= mol NO 2 1 mol Cu Now we use the ideal gas law: atm V = mol L atm/mol K 304 K

Size: px
Start display at page:

Download "= mol NO 2 1 mol Cu Now we use the ideal gas law: atm V = mol L atm/mol K 304 K"

Transcription

1 CHEM 101A ARMSTRONG SOLUTIONS TO TOPIC C PROBLEMS 1) This problem is a straightforward application of the combined gas law. In this case, the temperature remains the same, so we can eliminate it from the combined gas law equation, giving P 1 V 1 P V (Boyle s law) atm 451 ml 1.65 atm V Solving this equation gives V 364 ml. ) This is another application of the combined gas law. Be sure to convert all temperatures to kelvin! 744 torr 3.85 L 591 torr 4.13 L 304 K T Solving this equation gives T 59 K 14ºC. 3) This is an application of the ideal gas law. We are given T and P, and we can calculate n from the given mass of hydrogen. Our only unknown in the ideal gas law is then V. To begin, 5.3 g of hydrogen equals moles (note that elemental hydrogen is diatomic, so the molar mass is.016 g/mol). When we substitute into the ideal gas law, the temperature must be in kelvins, as it is in all gas law equations. Also, we must use the correct value of R. In this case, our pressure is given in atm, so we use R L atm/mol K atm V mol L atm/mol K 304 K Solving this equation gives V 80.7 L. 4) We can use the ideal gas law to calculate the number of moles of methane, then convert moles into grams. However, we must first account for the fact that the methane is collected over water. P methane + (vapor pressure of water) P total Now we can calculate the pressure that the methane exerts: P methane torr 731 torr P methane 711. torr Next, we substitute into the ideal gas law, being sure to use a kelvin temperature. Since our pressure is in torr, we use R 6.36 L torr/mol K. Using this value of R in turn means that our volume must be in liters. (Keeping the units straight in the ideal gas law isn t actually hard at all just be sure that the units in R match the units you use for P, V, n and T!) 711. torr L n 6.36 L torr/mol K 95 K Solving this equation gives n mol. Finally, we convert moles into grams, using the molar mass of CH 4 (16.04 g/mol). Doing so gives us g of methane. 5) First, we use stoichiometry to calculate the number of moles of NO that will be formed. Then we use the ideal gas law to determine the volume of the NO. The stoichiometry gives us: 1 mol Cu 4.17 g Cu g Cu mol NO mol NO 1 mol Cu Now we use the ideal gas law: 1.0 atm V mol L atm/mol K 304 K

2 Solving this equation gives V 3.0 L. 6) Each gas behaves as if the other gas was not present. When the valve is opened, the He will expand to fill the entire system, which has a volume of 4 ml ml 1183 ml: 817 torr 4 ml P 1183 ml P torr (final pressure of He) The Ar will also expand to fill the entire system: 685 torr 761 ml P 1183 ml P torr (final pressure of Ar) The total pressure in the container is the sum of these two partial pressures: torr torr 73 torr 7) a) We can set up an ICE table using partial pressures whenever all of the reactants and products are gases and the volume and temperature do not change. Doing so here gives: C H Cl CCl HCl Initial 35.5 torr 318 torr 0 torr 0 torr Change 35.5 torr 48.5 torr torr torr End 0 torr 69.5 torr 71.0 torr 13.0 torr The initial values are given in the problem. To find the changes, use the mole ratios. For instance, to find the amount of Cl that is used up: 7 torr Cl 35.5 torr C H 6 1 torr C H torr Cl (Note that this calculation assumes that C H 6 is the limiting reactant. Since the amount of Cl used up is less than the amount we start with, the assumption is correct.) To find the amount of CCl 4 that is formed: 35.5 torr C H 6 torr CCl 4 1 torr C H torr CCl 4 The amount of HCl that is formed is calculated similarly. Finally, we calculate the final amounts doing arithmetic. The final mixture contains 69.5 torr Cl, 71.0 torr CCl 4, and 13.0 torr HCl. b) The total pressure is the sum of the partial pressures, which equals torr. 8) Since there is oxygen in the container after the reaction, we can conclude that Cr + is the limiting reactant. This allows us to calculate the amount of oxygen that was consumed in the reaction using stoichiometry mol Cr+ 1 mol O L 1 L 4 mol Cr mol O consumed Now, let s calculate the number of moles of O in the container after the reaction, using the ideal gas law. The volume of the oxygen is not equal to the container volume (56 ml), because the solution occupies 1.3 ml, so the volume of the oxygen is only ml L. 119 torr L n 6.36 L torr/mol K 94 K Solving this equation gives n mol O left over The total number of moles of oxygen in the container before the reaction was: mol x 10-4 mol mol O

3 Finally, we can use the ideal gas law to calculate the original pressure of the oxygen. Before the Cr + solution was added, the O occupied the entire container (56 ml 0.56 L). P 0.56 L mol 6.36 L atm/mol K 94 K Solving this equation gives P 137 torr. 9) In this problem, we must consider two processes: each gas expands to fill the entire system, and the gases react with one another. These two processes actually occur at the same time, but we can treat them as if they occur one at a time, with the expansion occurring first. The C 3 H 6 expands from 367 ml to 969 ml (367 ml + 60 ml): 663 torr 367 ml P 969 ml P 51.1 torr The O expands from 60 ml to 969 ml: 16 torr 60 ml P 969 ml P torr Now, we consider the reaction. Since the volume of the entire system and the temperature are constant, we can do our stoichiometry using partial pressures. (Note that we can not use the pressures given in the problem to do stoichiometry!) Here is an ICE table for the reaction. C 3 H O 6 CO + 6 H O Initial 51.1 torr torr 0 torr 0 torr Change 51.1 torr torr torr torr End 0 torr 46.7 torr torr torr As in problem 7, I assumed that C 3 H 6 was the limiting reactant and calculated the amount of O that would react: 9 torr O 51.1 torr C 3 H torr O consumed torr C 3 H 6 Since this amount is less than the initial O pressure, my assumption was correct. The amounts of CO and H O are calculated similarly. Finally, we calculate the total pressure in the system: 46.7 torr torr torr 1753 torr 10) This problem is similar to #9, except that we do not know the initial pressure of the oxygen. Let s assign it a variable: the initial pressure of oxygen is x torr. Now we proceed just as we did in problem 9. The C 3 H 6 expands from 617 ml to 1550 ml: 317 torr 617 ml P 1550 ml P 16. torr The O expands from 933 ml to 1550 ml: x torr 933 ml P 1550 ml P x torr

4 Note that the final pressure of oxygen contains the variable x. Now we consider the reaction. In this case, we have no obvious way of determining the limiting reactant, so let s just assume that it is C 3 H 6 and see what we get C 3 H O 6 CO + 6 H O Initial 16. torr x torr 0 torr 0 torr Change 16. torr torr torr torr End 0 torr ( x 567.8) torr torr torr The problem tells us that the total pressure after the reaction is 1133 torr. Therefore, the expressions in the End row of our ICE table must add up to 1133 torr. ( x 567.8) Solving this equation gives x 1568 torr. Is this a reasonable value? If so, it should give positive pressures for every gas in the final mixture. The pressures of CO and H O are obviously positive, but is x a positive number? Let s check: (1568) torr This is a positive value, so 1568 torr is a reasonable answer. Now let s see what we get if we assume that O is the limiting reactant. In this case, all of the Change entries are going to contain x, because they will be based on the O pressure. For instance, the amount of C 3 H 6 we consume will be: x torr O torr C H torr O x torr C 3 H 6 Here is the ICE table we get when we assume that O is the limiting reactant: C 3 H O 6 CO + 6 H O Initial 16. torr x torr 0 torr 0 torr Change x torr x torr x torr x torr End ( x) torr 0 torr x torr x torr Once again, the expressions in the End row must add up to 1133 torr. ( x) x x 1133 Solving this equation gives x 1505 torr. This is a different value from our previous answer (1568 torr). Are there two possible answers? We must again make sure that our final pressures are all positive. The pressures of CO and H O will clearly be positive, but what about the pressure of C 3 H 6? (1505) 75.1 torr The final pressure of C 3 H 6 turns out to be a negative number, which is an absurd result. This is an example of a situation where we get two answers that satisfy the mathematics, but only one of the two is physically reasonable. (The final pressure turns out to be negative because we use more C 3 H 6 than we have: (1505) 01.3 torr used up, which is greater than the 16. torr we started with).

