Some common molecular compounds that are gases at STP are the following: HF, HCl, HBr, HI, CO, CO2, NH3, NO, NO2, N2O, SO2, H2S, HCN Most of these gases are colorless, except F2 (greenish yellow), Cl2 (greenish yellow), and NO2 (brown)

5
**Physical Characteristics of Gases**

Gases assume the volume and shape of their containers. Gases are the most compressible of the states of matter. Gases will mix evenly and completely when confined to the same container. Gases have much lower densities than liquids and solids.

6
The SI Unit of Pressure We need to derive pressure from the 7 SI base units. First, we derive force: Force = mass x acceleration Force = kilogram x m/s2 = 1 newton (N) = the SI unit for force Next, we derive pressure: Pressure = force / area = N / m2 = (kg x m/s2)/m2 = 1 pascal (Pa) Pascal is the SI unit for pressure A sample of gas in a closed container will have “pressure”. This is the same thing as the force that is exerted (from the movement of the molecules) on the walls of the container (the area of the walls of the container). The molecules move so fast that the molecules will exert a constant average amount of force on the container walls.

7
**Instrument used to measure the pressure of the air in the atmosphere.**

= 760 mm Hg A barometer is used to measure the atmospheric pressure. Or the pressure of the gases in the atmosphere surrounding us. At sea level, the pressure of the atmosphere pushes down on the mercury and will exert a force onto the mercury until it rises inside of the inverted tube. The mercury will rise exactly 760 mm because of the pressure of the gases in the atmosphere surrounding the mercury. Therefore, we say that the standard atmosphere (which equals sea level) is equal to 760 mm Hg. This is also equal to 1 atm and is also equal to 760 torr. Because the molecules in the air are subject to the gravitational pull of the earth, the air closer to the earth is denser than the air further away from the earth. This is because gravity pulls down on the molecules of gas in the air. This is why airplanes must be pressurized. Because the air is less dense at higher altitudes and it is too thin to breathe (too little mass of molecules per volume). Of course the denser the air is, the more pressure it will exert (more mass = more pressure).

8
**Used to measure pressures below atmospheric pressure.**

The manometer is the instrument that we use in order tot measure the pressure of any gas other than the atmosphere gas. There are 2 types of manometers: The closed-tube manometer: which is normally used to measure pressures below atmospheric pressure. The open-tube manometer: which is better suited for measuring pressures equal to or greater than atmospheric pressure. Almost all manometers and barometers use mercury as the working fluid even though it is toxic. This is because it has a very high density (13.6 g/mL) compared to other liquids. Because the height of the liquid in a column is inversely proportional to the liquid’s density, this property enables us to construct small barometers and manometers. The closed-tube manometer The open-tube manometer Used to measure pressures below atmospheric pressure. Used for measuring pressures equal to or greater than atmospheric pressure.

9
This type of apparatus was used by Robert Boyle in order to study the behavior of gases. In (a) the pressure exerted on the gas by the mercury added to the tube is equal to the atmospheric pressure. In (b) an increase in pressure due to the addition of more mercury results in a decrease in the volume of the gas and in unequal levels of mercury in the tube. You can then see that in (c) and in (d), the more mercury that is added (which creates more pressure added) creates a decrease in the volume of the gas. From this apparatus, Boyle was able to determine that when temperature is held constant, the volume of a given amount of gas decreases as the total applied pressure – (atmospheric + pressure from mercury) – is increased. This can overall be summed by the mathematical relationship: P is proportional to 1/V, or that the pressure of a fixed amount of gas is inversely proportional to the volume of the gas. P = k/V and PV =k. Therefore k=nRT

10
**(k is the proportionality constant k1 = nRT)**

Boyle’s Law: P a 1/V P = k1 / V and PV = k1 (k is the proportionality constant k1 = nRT) Click on the box to view Boyle’s Law as a movie. At constant temperature

11
Using Boyle’s Law In Boyle’s Law: PV=k1 and k1 is the proportionality constant therefore it will be equal to nRT. For this reason, at constant temperature, a given sample of gas (with the same number of moles = n) at two different sets of conditions (different volumes and pressures) will follow the following equation: P1V1= P2V2 Example: An inflated balloon has a volume of 0.55 L at sea level (1.0 atm) and is allowed to rise to a height of 6.5 km, where the pressure is about 0.40 atm. Assuming that the temperature remains constant, what is the final volume of the balloon? P1V1= P2V2 => V2 = (P1 / P2) x V1 (1.0 atm / 0.40 atm) x 0.55 L = 1.4 L

12
**At Constant Pressure V a T**

This is an example of the effects of variation of temperature on a gas with constant pressure. Notice that the volume of the gas varies when the pressure is constant and the temperature varies. At higher temperatures, a sample of gas will expand and at lower temperatures a sample of gas will contract. Click on the grey circle to view a movie of this as well. Click on the yellow square to view this.