5 Therefore, we can conclude that the initial pressure of O was 1568 torr. 11) If you want to memorize the formula in Zumdahl, you can, but you will not be given this formula on your exam data sheet, because you can solve gas density problems using the ideal gas law. The key is to recognize that the density is the number of grams of gas in one liter, and we can calculate the number of grams of any gas in one liter using PVnRT and then converting moles to grams. Start by calculating the number of moles of CO in 1 L: 855 torr 1 L n 6.36 L torr/mol K 34 K n mol Then convert this number of moles to grams: mol g g 1 mol Since this is the mass of 1 L of CO, the density is 1.86 g/l. 1) First, we can use the percent composition to find the empirical formula of the compound. You did this kind of problem in topic A, so I will just outline the solution here: 100 g of compound contains 88.8 g of carbon mol C 11. g of hydrogen mol H Dividing both of these numbers by the smaller (7.394) gives: 1 mol C/mol C 1.50 mol H/mol C Multiplying both of these by mol C converts them to whole numbers: mol C 3 mol H So the empirical formula of the compound is C H 3. Now we use the vapor density to determine the molar mass. The density tells us that 1 L of the compound weighs 1.91 g. We use the ideal gas law to determine the number of moles in 1 L. 675 torr 1 L n 6.36 L torr/mol K K n mol So moles of the compound weighs 1.91 grams. The molar mass of the compound is therefore: 1.91 g 54.1 g/mol mol If the compound were C H 3, the molar mass would be about 7 g/mol. The actual molar mass is twice this number, so the molecular formula must be C 4 H 6. 13) There are several ways to approach this problem; here is one of them Let s start by using the ideal gas law to get an expression for the number of moles of oxygen in 1 liter:

6 n PV 1.00 atm 1 L 1 L atm RT RT RT The temperature is unknown, so we cannot insert a value for T here. We can insert a value for R (we would use L atm/mol K), but there is no need to do so yet. Next, let s convert this number of moles into a mass, using the molar mass of oxygen (3.00 g/mol): 1 L atm 3.00 g 3 g L atm/mol RT 1 mol RT This expression tells us that the mass (in grams) of O in 1 L of the sample is 3 divided by RT. The g L atm/mol in the numerator is required to make the units cancel. Now, the nitrogen and the oxygen have the same density, so the mass of N in a 1 L sample is also 3/RT. Let s convert that to moles, using the molar mass of N (8.0 g/mol): 3 g L atm/mol 1 mol 1.14 L atm RT 8.0 g RT This expression tells us that the number of moles of N in 1 L of the sample is 1.14 divided by RT. Finally, we can substitute this expression into the ideal gas law to calculate a pressure for the N : 1.14 L atm P 1 L RT P 1 L 1.14 L atm R T 1.14 L atm P 1.14 atm 1 L Note that since the two gases are at the same temperature, the value of T cancels out here (as does the value of R). The pressure of N must be higher because N is a lighter gas than O ; in order to have equal masses, we need more molecules of N, which gives us a higher pressure. 14) First, let s calculate the number of moles of gaseous H S in the original container, using the ideal gas law: torr.50 L n 6.36 L torr/mol K K n mol Next, let s calculate the number of moles of gaseous H S in the container after the system has reached equilibrium. The gas volume is now.50 L L.30 L, and the partial pressure of the H S is torr 11. torr torr torr.30 L n 6.36 L torr/mol K K n mol Now we can calculate the number of moles of H S that dissolved; it is the difference between the initial and final moles of gaseous H S mol mol mol

7 The molar concentration of H S in the water is therefore: mol L 0.05 M 15) a) For a moving object, KE 1 / mv. In the metric system, the unit of energy is the joule, which equals a kg m /sec. Therefore, the mass must be in kg and the velocity must be in m/sec. We must start by converting Janice s mass from pounds to kilograms. 454 g 131 pounds 1 pound 1 kg kg 1000 g Now we can calculate Janice s speed. In order to make the units cancel, we must make the substitution J kg m /sec KE 1 mv 1.00 J 1 ( kg)v 1.00 kg m /sec 1 ( kg)v m /sec v v m /sec m/sec (This is a very slow walking speed. At this speed, it would take Janice about ten minutes to walk the length of a football field, including the end zones.) b) Again, we use KE 1 / mv. KE kg ( ) 1.34 m/sec 53.4 kg m /sec 53.4 J ( ) 16) a) If one mole (6.0 x 10 3 ) atoms has an energy of 615 joules, then one atom must have 615 J divided by Avogadro s number. Writing this as a formal unit conversion looks like this: 1 mol 1 atom atoms 615 J J 1 mol b) Kinetic energy equals 1 / mv, but we cannot calculate the velocity unless we know the mass. We have two options here Option 1: do everything on a per mole basis. The kinetic energy of 1 mole is 615 J, and the mass of 1 mole of neon is 0.18 g kg. 615 J 1 ( kg)v 615 kg m /sec 1 ( kg)v m /sec v v m /sec m/sec

8 Option : do everything based on a single atom of neon. We already know that one neon atom has a kinetic energy of 1.03 x 10-0 J. A single neon atom weighs 1 mol 1 atom atoms 0.18 g 1 kg 1 mol 1000 g kg Now we can use KE 1 / mv J 1 ( kg)v kg m /sec 1 ( kg)v m /sec v v m /sec We end up with the same velocity, as we must m/sec c) To get the total kinetic energy, we multiply the average KE (in J/mol) by the number of moles of neon g of Ne equals mol, so we have: mol 615 J 1 mol J Strictly, this only works if we have a very large number of atoms. In this case, moles equals about 37,300,000,000,000,000,000,000 atoms, which is certainly a very large number! d) For any gas, the average KE equals 3 / RT, where R J/mol K: 615 J/mol 3 T K (5. C) ( J/mol K) T e) The most probable KE for a gas equals 1 / RT. Now that we know the temperature, we can calculate KE mp : KE mp 1 ( J/mol K) K 07 J/mol f) For the root-mean-square speed, we have:

9 v rms 3RT M ( )( K) J/mol K kg/mol J/kg Before we take the final square root, let s make sense of the unit. A joule is a kg m /sec, so a J/kg equals a m /sec (because the kg units cancel out). v rms m /sec m/sec g) For the average speed, we have: v ave 8RT πm ( )( K) ( kg/mol) J/mol K J/kg m /sec 73.1 m/sec h) For the most probable speed, we have: v mp RT M ( )( K) J/mol K kg/mol J/kg m /sec m/sec 17) a) First, we need to calculate the mass of a single water molecule. One mole of H O weighs g, so one molecule weighs g divided by Avogadro s number, which equals.9917 x 10-3 g.9917 x 10-6 kg. Now we can calculate the kinetic energy of the molecule.