13
**V a T V = k2T V/T = k2 k2 = nR/P P a T P = k3 T P/T = k3 k3 = nR/V**

The relationship between volume and temperature (at constant pressure) is something that the French scientists Jacques Charles and Joseph Gay-Lussac studied. Therefore, this relationship is known as Charles’s and Gay-Lussac’s Law or just simply Charles’s Law. Charles’s law states that: The volume of a fixed amount of gas maintained at constant pressure is directly proportional to the absolute temperature of the gas. You can also say that if you keep the volume constant, the Pressure of a gas is directly proportional to the temperature. Click on the red box to view this in a movie.

14
**Absolute Zero = -273.15 oC = 0 Kelvin**

At Absolute Zero, there is no volume for the gases. Volume is directly proportional to the Temperature in Kelvin. Meaning if you doubled the temp, the volume doubles!

15
Using Charles’s Law In Charles’s Law: V1 / T1 =k2 and k2 is the proportionality constant therefore it will be equal to nR / P. For this reason, at constant pressure, a given sample of gas (with the same number of moles = n) at two different sets of conditions (different volumes and temperatures) will follow the following equation: V1 / T1 = k2 = V2 / T2 Example: A 452-mL sample of fluorine gas is heated from 22 degrees Celsius to 187 degrees Celsius at constant pressure. What is its final volume? V2 = V1 x (T2 / T1) = 452 mL x (460 K / 295 K) = 705 mL

16
V a n V = k4n k4 = RT/P The work of the Italian scientist Amedeo Avogadro complemented the studies of Boyle, Charles and Gay-Lussac. In 1811 he published a hypothesis stating that at the same temperature and pressure, equal volumes of different gases contain the same number of molecules (or atoms if the gas is monatomic). It follows that the volume of any given gas must be proportional to the number of moles of molecules present; that is : Volume is proportional to the number of moles or V = k4n. Click on the red box and then the green circle to visualize this in a movie.

17
**This also demonstrates Avogadro’s Law.**

18
**Gas Law Summary So Far: Boyle’s Law: V a 1/P (at constant n and T)**

Charles’s Law: V a T (at constant n and P) Avogadro’s Law: V a n (at constant P and T) We can combine all three expressions to form a single master equation for the behavior of gases: V a nT / P = R (nT / P) or PV = nRT where R is the proportionality constant called the gas constant. The equation PV = nRT is called the ideal gas equation or ideal gas law. It describes the relationship among the four variables P, V, T and n.

19
**The Ideal Gas Law PV = nRT**

An ideal gas is a hypothetical gas whose pressure-volume-temperature behavior can be completely accounted for by the ideal gas equation. The molecules of an ideal gas do not attract or repel one another, and their volume is negligible compared with the volume of the container. Although there is no such thing as an ideal gas, discrepancies in the behavior of real gases over reasonable temperature and pressure ranges do not significantly affect calculations. Therefore, we can safely use the ideal gas law equation to solve many gas problems.

20
**The Ideal Gas Law Constant, R**

At Standard Temperature and Pressure (STP) – O oC ( K) and 1 atm pressure, many real gases behave like an ideal gas. Experiments show that under these conditions, 1 mole of gas occupies L, which is somewhat greater than the volume of a basketball. From PV=nRT we can write: R = PV / nT R = (1 atm)( L) (1 mol)( K) R = L x atm K x mol

22
**P = (1.82 mol) (0.0821 L x atm/K x mol) (69.5 + 273)K**

Using the Ideal Gas Law Example: Sulfur hexafluoride (SF6) is a colorless, odorless, very unreactive gas. Calculate the pressure (in atm) exerted by 1.82 moles of the gas in a steel vessel of volume 5.43 L at 69.5 oC. P = nRT / V P = (1.82 mol) ( L x atm/K x mol) ( )K 5.43 L P = 9.42 atm Try this problem on your own: Calculate the volume (in liters) occupied by 2.12 moles of nitric oxide (NO) at 6.54 atm and 76 oC.