10 ( ) 46 m/sec KE kg kg m /sec J ( ) b) One mole equals 6.0 x 10 3 molecules, so the kinetic energy expressed in J/mol is: J mol J/mol c) Calculating the average speed of water molecules at 5 C gives us v ave 8RT πm ( )( K) ( kg/mol) J/mol K J/kg m /sec m / sec Our original water molecule was moving at 46 m/sec, so it is moving slower than the average speed of water molecules at 5 C. 18) a) The average kinetic energy of a gas depends only on the temperature, so the krypton would have to be at 5ºC (the same temperature as the argon). b) The molecular speed depends on both the temperature and the molar mass. The most straightforward way (although not the shortest way) to answer this question is to calculate the average molecular speed of the argon, and then use that speed to calculate the temperature of the krypton. The average speed of the argon at 5ºC is: v ave 8RT πm 8(8.314 J/mol K)(98 K) ( kg/mol) m/sec Using this speed, we can calculate the necessary temperature for the krypton: m/sec ( m/sec) T 8(8.314 J/mol K)T ( kg/mol) 8(8.314 J/mol K)T ( kg/mol) ( m/sec) ( )( kg/mol) 8(8.314 J/mol K) 65 K (35ºC)

11 The quick way to answer this question is to recognize that if the molecular speeds are equal, we can say: 8RT Kr 8RT Ar πm Kr πm Ar Since the factor 8R/π is the same for both expressions, this reduces to T Kr T Ar M Kr M Ar Substituting in the molar masses of Kr and Ar and the temperature of the Ar allows us to calculate the temperature of the Kr. 19) a) The two gases have the same average kinetic energy, because kinetic energy depends only on temperature, and both gases are at the same temperature. b) Average speed is directly related to temperature and inversely related to molar mass, but since the gases are at the same temperature, we need only consider the molar mass. The CO has the higher average speed, because the molar mass of CO is lower than the molar mass of CO. c) The two gases have the same fraction of molecules with KE > 5 kj/mol. As in part a, the kinetic energy distribution depends only on temperature. d) The CO has the higher fraction of molecules with speeds greater than 500 m/sec, for the same reason that it has the higher average speed. If two gases are at the same temperature, the lighter gas will always have the greater fraction of fast-moving molecules, regardless of what cutoff we use to define fast. e) Since the gases are have the same temperature, pressure, and volume, they must contain the same number of moles. However, a mole of CO weighs more than a mole of CO, so the CO weighs more. 0) a) The O has the higher average kinetic energy, because it is at a higher temperature. b) For this question, we need to calculate the average speeds, because the two relevant factors (molar mass and temperature) are opposing each other; the O is hotter (which would give it a higher average speed), but it also has a higher molar mass (which would give it a lower average speed). Calculating the average speeds gives the following values: For O : 480 m/sec For NO: 459 m/sec So the O has the higher average speed. c) The NO has a higher fraction of molecules with KE < 5 kj/mol. Since the NO is at the lower temperature, it has more low-energy molecules, regardless of the cutoff we use to define low. d) The NO has a higher fraction of molecules with speeds below 500 m/sec. In part b, we found that O has the higher average speed, so O must also have the greater fraction of fastmoving molecules. Therefore, NO must have the higher fraction of slow-moving molecules. e) We need to use PVnRT to calculate the moles of each gas, then convert to grams. The volume is not given, but it is the same for both gases, so we can just leave it as a variable. We can also leave R as a variable (although if you want to plug it in, you can). For the O : n PV RT 1 atm V R 348 K V R

12 mass V R 3.00 g 1 mol V R For the NO: n PV RT 1 atm V R 98 K V R mass V R g 1 mol V R Since 0.101V/R is larger than 0.090V/R, the NO weighs more. 1) a) The most probable kinetic energy is around 700 J/mol. This is the kinetic energy that corresponds to the highest point on the curve. (This is a rough estimate anything within 100 J/mol of this value is a reasonable answer.) b) Since KE mp 1 / RT, we can calculate a rough value for the temperature from the most probable KE we estimated in part a. You should get somewhere around 170 K. (Using 700 J/mol gives T 168 K.) c) When x 000 J/mol, y This is the fraction of molecules that have kinetic energies between J/mol and J/mol. (A fraction of equals 0.03%.) d) This value tells us that the fraction of molecules that have kinetic energies between 000 J/mol and 4000 J/mol is Another way to say this is that 5.9% of the molecules have kinetic energies in this range. ) a) Curve B represents the KE distribution for Ar(g) at 300 K. Neon at 300 K and argon at 300 K have exactly the same KE distribution, because they are at the same temperature. b) Curve C could represent the KE distribution for Ne(g) at 600 K. As the temperature increases, the curve shifts toward higher kinetic energies. 3) a) Curve A could represent the speed distribution for H (g) at 15ºC. As the temperature decreases, the distribution shifts toward lower speeds. b) Curve A could represent the speed distribution for N (g) at 5ºC. Comparing the distributions for two different gases at the same temperature (H and N in this case), we expect the heavier gas to have slower speeds. c) Curve B could represent the speed distribution for He(g) at 319ºC. In this case, we need to compare the average speeds for the two gases. When we calculate the average speed for each gas, we get: H at 5ºC: v ave m/sec He at 319ºC: v ave m/sec The two gases have essentially the same average speed. Therefore, they must have the same distribution of speeds. 4) a) The most probable speed is around 40 m/sec. This is the speed that corresponds to the highest point on the curve. (This is a rough estimate anything within 10 m/sec of this value is a reasonable answer.)

13 RT b) Since v mp, we can calculate a rough value for the molar mass from the most M probable speed we estimated in part a. Using v mp 40 m/sec and T 473 K gives M kg/mol 137 g/mol. (Did you remember that M is in kg/mol in this kind of formula???) This value is closest to the molar mass of xenon (131.3 g/mol). c) When x 00 m/sec, y This is the fraction of molecules that have speeds between J/mol and 00.5 m/sec. (A fraction of equals 0.33%.) d) This value tells us that the fraction of molecules that have speeds above 400 m/sec is Another way to say this is that 11.1% of the molecules have speeds in this range. e) The entire area under the curve is 1 ( 100%), so the area of the region that isn t shaded must be This value tells us that the fraction of molecules having speeds below 400 m/sec is ) We can use Graham s law of effusion to calculate the molar mass of the gas: rate 1 rate M M 1 Let s let gas 1 be argon and gas be the unknown substance. (You can do it the other way, too.) To calculate the rates, we need to express the times in a single unit, either minutes or seconds. I ll express the times in seconds. For the argon: time of effusion (5 min x 60 sec/min) + 13 sec 313 sec rate ml/313 sec ml/sec M g/mol (it s okay to use grams per mole in Graham s law) For the unknown: time of effusion (6 min x 60 sec/min) + 10 sec 370 sec rate 5.00 ml/370 sec ml/sec M is unknown Squaring both sides gives ml/sec ml/sec M g/mol ml/sec ml/sec M g/mol ml/sec ml/sec g/mol M M 55.9 g/mol

14 The empirical formula is CH, which has a molar mass of about 14 g/mol. 55.9/14 4 So the molecular formula is C 4 H 8. 6) a) Water has a higher a value because the attraction between H O molecules is stronger than the attraction between N molecules. This agrees with the fact that H O is a liquid at room temperature while N is a gas; the attraction between H O molecules is strong enough to keep them in contact with one another at room temperature. b) Nitrogen has a higher b value because N molecules have a larger volume than H O molecules. This is not surprising; it is reasonable that a nitrogen atom should be larger than two hydrogen atoms. (N and O should be similar sizes.) 7) a) Using the ideal gas law gives us: 150 atm 53 L n L atm/mol K 53 K n mol 884 moles of O (to three sig figs) b) The van der Waals equation for gases is P + a n V V - nb ( ) nrt For oxygen, a 1.36 atm L /mol and b L/mol. We already have n mol, V 53 L, and T 53 K. Substituting these values into the equation gives us: P atm L /mol mol 53 L mol L/mol 53 L ( ) mol L atm/mol K 53 K Doing all of the arithmetic and canceling units gives us:

15 ( P atm) ( L) L atm P atm L atm L P atm atm P atm 15 atm (to 3 sig figs) This is slightly above the safety limit for the reactor (150 atm), so the engineer should decrease the maximum number of moles of O accordingly. 8) a) The actual pressure is lower than the ideal gas prediction when the volume is greater than about 0.1 L. Over this range, P real /P ideal is less than 1, which means that P real is less than P ideal. In this range, the attraction between O molecules reduces the force with which the molecules collide with the container walls, giving a lower pressure than we expect based on the ideal gas model. b) The actual pressure is higher than the ideal gas prediction when the volume is less than about 0.1 L. In this range, the O molecules are so tightly packed together that there is almost no empty space between them. Reducing the volume requires extremely high pressures, because you must compress the molecules themselves. 9) a) The kinetic energy of the molecules in a gas is proportional to the temperature (by the kinetic theory); raising the temperature increases the kinetic energy. Since the kinetic energy is related to the speed of the molecules by KE 1 / mv, increasing the kinetic energy is equivalent to increasing the molecular speeds. The molecules therefore collide with the container walls more often and at higher speeds. Since pressure is simply the collective result of molecules colliding with the walls, the pressure increases. b) Gases are mostly empty space (by the kinetic theory), so compressing a gas simply involves moving the molecules closer to one another. c) The kinetic energy of a mole of a gas is related only to its temperature (by the kinetic theory), and is not a function of the molar mass of the gas. Therefore, the amount of energy you must put into a mole of a gas in order to raise the temperature is likewise a function only of the temperature; it is the difference between the energy of the gas when it s cool and the energy of the gas when it s hot. Note: strictly, this is only valid for monatomic gases. For gases that are molecules, the kinetic energy of the gas depends on the structure of the gas; the more complex the molecules, the higher the energy.