23
**Molar Volume at STP Gas Calculations**

Example: Calculate the volume (in liters) occupied by 7.40 g of NH3 at STP. Remember that 1 mole of an ideal gas at STP occupies 22.4 Liters. V = 7.40 g NH3 x 1 mol NH3 x L 17.03 g NH3 1mol NH3 V = 9.74 L

24
Modified Ideal Gas Law When conditions change for P, V, T and n we must use a modified form of the ideal gas equation that includes the final and initial conditions. R = P1V (before change) n1T1 R = P2V (after change) n2T2 So that… P1V1 = P2V2 n1T n2T2 If n1=n2 then, P1V1 = P2V2 T T2

25
**Using the Modified Ideal Gas Law**

Example: A small bubble rises from the bottom of a lake, where the temperature and pressure are 8 oC and 6.4 atm, to the water’s surface, where the temperature and pressure is 25 oC and 1.0 atm. Calculate the final volume (in mL) of the bubble if its initial volume was 2.1 mL. Note that the moles do not change in this question. Therefore we can use the modified ideal gas law. P1V1 = P2V rearranged => V1P1 T2 = V2 T T P2T1 Therefore… 2.1 mL x 6.4 atm x 298 K 1.0 atm x 281 K V2 = 14 mL

26
**Using the Modified Ideal Gas Law**

Example: Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain light bulb containing argon at 1.2 atm and 18 oC is heated to 85 oC at constant volume. Calculate its final pressure (in atm). P1V1 = P2V2 rearranged => P1 = P2 T1 T2 T T2 rearranged => P2 = P1x T2 T1 P2 = 1.20 atm x 358 K 291 K P2 = 1.48 atm Electric Lightbulbs are usually filled with argon gas.

27
**We can find the number of moles (n) by:**

Density Calculations If we rearrange the ideal gas equation, we can calculate the density of the gas: n = P V RT We can find the number of moles (n) by: n = mass (g = m) molar mass (g/mol =M ) Therefore, plugging the above in for n in the ideal gas law, we get: m = P M x V RT Which rearranges into: d = m = P x M V R x T To check yourself on this type of a problem: Calculate the density of carbon dioxide in grams per liter (g/L) at 752 mmHg and 55 degrees Celsius. Your answer should be 1.62 g/L. Note that because gas samples are separated by distances that a large compared to their size, the densities of gas are very low under atmospheric conditions and are usually expressed in g/L.

28
**Finding the Molar Mass, M, of a Gas**

Many times we do not know the molar mass, M, of a gas. From our previous slide, we know that we can find the density of a gas from: density = m = P x M V R x T We can rearrange this equation in order to solve for the molar mass, M, of a gas: M = d x R x T P With the tool shown in the slide, chemists can figure out the density of a gas because the apparatus has a known volume. Therefore all they need to calculate is the mass of the gas. The answer to the example question in 67.9 g/mol. From this molar mass, we can calculate the molecular formula by trial and error from guessing. The molecular formula has to be ClO2 because this molar mass is equal to g/mol. Example: A chemist has synthesized a greenish-yellow gaseous compound of chlorine and oxygen and finds that its density is 7.71 g/L at 36 oC and 2.88 atm. Calculate the molar mass of the compound and determine its molecular formula.

29
**A Combined Gas Sample Problem**

Chemical analysis of a gaseous compound showed that it contained 33.0 % silicon and 67.0 % fluorine by mass. At 35 oC, L of the compound exerted a pressure of 1.70 atm. If the mass of L of the compound was 2.38 g, calculate the molecular formula of the compound. (R = L x atm/K x mol) Calculate the empirical formula: SiF3 Calculate the molecular molar mass: M = d x R x T = 169 g/mol P The empirical molar mass of SiF3 = g/mol. 169 / 85 = 2 Therefore the molecular formula is Si2F6

30
Gas Stoichiometry When you notice that you have a gas involved in a chemical reaction, the rules for stoichiometry change. You will still convert the amount of reactants to the amount of products using moles, but remember that moles (n) vary depending on PV = nRT for gases. To solve Gas Stoichiometry problems you must read the question and: 1) Identify whether the problem states that the gas is at STP = 0oC and 1 atm. If this is the case remember that 1 mole (n) of any gas will occupy 22.4 Liters. 2) If the problem states that each gas is not at STP, then you will have to determine the moles for each gas given based on PV = nRT for each gas.