CHEM 101A EXAM 1 SOLUTIONS TO VERSION 1

CHEM 101A EXAM 1 SOLUTIONS TO VERSION 1 CHEM 101A EXAM 1 SOLUTIONS TO VERSION 1 Multiple-choice questions (3 points each): Write the letter of the best answer on the line beside the question. Give only one answer for each question. B 1) If 0.1

More information

Chapter Elements That Exist as Gases at 25 C, 1 atm. 5.2 Pressure basic physics. Gas Properties

Chapter Elements That Exist as Gases at 25 C, 1 atm. 5.2 Pressure basic physics. Gas Properties 5.1 Elements That Exist as Gases at 25 C, 1 atm Chapter 5 The Gaseous State YOU READ AND BE RESPONSIBLE FOR THIS SECTION! Gaseous compounds include CH 4, NO, NO 2, H 2 S, NH 3, HCl, etc. Gas Properties

More information

Gases and Kinetic Molecular Theory

Gases and Kinetic Molecular Theory 1 Gases and Kinetic Molecular Theory 1 CHAPTER GOALS 1. Comparison of Solids, Liquids, and Gases. Composition of the Atmosphere and Some Common Properties of Gases 3. Pressure 4. Boyle s Law: The Volume-Pressure

More information

A Gas Uniformly fills any container. Easily compressed. Mixes completely with any other gas. Exerts pressure on its surroundings.

A Gas Uniformly fills any container. Easily compressed. Mixes completely with any other gas. Exerts pressure on its surroundings. Chapter 5 Gases Chapter 5 A Gas Uniformly fills any container. Easily compressed. Mixes completely with any other gas. Exerts pressure on its surroundings. Copyright Cengage Learning. All rights reserved

More information

Standard T & P (STP) At STP, 1 mol of any ideal gas occupies 22.4 L. The standard temperature and pressure for gases is:

Standard T & P (STP) At STP, 1 mol of any ideal gas occupies 22.4 L. The standard temperature and pressure for gases is: Standard T & P (STP) The standard temperature and pressure for gases is: At STP, 1 mol of any ideal gas occupies 22.4 L T = 273 K (0 o C) P = 1 atm = 101.325 kpa = 1.01325 bar 22.4 L Using STP in problems

More information

Chapter 5. The Gas Laws

Chapter 5. The Gas Laws Chapter 5 The Gas Laws 1 Pressure Force per unit area. Gas molecules fill container. Molecules move around and hit sides. Collisions are the force. Container has the area. Measured with a barometer. 2

More information

Although different gasses may differ widely in their chemical properties, they share many physical properties

Although different gasses may differ widely in their chemical properties, they share many physical properties IV. Gases (text Chapter 9) A. Overview of Chapter 9 B. Properties of gases 1. Ideal gas law 2. Dalton s law of partial pressures, etc. C. Kinetic Theory 1. Particulate model of gases. 2. Temperature and

More information

GASES (Chapter 5) Temperature and Pressure, that is, 273 K and 1.00 atm or 760 Torr ) will occupy

GASES (Chapter 5) Temperature and Pressure, that is, 273 K and 1.00 atm or 760 Torr ) will occupy I. Ideal gases. A. Ideal gas law review. GASES (Chapter 5) 1. PV = nrt Ideal gases obey this equation under all conditions. It is a combination ofa. Boyle's Law: P 1/V at constant n and T b. Charles's

More information

Gas Laws. Gas Properties. Gas Properties. Gas Properties Gases and the Kinetic Molecular Theory Pressure Gas Laws

Gas Laws. Gas Properties. Gas Properties. Gas Properties Gases and the Kinetic Molecular Theory Pressure Gas Laws Gas Laws Gas Properties Gases and the Kinetic Molecular Theory Pressure Gas Laws Gas Properties 1) Gases have mass - the density of the gas is very low in comparison to solids and liquids, which make it

More information

Gases. A gas. Difference between gas and vapor: Why Study Gases?

Gases. A gas. Difference between gas and vapor: Why Study Gases? Gases Chapter 5 Gases A gas Uniformly fills any container. Is easily compressed. Mixes completely with any other gas. Exerts pressure on its surroundings. Difference between gas and vapor: A gas is a substance

More information

Part One: The Gas Laws. gases (low density, easy to compress)

Part One: The Gas Laws. gases (low density, easy to compress) CHAPTER FIVE: THE GASEOUS STATE Part One: The Gas Laws A. Introduction. 1. Comparison of three states of matter: fluids (flow freely) solids condensed states liquids (high density, hard to compress) gases

More information

Exercises. Pressure. CHAPTER 5 GASES Assigned Problems

Exercises. Pressure. CHAPTER 5 GASES Assigned Problems For Review 7. a. At constant temperature, the average kinetic energy of the He gas sample will equal the average kinetic energy of the Cl 2 gas sample. In order for the average kinetic energies to be the

More information

Chapter 5 The Gaseous State

Chapter 5 The Gaseous State Chapter 5 The Gaseous State Contents and Concepts Gas Laws We will investigate the quantitative relationships that describe the behavior of gases. 1. Gas Pressure and Its Measurement 2. Empirical Gas Laws

More information

= (25.0 g)(0.137 J/g C)[61.2 C - (-31.4 C)] = 317 J (= kj)

= (25.0 g)(0.137 J/g C)[61.2 C - (-31.4 C)] = 317 J (= kj) CHEM 101A ARMSTRONG SOLUTIONS TO TOPIC D PROBLEMS 1) For all problems involving energy, you may give your answer in either joules or kilojoules, unless the problem specifies a unit. (In general, though,

More information

4. 1 mole = 22.4 L at STP mole/volume interconversions at STP

4. 1 mole = 22.4 L at STP mole/volume interconversions at STP Ch. 10 Gases and the Ideal Gas Law(s) Chem 210 Jasperse Ch. 10 Handouts 1 10.1 The Atmosphere 1. Earth surrounded by gas 2. Major components: Nitrogen 78% Oxygen 21% Miscellaneous: All

More information

SAMPLE EXERCISE 10.1 Converting Units of Pressure. SAMPLE EXERCISE 10.1 continued

SAMPLE EXERCISE 10.1 Converting Units of Pressure. SAMPLE EXERCISE 10.1 continued SAMPLE EXERCISE 10.1 Converting Units of Pressure (a) Convert 0.357 atm to torr. (b) Convert 6.6 10 2 torr to atm. (c) Convert 147.2 kpa to torr. Analyze: In each case we are given the pressure in one

More information

Gases: Their Properties & Behavior. Chapter 09 Slide 1

Gases: Their Properties & Behavior. Chapter 09 Slide 1 9 Gases: Their Properties & Behavior Chapter 09 Slide 1 Gas Pressure 01 Chapter 09 Slide 2 Gas Pressure 02 Units of pressure: atmosphere (atm) Pa (N/m 2, 101,325 Pa = 1 atm) Torr (760 Torr = 1 atm) bar

More information

Example Problems: 1.) What is the partial pressure of: Total moles = 13.2 moles 5.0 mol A 7.0 mol B 1.2 mol C Total Pressure = 3.

Example Problems: 1.) What is the partial pressure of: Total moles = 13.2 moles 5.0 mol A 7.0 mol B 1.2 mol C Total Pressure = 3. 5.6 Dalton s Law of Partial Pressures Dalton s Law of Partial Pressure; The total pressure of a gas is the sum of all its parts. P total = P 1 + P + P 3 + P n Pressures are directly related to moles: n

More information

Quick Review 1. Properties of gases. 2. Methods of measuring pressure of gases. 3. Boyle s Law, Charles Law, Avogadro s Law. 4. Ideal gas law.

Quick Review 1. Properties of gases. 2. Methods of measuring pressure of gases. 3. Boyle s Law, Charles Law, Avogadro s Law. 4. Ideal gas law. Quick Review 1. Properties of gases. 2. Methods of measuring pressure of gases. 3. Boyle s Law, Charles Law, Avogadro s Law. 4. Ideal gas law. 5. Dalton s law of partial pressures. Kinetic Molecular Theory

More information

Gases and Kinetic Theory

Gases and Kinetic Theory Gases and Kinetic Theory Chemistry 35 Fall 2000 Gases One of the four states of matter Simplest to understand both physically and chemically Gas Properties Low density Fluid Can be defined by their: 1.

More information

The Gaseous State of Matter

The Gaseous State of Matter The Gaseous State of Matter Chapter 12 Hein and Arena Version 1.1 Dr. Eugene Passer Chemistry Department Bronx Community 1 College John Wiley and Company The Kinetic- Molecular Theory 2 The Kinetic-Molecular

More information

Comparison of Solids, Liquids, and Gases

Comparison of Solids, Liquids, and Gases CHAPTER 8 GASES Comparison of Solids, Liquids, and Gases The density of gases is much less than that of solids or liquids. Densities (g/ml) Solid Liquid Gas H O 0.97 0.998 0.000588 CCl 4.70.59 0.00503

More information

Gases and the Kinetic Molecular Theory

Gases and the Kinetic Molecular Theory Gases and the Kinetic olecular Theory Importance in atmospheric phenomena, gas phase reactions, combustion engines, etc. 5.1 The hysical States of atter The condensed states liquid and solid The gaseous

More information

Slide 1 / A gas at a pressure of 10.0 Pa exerts a force of N on an area of 5.5 m 2 A 55 B 0.55 C 5.5 D 1.8 E 18

Slide 1 / A gas at a pressure of 10.0 Pa exerts a force of N on an area of 5.5 m 2 A 55 B 0.55 C 5.5 D 1.8 E 18 Slide 1 / 76 1 A gas at a pressure of 10.0 Pa exerts a force of N on an area of 5.5 m 2 A 55 B 0.55 C 5.5 D 1.8 E 18 Slide 2 / 76 2 A pressure of 1.00 atm is the same as a pressure of of mm Hg. A 193 B

More information

This should serve a s a study guide as you go on to do the problems in Sapling and take the quizzes and exams.

This should serve a s a study guide as you go on to do the problems in Sapling and take the quizzes and exams. CHM 111 Chapter 9 Worksheet and Study Guide Purpose: This is a guide for your as you work through the chapter. The major topics are provided so that you can write notes on each topic and work the corresponding

More information

Chapter 5 The Gaseous State

Chapter 5 The Gaseous State Chapter 5 The Gaseous State Contents and Concepts Gas Laws We will investigate the quantitative relationships that describe the behavior of gases. 1. Gas Pressure and Its Measurement 2. Empirical Gas Laws

More information

Unit Outline. I. Introduction II. Gas Pressure III. Gas Laws IV. Gas Law Problems V. Kinetic-Molecular Theory of Gases VI.

Unit Outline. I. Introduction II. Gas Pressure III. Gas Laws IV. Gas Law Problems V. Kinetic-Molecular Theory of Gases VI. Unit 10: Gases Unit Outline I. Introduction II. Gas Pressure III. Gas Laws IV. Gas Law Problems V. Kinetic-Molecular Theory of Gases VI. Real Gases I. Opening thoughts Have you ever: Seen a hot air balloon?

More information

Chapter 11. Molecular Composition of Gases

Chapter 11. Molecular Composition of Gases Chapter 11 Molecular Composition of Gases PART 1 Volume-Mass Relationships of Gases Avogadro s Law Equal volumes of gases at the same temperature and pressure contain equal numbers of molecules. Recall

More information

Gas Density. Standard T & P (STP) 10/29/2011. At STP, 1 mol of any ideal gas occupies 22.4 L. T = 273 K (0 o C) P = 1 atm = kpa = 1.

Gas Density. Standard T & P (STP) 10/29/2011. At STP, 1 mol of any ideal gas occupies 22.4 L. T = 273 K (0 o C) P = 1 atm = kpa = 1. Standard T & P (STP) T = 73 K (0 o C) P = 1 atm = 101.35 kpa = 1.0135 bar At STP, 1 mol of any ideal gas occupies.4 L.4 L Gas Density We can use PV = nrt to determine the density of gases. What are the

More information

Gases. Which elements exist as gases at ordinary temperature and pressure? Gases: Have simple molecular formulas. Chapter 10 part 1: Ideal Gases

Gases. Which elements exist as gases at ordinary temperature and pressure? Gases: Have simple molecular formulas. Chapter 10 part 1: Ideal Gases Chapter 10 part 1: Ideal Gases Read: BLB 10.1 5 HW: BLB 10.2,19a,b, 23, 26, 30, 39, 41, 45, 49 Sup 10:1 6 Know: What is pressure? Gases Which elements exist as gases at ordinary temperature and pressure?

More information

Chapter 10 Gases Characteristics of Gases Elements that exist as gases: Noble gases, O 2, N 2,H 2, F 2 and Cl 2. (For compounds see table 10.

Chapter 10 Gases Characteristics of Gases Elements that exist as gases: Noble gases, O 2, N 2,H 2, F 2 and Cl 2. (For compounds see table 10. Chapter 10 Gases 10.1 Characteristics of Gases Elements that exist as gases: Noble gases, O 2, N 2,H 2, F 2 and Cl 2. (For compounds see table 10.1) Unlike liquids and solids, gases expand to fill their

More information

Chapter 10 Practice. Name: Multiple Choice Identify the choice that best completes the statement or answers the question.

Chapter 10 Practice. Name: Multiple Choice Identify the choice that best completes the statement or answers the question. Name: Score: 0 / 18 points (0%) [3 open ended questions not graded] Chapter 10 Practice Multiple Choice Identify the choice that best completes the statement or answers the question. 1. A sample of gas

More information

Chapter 11 Gases 1 Copyright McGraw-Hill 2009

Chapter 11 Gases 1 Copyright McGraw-Hill 2009 Chapter 11 Gases Copyright McGraw-Hill 2009 1 11.1 Properties of Gases The properties of a gas are almost independent of its identity. (Gas molecules behave as if no other molecules are present.) Compressible

More information

Chapter 10 Gases. Dr. Ayman Nafady. Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E.