31
**Gas Stoichiometry Sample Problem**

Calculate the volume of O2 (in liters) at STP required for the complete combustion of 2.64 L of acetylene (C2H2) at STP: 2C2H2(g) O2(g) CO2(g) H2O(l) 2.64 L C2H2 ? L O2 2.64 L C2H mol C2H2 = .12 mol C2H2 22.4 L C2H2 .12 mol C2H mol O L O2 = 6.6 L O2 2 mol C2H mol O2

32
**Click on the red box to see how we would use gas stoichiometry to help us solve every day problems.**

33
**(Note : 760 mmHg = 1 atm and R = 0.0821 L x atm / K x mol)**

Sodium azide (NaN3) is used in some automobile air bags. The impact of a collision triggers the decomposition of NaN3 as follows: 2 NaN3(s) Na(s) + 3N2(g) The nitrogen gas produced quickly inflates the bag between the driver and the windshield. Calculate the volume of N2 generated at 80 oC and 823 mmHg by the decomposition of 60.0 g of NaN3. 60.0g NaN ? Volume N2 60.0 g NaN3 1 mol NaN mol N = mol N2 65.02 g NaN3 2 mol NaN3 V = nRT = (1.38 mol) x ( L x atm) x ( )K = 36.9 L N2 P (K x mol) x (1.08 atm) (Note : 760 mmHg = 1 atm and R = L x atm / K x mol) Click on black box to view the air bag movie.

35
**Dalton’s Law of Partial Pressures**

The total pressure of a mixture of gases is just the sum of the pressures that each gas would exert if it were present alone. Therefore: The pressure exerted by gas A and B is: PA = nART and PB = nBRT V V PT = PA + PB = nART + nBRT = RT (nA + nB) V V V Xi is the mole fraction of gas i, within a mixture of gases: Xi = ni ntotal The partial pressure, Pi, of any substance, i, is equal to the mole fraction of the substance multiplied by the total pressure of a mixture of gases: Pi = (Xi)(PT)

36
Whenever Gas is collected over Water Use Dalton’s Law of Partial Pressure to Figure out the Pressure of the Gas Being Collected. Always look for this in problems!!!! 2KClO3(s) KCl(s) + 3O2(g) PT = PO2 + PH20 PO2 = PT - PH20 The inverted bottle actually starts totally filled with water and as the oxygen bubbles in, it displaces the water. The reason you use dalton’s law of partial pressures is because the water also have vapors inside of the gas and the gas pressure therefore is a mixture of the water vapor + the oxygen gas.

37
P a 1 V As pressure goes up, volume goes down. As pressure goes down, volume goes up. This is a serious law in regards to scuba diving. The pressure in the water increases with increasing depth, this means that the deeper you go in the water, the more pressure there will be surrounding you and the less volume the air inside your lungs will be filling. But what happens if you decide to surface? The higher you go, in order to reach the surface, the less pressure your body will have surrounding it and therefore the more volume the air in your lungs will take up. What then would happen to your lungs if you did not breathe as you went up? If you close your mouth and you do not allow the excess volume to escape, then the air in your lungs will expand and can rupture the membranes of your lungs. This is why scuba divers always breathe as they surface from a dive, this is also why divers surface slowly and why they stop at intervals in order to equalize the air out that is trapped inside of their lungs.

38
**Scuba Diving and Partial Pressures**

The composition of the air that we breathe is 20% oxygen and 80% nitrogen gas by volume. To much oxygen can actually be harmful to us. At sea level, the pressure of the air around us is 1 atm. This means that the pressure of the oxygen we breathe is : PO= moles x 1 atm = 0.2 atm = PO ( ) moles (Remember that moles (n) a V therefore we can replace the volume with moles.) Our bodies are therefore used to breathing in 0.2 atm of pressure of oxygen. But, as a diver swims deeper into the ocean, at 100 feet the pressure increases to 4 atm and the oxygen pressure now becomes: PO = moles x 4 atm = 0.8 atm = PO This pressure of oxygen is much too high, therefore the oxygen needs to be diluted in order to keep it at a level of 0.2 atm. For this to happen, the oxygen must be diluted to a level of 5%. Nitrogen gas is not used to dilute the oxygen down because at a pressure of 1 atm, nitrogen gas in the lungs will start to dissolve into the blood. This causes nitrogen narcosis. This is also responsible for the bends. As a diver surfaces, the nitrogen dissolved in the blood transforms into bubbles of nitrogen gas and can stop blood flow and impair the nervous system. Helium is used instead to dilute oxygen down and a special valve automatically adjusts for the pressure of oxygen. Helium is used because it is an inert gas and it has a low solubility in blood.

39
**The Kinetic Molecular Theory of an Ideal Gas**

The particles are so small compared with the distances between them that the volume of the individual particles can be assumed to be negligible (zero). They can be considered as “points”, where they posses mass but have zero volume. The particles are in constant motion. The collisions of the particles with the walls of the container are the cause of the pressure exerted by the gas. Collisions among molecules are perfectly elastic, in other words, energy can not be transferred from one molecule to another as a result of a collision, the total energy of all the molecules in a system remains the same. Gas molecules exert neither attractive nor repulsive forces on one another. The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas. Any two gases at the same temperature will have the same average kinetic energy.