Chapter 10 Gases. Dr. Ayman Nafady. Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Chemistry, The Central Science, 11th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 10 Gases Dr. Ayman Nafady 2009, Prentice-Hall, 10.1. Characteristics of Gases Unlike liquids

More information

Test Bank for Chemistry 9th Edition by Zumdahl

Test Bank for Chemistry 9th Edition by Zumdahl Test Bank for Chemistry 9th Edition by Zumdahl 1. Gases generally have A) low density B) high density C) closely packed particles D) no increase in volume when temperature is increased E) no decrease in

More information

1. What is the value of the quantity PV for one mole of an ideal gas at 25.0 C and one atm?

1. What is the value of the quantity PV for one mole of an ideal gas at 25.0 C and one atm? Real Gases Thought Question: How does the volume of one mole of methane gas (CH4) at 300 Torr and 298 K compare to the volume of one mole of an ideal gas at 300 Torr and 298 K? a) the volume of methane

More information

CHAPTER 5 GASES AND THE KINETIC- MOLECULAR THEORY

CHAPTER 5 GASES AND THE KINETIC- MOLECULAR THEORY CHAPTER 5 GASES AND THE KINETIC- MOLECULAR THEORY FOLLOW UP PROBLEMS 5.1A Plan: Use the equation for gas pressure in an open-end manometer to calculate the pressure of the gas. Use conversion factors to

More information

HOMEWORK 11-1 (pp )

HOMEWORK 11-1 (pp ) CHAPTER 11 HOMEWORK 11-1 (pp. 333 335) VOCABULARY Define. 1. Gay-Lussac s law of combining volumes of gases 2. Avogadro s law Answer each question. 3. Write and explain the equation that expresses the

More information

Chapter 10 Notes: Gases

Chapter 10 Notes: Gases Chapter 10 Notes: Gases Watch Bozeman Videos & other videos on my website for additional help: Big Idea 2: Gases 10.1 Characteristics of Gases Read p. 398-401. Answer the Study Guide questions 1. Earth

More information

Gases. and all other variables are directly proportional. a. From what laws is this equation derived?

Gases. and all other variables are directly proportional. a. From what laws is this equation derived? Gases 1. What is the Ideal Gas Law Equation? and all other variables are directly proportional a. From what laws is this equation derived? i. Boyle s Law relationship between pressure and volume ii. Charles

More information

OUTLINE. States of Matter, Forces of Attraction Phase Changes Gases The Ideal Gas Law Gas Stoichiometry

OUTLINE. States of Matter, Forces of Attraction Phase Changes Gases The Ideal Gas Law Gas Stoichiometry UNIT 6 GASES OUTLINE States of Matter, Forces of Attraction Phase Changes Gases The Ideal Gas Law Gas Stoichiometry STATES OF MATTER Remember that all matter exists in three physical states: Solid Liquid

More information

B 2, C 2, N 2. O 2, F 2, Ne 2. Energy order of the p 2p and s 2p orbitals changes across the period.

B 2, C 2, N 2. O 2, F 2, Ne 2. Energy order of the p 2p and s 2p orbitals changes across the period. Chapter 11 Gases Energy order of the p p and s p orbitals changes across the period. Due to lower nuclear charge of B, C & N there is no s-p orbitals interaction Due to high nuclear charge of O, F& Ne

More information

Why study gases? A Gas 10/17/2017. An understanding of real world phenomena. An understanding of how science works.

Why study gases? A Gas 10/17/2017. An understanding of real world phenomena. An understanding of how science works. Kinetic Theory and the Behavior of Ideal & Real Gases Why study gases? n understanding of real world phenomena. n understanding of how science works. Gas Uniformly fills any container. Mixes completely

More information

Chapter 5 Gases and the Kinetic-Molecular Theory

Chapter 5 Gases and the Kinetic-Molecular Theory Chapter 5 Gases and the Kinetic-Molecular Theory Name (Formula) Methane (CH 4 ) Ammonia (NH 3 ) Chlorine (Cl 2 ) Oxygen (O 2 ) Ethylene (C 2 H 4 ) Origin and Use natural deposits; domestic fuel from N

More information

Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Chapter 10. Gases.

Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten. Chapter 10. Gases. Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Chapter 10 Characteristics of Unlike liquids and solids, they Expand to fill their containers.

More information

Unit 08 Review: The KMT and Gas Laws

Unit 08 Review: The KMT and Gas Laws Unit 08 Review: The KMT and Gas Laws It may be helpful to view the animation showing heating curve and changes of state: http://cwx.prenhall.com/petrucci/medialib/media_portfolio/text_images/031_changesstate.mov

More information

Gases. Chapter 5. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Gases. Chapter 5. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Gases Chapter 5 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Elements that exist as gases at 25 0 C and 1 atmosphere 2 3 Physical Characteristics of Gases

More information

Gases. Chapter 5. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

Gases. Chapter 5. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Gases Chapter 5 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Elements that exist as gases at 25 0 C and 1 atmosphere 2 3 Physical Characteristics of Gases

More information

10/15/2015. Why study gases? An understanding of real world phenomena. An understanding of how science works.

10/15/2015. Why study gases? An understanding of real world phenomena. An understanding of how science works. 0/5/05 Kinetic Theory and the Behavior of Ideal & Real Gases Why study gases? An understanding of real world phenomena. An understanding of how science works. 0/5/05 A Gas fills any container. completely

More information

Substances that are Gases under Normal Conditions

Substances that are Gases under Normal Conditions Chapter 5: Gases 5.1 Early Experiments 5.2 The gas laws of Boyle, Charles, and Avogadro 5.3 The Ideal Gas Law 5.4 Gas Stiochiometry 5.5 Dalton s Law of Partial Pressures 5.6 The Kinetic molecular Theory

More information

Apparatus for Studying the Relationship Between Pressure and Volume of a Gas

Apparatus for Studying the Relationship Between Pressure and Volume of a Gas The Gas Laws Apparatus for Studying the Relationship Between Pressure and Volume of a Gas As P (h) increases V decreases Boyle s Law P x V = constant P 1 x V 1 = P 2 x V 2 Constant temperature Constant

More information

Chapter Ten- Gases. STUDY GUIDE AP Chemistry

Chapter Ten- Gases. STUDY GUIDE AP Chemistry STUDY GUIDE AP Chemistry Chapter Ten- Gases Lecture Notes 10.1 Characteristics of Gases All substances have three phases: solid, liquid and gas. Substances that are liquids or solids under ordinary conditions

More information

Chapter 6: The States of Matter

Chapter 6: The States of Matter Spencer L. Seager Michael R. Slabaugh www.cengage.com/chemistry/seager Chapter 6: The States of Matter PHYSICAL PROPERTIES OF MATTER All three states of matter have certain properties that help distinguish

More information

Section Using Gas Laws to Solve Problems

Section Using Gas Laws to Solve Problems Gases and Gas Laws Section 13.2 Using Gas Laws to Solve Problems Kinetic Molecular Theory Particles of matter are ALWAYS in motion Volume of individual particles is zero. Consists of large number of particles

More information

Homework Problem Set 3 Solutions

Homework Problem Set 3 Solutions Chemistry 360 Dr. Jean M. Standard Homework Problem Set 3 Solutions 1. A bicycle tire was inflated to 9.3 bar total pressure in the cool of the morning when the temperature was 50 F. Later, the bicycle

More information

The Gaseous State. Definition

The Gaseous State. Definition The Gaseous State Lecture Material Basic Chemistry 1 2013/2014 Inneke Hantoro Definition A gas is a substance that is normally in the gaseous state at ordinary temperatures and pressures. A vapor is the

More information

3. Which of the following elements is primarily responsible for the photochemical smog? Chemistry 12, Exam III, Form A, April 4, 2001

3. Which of the following elements is primarily responsible for the photochemical smog? Chemistry 12, Exam III, Form A, April 4, 2001 Chemistry 12, Exam III, Form A, April 4, 2001 In all questions involving gases, assume that the ideal-gas laws hold, unless the question specifically refers to the non-ideal behavior. 1. It takes 21.3

More information

Centimeters of mercury

Centimeters of mercury CHAPTER 11 PROPERTIES OF GASES Gases have an indefinite shape: a gas takes the shape of its container and fills it uniformly. If the shape of the container changes, so does the shape of the gas. Gases

More information

L = 6.02 x mol Determine the number of particles and the amount of substance (in moles)

L = 6.02 x mol Determine the number of particles and the amount of substance (in moles) 1.1 The Mole 1.1.1 - Apply the mole concept to substances A mole is the name given to a certain quantity. It represents 6.02 x 10 23 particles. This number is also known as Avogadro's constant, symbolised

More information

CHAPTER 12 GASES AND KINETIC-MOLECULAR THEORY

CHAPTER 12 GASES AND KINETIC-MOLECULAR THEORY . Pressure CHAPER GASES AND KINEIC-MOLECULAR HEORY. Boyle s Law: he -P Relationship 3. Charles Law: he - Relationship 4. Standard &P 5. he Combined Gas Law Equation 6. Avogadro s Law and the Standard Molar

More information

SUPeR Chemistry CH 222 Practice Exam

SUPeR Chemistry CH 222 Practice Exam SUPeR Chemistry CH 222 Practice Exam This exam has been designed to help you practice working multiple choice problems over the material that will be covered on the first CH 222 midterm. The actual exams

More information

The Kinetic-Molecular Theory of Gases

The Kinetic-Molecular Theory of Gases The Kinetic-Molecular Theory of Gases kinetic-molecular theory of gases Originated with Ludwig Boltzman and James Clerk Maxwell in the 19th century Explains gas behavior on the basis of the motion of individual

More information

Chapter 5. Gases and the Kinetic-Molecular Theory

Chapter 5. Gases and the Kinetic-Molecular Theory Chapter 5 Gases and the Kinetic-Molecular Theory Macroscopic vs. Microscopic Representation Kinetic Molecular Theory of Gases 1. Gas molecules are in constant motion in random directions. Collisions among

More information

Chapter 5. Question. Question. Answer. Answer. Question (continued) The Gaseous State

Chapter 5. Question. Question. Answer. Answer. Question (continued) The Gaseous State Chapter 5 CRS s The Gaseous State Equal volumes of propane, C 3 H 8, and carbon monoxide at the same temperature and pressure have the same a. density. b.. c. number of atoms. 1) a only 2) b only 3) c

More information

(E) half as fast as methane.

(E) half as fast as methane. Name AP Chem / / AP Chem Practice Exam #2 Part I: 40 Questions, 40 minutes, Multiple Choice, No Calculator Allowed Bubble the correct answer on the BLUE SIDE of your scantron for each of the following.

More information

Calculate the mass of L of oxygen gas at 25.0 C and 1.18 atm pressure.

Calculate the mass of L of oxygen gas at 25.0 C and 1.18 atm pressure. 142 Calculate the mass of 22650 L of oxygen gas at 25.0 C and 1.18 atm pressure. Volume of a 10'x10'x8' room 1) First, find the MOLES of gas using the ideal gas equation and the information given. 2) Convert

More information

AP Chemistry Unit 5 - Gases

AP Chemistry Unit 5 - Gases Common Gases at Room Temperature AP Chemistry Unit 5 - Gases Know these! HCN toxic slight odor of almonds HS toxic odor of rotten eggs CO toxic odorless CO odorless CH4 methane odorless, flammable CH4

More information

Gases. Chapter 5. Elements that exist as gases at 25 0 C and 1 atmosphere

Gases. Chapter 5. Elements that exist as gases at 25 0 C and 1 atmosphere Gases Chapter 5 Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. 1 Elements that exist as gases at 25 0 C and 1 atmosphere 2 3 1 Physical Characteristics of Gases

More information

IMPORTANT CONCEPTS. 5.1 Pressure Units for pressure STP. 5.6 Kinetic Molecular Theory. 5.3 Ideal Gas Law. 5.4 Gas Stoichiometry Gas density Molar mass

IMPORTANT CONCEPTS. 5.1 Pressure Units for pressure STP. 5.6 Kinetic Molecular Theory. 5.3 Ideal Gas Law. 5.4 Gas Stoichiometry Gas density Molar mass TOPICS 1. Intermolecular Forces 2. Properties of Gases 3. Pressure 4. Gas Laws Boyle, Charles, Lussac 5. Ideal Gas Law 6. Gas Stoichiometry 7. Partial Pressure 8. Kinetic Molecular Theory 9. Effusion &

More information

vapors: gases of substances that are normally liquids or solids 1 atm = 760 mm Hg = 760 torr = kpa = bar

vapors: gases of substances that are normally liquids or solids 1 atm = 760 mm Hg = 760 torr = kpa = bar Gases A Chemistry Lecture Outline Name: Basics on Gases composition of the atmosphere: properties of gases: vapors: gases of substances that are normally liquids or solids Equation for pressure: 1 atm

More information

1 atm 760 mmhg =.454 atm (3 points/6) 14.7 psi =.816 atm (3 points/9)

1 atm 760 mmhg =.454 atm (3 points/6) 14.7 psi =.816 atm (3 points/9) Chapter 5 Homework acket 1. Gases generally have a) low density b) high density c) closely packed particles d) no increase in volume when temperature is increased e) no decrease in volume when pressure

More information

Lecture 2 PROPERTIES OF GASES

Lecture 2 PROPERTIES OF GASES Lecture 2 PROPERTIES OF GASES Reference: Principles of General Chemistry, Silberberg Chapter 6 SOME FUNDAMENTAL DEFINITIONS: SYSTEM: the part of the universe being the subject of study 1 SOME FUNDAMENTAL

More information

CHEMISTRY 102A Spring 2012 Hour Exam II. 1. My answers for this Chemistry 102 exam should be graded with the answer sheet associated with:

CHEMISTRY 102A Spring 2012 Hour Exam II. 1. My answers for this Chemistry 102 exam should be graded with the answer sheet associated with: . My answers for this Chemistry 0 exam should be graded with the answer sheet associated with: a) Form A b) Form B c) Form C d) Form D e) Form E. A sample of LSD (D-lysergic acid diethylamide, C 4 H 30

More information

Chemistry 11. Unit 11 Ideal Gas Law (Special Topic)

Chemistry 11. Unit 11 Ideal Gas Law (Special Topic) Chemistry 11 Unit 11 Ideal Gas Law (Special Topic) 2 1. States of substances It has been studied in Unit 3 that there exist 3 states of matter in nature: gas, liquid and solid. (Technically there is the

More information

10/16/2018. Why study gases? An understanding of real world phenomena. An understanding of how science works.

10/16/2018. Why study gases? An understanding of real world phenomena. An understanding of how science works. 10/16/018 Kinetic Theory and the Behavior of Ideal & Real Gases Why study gases? An understanding of real world phenomena. An understanding of how science works. 1 10/16/018 A Gas Uniformly fills any container.

More information

Gases. Characteristics of Gases. Unlike liquids and solids, gases

Gases. Characteristics of Gases. Unlike liquids and solids, gases Gases Characteristics of Gases Unlike liquids and solids, gases expand to fill their containers; are highly compressible; have extremely low densities. 1 Pressure Pressure is the amount of force applied

More information

Name AP Chemistry / / Chapter 5 Collected AP Exam Free Response Questions Answers

Name AP Chemistry / / Chapter 5 Collected AP Exam Free Response Questions Answers Name AP Chemistry / / Chapter 5 Collected AP Exam Free Response Questions 1980 2010 - Answers 1982 - #5 (a) From the standpoint of the kinetic-molecular theory, discuss briefly the properties of gas molecules

More information

C H E M 1 CHEM 101-GENERAL CHEMISTRY CHAPTER 5 GASES INSTR : FİLİZ ALSHANABLEH

C H E M 1 CHEM 101-GENERAL CHEMISTRY CHAPTER 5 GASES INSTR : FİLİZ ALSHANABLEH C H E M 1 CHEM 101-GENERAL CHEMISTRY CHAPTER 5 GASES 0 1 INSTR : FİLİZ ALSHANABLEH CHAPTER 5 GASES Properties of Gases Pressure History and Application of the Gas Laws Partial Pressure Stoichiometry of

More information

Chapter 5 Gases - 4 Gas Stoichiometry. Dr. Sapna Gupta

Chapter 5 Gases - 4 Gas Stoichiometry. Dr. Sapna Gupta Chapter 5 Gases - 4 Gas Stoichiometry Dr. Sapna Gupta Stoichiometry in Gases Amounts of gaseous reactants and products can be calculated by utilizing The ideal gas law to relate moles to T, P and V. Moles