42
**The Kinetic Molecular Theory and It’s Applications to the Gas Laws**

Boyle’s Law: V a 1/P (at constant n and T) Charles’s Law: V a T (at constant n and P) P a T (at constant n and V) Avogadro’s Law: V a n (at constant P and T)

43
**The Average Kinetic Energy of a Molecule**

Remember the term Kinetic Energy (KE)? It means molecules in motion. And do you remember that temperature is the average kinetic energy? The average KE for one molecule in a gas is represented by: KE = ½ mu2 Where m = mass of one molecule and u = is speed. The term u2 = mean square speed. It is the average of the square of the speeds of all of the molecules: u2 = u21 + u22 + u23 + …. + u2n N N = the number of molecules.

44
**Maxwell Speed Distribution Curves**

The kinetic theory of gases allows us to investigate molecular motion in more detail. As long as we hold the temperature of a gas constant, the average KE and the means square speed will stay constant for the same length of time that the temperature is held constant. The motion of the molecules are totally random and unpredictable. At a given instant, different molecules within a sample are moving at different speeds. How many molecules are moving at a particular speed? Maxwell Boltzman analyzed the behavior of gas molecules at different temperatures to answer this question. Maxwell speed distribution curves show gas at different temperatures and graph the number of molecules that move at a particular molecular speed.

46
**Root-Mean-Square Speed**

How fast does a molecule move, on the average, at any temperature T? One way to estimate molecular speed is to calculate the root-mean-square (rms) speed (urms), which is an average molecular speed. The Total KE of a mole of any gas = 3/2 RT Because Average KE is only for one molecule of a gas: NA (1/2mu2) = 3/2 RT Because NAm = M , we can write: u2 = 3RT / M Taking the square root of both sides gives us: u2 = urms = 3RT / M The total KE of a mole of any gas = 3/2 RT comes from the kinetic theory of gases.

48
Diffusion A direct demonstration of random motion is provided by diffusion, the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. Diffusion always proceeds from areas of higher concentrations to areas of lower concentrations. The diffusion process takes a long time to complete even if the molecular speeds are great. This is because molecules experience numerous collisions while moving from one end of the bench to the other. Also, because the root mean square speed of a light gas is greater than that of a heavier gas, a lighter gas will diffuse through a certain space more quickly than will a heavier gas. Because NH3 is lighter and therefore diffuses heavier than HCl, the solid NH4Cl first appears nearer the HCl bottle on the right. Calculating the ratio of the effusion rates of two gases or the distance traveled for two gases: (Distance traveled/Effusion rate) for gas 1: urms = 3RT / M 1 = M 1 (Distance traveled/Effusion rate) for gas 2: urms = 3RT / M M 2

49
A direct demonstration of random motion is provided by diffusion, the gradual mixing of molecules of one gas with molecules of another by virtue of their kinetic properties. Diffusion always proceeds from areas of higher concentrations to areas of lower concentrations. The diffusion process takes a long time to complete even if the molecular speeds are great. This is because molecules experience numerous collisions while moving from one end of the bench to the other. Also, because the root mean square speed of a light gas is greater than that of a heavier gas, a lighter gas will diffuse through a certain space more quickly than will a heavier gas. Because NH3 is lighter and therefore diffuses heavier than HCl, the solid NH4Cl first appears nearer the HCl bottle on the right.

50
**Deviation from Ideal Behavior**

Ideal gases are said to: Not exert any attractive or repulsive forces upon one another. Have negligibly small volumes compared to that of the container. To study real gases, the van der Waals equation is used: (P + an2/V2) (V – nb) = nRT Where a and b are van der Waals constants that can be looked up in an appendix or table. Corrected Pressure Corrected Volume

Similar presentations

OK

Chapter Five Gases. Chapter Five / Gases Substances That Exist as Gases Element in blue are Gases Noble gases are monatomic All other gases (H 2, N 2,

Chapter Five Gases. Chapter Five / Gases Substances That Exist as Gases Element in blue are Gases Noble gases are monatomic All other gases (H 2, N 2,

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Ppt on question tags worksheet Ppt on solar energy usage Ppt on office 365 Ppt on computer virus and antivirus Ppt on first conditional worksheets Ppt on electric power grid Ppt on materials metals and nonmetals Ppt on service design and development Ppt on production function Ppt on marie curie nobel