More information

a) 1.3 x 10 3 atm b) 2.44 atm c) 8.35 atm d) 4.21 x 10-3 atm e) 86.5 atm

a) 1.3 x 10 3 atm b) 2.44 atm c) 8.35 atm d) 4.21 x 10-3 atm e) 86.5 atm 1. (6 pts) A sample of gas with a volume of 750 ml exerts a pressure of 756 mm Hg at 30.0 0 C. What pressure (atm) will the sample exert when it is compressed to 250 ml and cooled to -25.0 0 C? a) 1.3

More information

Substances that Exist as Gases

Substances that Exist as Gases Gases Properties of Gases assume the volume and shape of their containers most compressible of the states of matter mix evenly and completely with other gases much lower density than other forms of matter

More information

Engr. Yvonne Ligaya F. Musico Chemical Engineering Department

Engr. Yvonne Ligaya F. Musico Chemical Engineering Department GASEOUS STATE Engr. Yvonne Ligaya F. Musico Chemical Engineering Department TOPICS Objective Properties of Gases Kinetic Molecular Theory of Gases Gas Laws OBJECTIVES Determine how volume, pressure and

More information

CHAPTER 14: The Behavior of Gases

CHAPTER 14: The Behavior of Gases Name: CHAPTER 14: The Behavior of Gases Period: RELATIONSHIPS BETWEEN PRESSURE, VOLUME & TEMPERATURE OF A GAS Boyle s Law-Pressure and Volume Volume (ml) Pressure ( ) 60 50 40 30 20 10 Practice problem:

More information

1,2,8,9,11,13,14,17,19,20,22,24,26,28,30,33,38,40,43,45,46,51,53,55,57,62,63,80,82,88,94

1,2,8,9,11,13,14,17,19,20,22,24,26,28,30,33,38,40,43,45,46,51,53,55,57,62,63,80,82,88,94 CHAPTER 5GASES 1,,8,9,11,1,14,17,19,0,,4,6,8,0,,8,40,4,45,46,51,5,55,57,6,6,80,8,88,94 5.1 a) The volume of the liquid remains constant, but the volume of the gas increases to the volume of the larger

More information

Chapter 11. Preview. Lesson Starter Objectives Pressure and Force Dalton s Law of Partial Pressures

Chapter 11. Preview. Lesson Starter Objectives Pressure and Force Dalton s Law of Partial Pressures Preview Lesson Starter Objectives Pressure and Force Dalton s Law of Partial Pressures Section 1 Gases and Pressure Lesson Starter Make a list of gases you already know about. Separate your list into elements,

More information

5. What pressure (in atm) would be exerted by 76 g of fluorine gas in a 1.50 liter vessel at -37 o C? a) 26 atm b) 4.1 atm c) 19,600 atm d) 84 atm

5. What pressure (in atm) would be exerted by 76 g of fluorine gas in a 1.50 liter vessel at -37 o C? a) 26 atm b) 4.1 atm c) 19,600 atm d) 84 atm Test bank chapter (5) Choose the most correct answer 1. A sample of oxygen occupies 47.2 liters under a pressure of 1240 torr at 25 o C. What volume would it occupy at 25 o C if the pressure were decreased

More information

Chapter 13. Kinetic Theory (Kinetikos- Moving ) Based on the idea that particles of matter are always in motion

Chapter 13. Kinetic Theory (Kinetikos- Moving ) Based on the idea that particles of matter are always in motion Chapter 3 Kinetic Theory (Kinetikos- Moving ) Based on the idea that particles of matter are always in motion The motion has consequences Behavior of Gases Physical Properties of Gases Ideal Gas an imaginary

More information

KINETIC MOLECULAR THEORY

KINETIC MOLECULAR THEORY KINETIC MOLECULAR THEORY IMPORTANT CHARACTERISTICS OF GASES 1) Gases are highly compressible An external force compresses the gas sample and decreases its volume, removing the external force allows the

More information

mcdonald (pam78654) HW 4B: Gases and Kinetics laude (89560) 1

mcdonald (pam78654) HW 4B: Gases and Kinetics laude (89560) 1 mcdonald (pam78654) HW 4B: Gases and Kinetics laude (89560) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page find all choices before answering.

More information

AP Chemistry Ch 5 Gases

AP Chemistry Ch 5 Gases AP Chemistry Ch 5 Gases Barometer - invented by Evangelista Torricelli in 1643; uses the height of a column of mercury to measure gas pressure (especially atmospheric) Manometer- a device for measuring

More information

12.2. The Ideal Gas Law. Density and Molar Mass of Gases SECTION. Key Terms

12.2. The Ideal Gas Law. Density and Molar Mass of Gases SECTION. Key Terms SECTION 12.2 The Ideal Gas Law You have related the combined gas law to Avogadro s volume-mole gas relationship using two sets of conditions. This enabled you to make calculations of pressure, temperature,

More information

Lecture 24. Ideal Gas Law and Kinetic Theory

Lecture 24. Ideal Gas Law and Kinetic Theory Lecture 4 Ideal Gas Law and Kinetic Theory Today s Topics: Ideal Gas Law Kinetic Theory of Gases Phase equilibria and phase diagrams Ideal Gas Law An ideal gas is an idealized model for real gases that

More information

Forces between atoms/molecules

Forces between atoms/molecules Professor K gases Forces between atoms/molecules BONDS are the INTRAMOLECULAR FORCES holding the atoms in molecules together... What holds the molecules of a solid or liquid together?... INTERMOLECULAR

More information

Ideal Gas & Gas Stoichiometry

Ideal Gas & Gas Stoichiometry Ideal Gas & Gas Stoichiometry Avogadro s Law V a number of moles (n) V = constant x n Constant temperature Constant pressure V 1 /n 1 = V 2 /n 2 Ammonia burns in oxygen to form nitric oxide (NO) and water

More information

Chapter 5. The Properties of Gases. Gases and Their Properties. Why Study Gases? Gas Pressure. some very common elements exist in a gaseous state

Chapter 5. The Properties of Gases. Gases and Their Properties. Why Study Gases? Gas Pressure. some very common elements exist in a gaseous state Chapter 5 Gases and Their Properties Why Study Gases? some very common elements exist in a gaseous state our gaseous atmosphere provides one means of transferring energy and material throughout the globe

More information

CHEMISTRY XL-14A GASES. August 6, 2011 Robert Iafe

CHEMISTRY XL-14A GASES. August 6, 2011 Robert Iafe CHEMISTRY XL-14A GASES August 6, 2011 Robert Iafe Chemistry in the News 2 Polymer nicotine trap is composed of a porphyrin derivative (black), in which amide pincers (green) are attached to the zinc (violet)

More information

Gases. Chapter 11. Preview. 27-Nov-11

Gases. Chapter 11. Preview. 27-Nov-11 Chapter 11 Gases Dr. A. Al-Saadi 1 Preview Properties and measurements of gases. Effects of temperature, pressure and volume. Boyle s law. Charles s law, and Avogadro s law. The ideal gas equation. Gas

More information

Name: Score: /100. Part I. Multiple choice. Write the letter of the correct answer for each problem. 3 points each

Name: Score: /100. Part I. Multiple choice. Write the letter of the correct answer for each problem. 3 points each Name: Score: /100 Part I. Multiple choice. Write the letter of the correct answer for each problem. 3 points each 1. Which of the following contains the greatest number of moles of O? A) 2.3 mol H 2 O

More information

Properties of Gases. assume the volume and shape of their containers. most compressible of the states of matter

Properties of Gases. assume the volume and shape of their containers. most compressible of the states of matter Gases Properties of Gases assume the volume and shape of their containers most compressible of the states of matter mix evenly and completely with other gases much lower density than other forms of matter

More information

Gas Volumes and the Ideal Gas Law

Gas Volumes and the Ideal Gas Law SECTION 11.3 Gas Volumes and the Ideal Gas Law Section 2 presented laws that describe the relationship between the pressure, temperature, and volume of a gas. The volume of a gas is also related to the

More information