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

Chapter 5 Gases

Chapter 5 Variables used to describe a gas Pressure (P) collision of gas molecules with wall of container. Volume (V) Temperature (T) related to average speed of gas molecules. Must be in Kelvin Number of moles (n)

Chapter 5 Pressure force Pr essure = area SI units = Newton/meter 2 = 1 Pascal (Pa) Standard Pressure = 1 atm = 101,325 Pa = 101.3 kpa = 760 mm Hg = 760 torr Copyright Cengage Learning. All rights reserved 4

Chapter 5 Barometer Device used to measure atmospheric pressure. Hg flows out of the tube until the pressure of the column of Hg standing on the surface of the Hg in the dish is equal to the pressure of the air on the rest of the surface of the Hg in the dish. Copyright Cengage Learning. All rights reserved 5

Chapter 5 Pressure Conversions Example: The pressure of a gas is measured as 2.5 atm. Represent this pressure in both torr and pascals. 760 torr 3 2.5 atm = 1.9 10 torr 1 atm 101,325 Pa 5 2.5 atm = 2.5 10 Pa 1 atm Copyright Cengage Learning. All rights reserved 7

Chapter 5 The Kinetic Molecular Theory provides a model for gases. Postulates of the Kinetic Molecular Theory: 1) 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). Copyright Cengage Learning. All rights reserved 8

Chapter 5 Postulates of the Kinetic Molecular Theory 2) 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. Copyright Cengage Learning. All rights reserved 9

Chapter 5 Postulates of the Kinetic Molecular Theory 3) The particles are assumed to exert no forces on each other; they are assumed neither to attract nor to repel each other. Copyright Cengage Learning. All rights reserved 10

Chapter 5 Postulates of the Kinetic Molecular Theory 4) The average kinetic energy of a collection of gas particles is assumed to be directly proportional to the Kelvin temperature of the gas. Copyright Cengage Learning. All rights reserved 11

Chapter 5 CONCEPT CHECK! You are holding two balloons of the same volume. One contains He, the other H. Complete each of the following statements with different or the same and be prepared to justify your answer. Ne H 2 Copyright Cengage Learning. All rights reserved 13

Chapter 5 CONCEPT CHECK! The numbers of moles of the gas in the two balloons are. The temperatures of the gas in the two balloons are. The pressures of the gas in the two balloons are. The densities of the gas in the two balloons are. He H 2 Copyright Cengage Learning. All rights reserved 14

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Liquid Nitrogen and a Balloon What happened to the gas in the balloon? A decrease in temperature was followed by a decrease in the pressure and volume of the gas in the balloon. Mathematical relationships among the properties of a gas (Pressure, Volume, Temperature and Moles) were discovered from observations like these. Copyright Cengage Learning. All rights reserved 16

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Boyle s Law Pressure and volume are inversely related (constant T and n). The formula: P 1 V 1 = P 2 V 2 Copyright Cengage Learning. All rights reserved 17 (Plotting 1/V gives a straight line)

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Boyle s law To play movie you must be in Slide Show Mode PC Users: Please wait for content to load, then click to play Mac Users: CLICK HERE Copyright Cengage Learning. All rights reserved 19

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro EXERCISE! A sample of helium gas occupies 12.4 L at 23 C and 0.956 atm. What volume will it occupy at 1.20 atm assuming that the temperature stays constant? P 1 V 1 = P 2 V 2 (0.956 atm)(12.4 L) = (1.20 atm)(v 2 ) 9.88 L Copyright Cengage Learning. All rights reserved 20

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Charles s Law Volume and Temperature (in Kelvin) are directly related (constant P and n). K = C + 273 0 K is absolute zero. V T 1 2 = V T 1 2 Copyright Cengage Learning. All rights reserved 21

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro V T EXERCISE! Suppose a balloon containing 1.30 L of air at 24.7 C is placed into a beaker containing liquid nitrogen at 78.5 C. What will the volume of the sample of air become (at constant pressure)? Remember to convert the temperatures to Kelvin. = V T 1 2 1 2 1.30 L = V 2 (24.7+273.15) (-78.5 + 273.15) 0.849 L Copyright Cengage Learning. All rights reserved 24

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro CONCEPT CHECK! You have a sample of nitrogen gas (N 2 ) in a container fitted with a piston that maintains a pressure of 6.00 atm. Initially, the gas is at 45 C in a volume of 6.00 L. You then cool the gas sample. Copyright Cengage Learning. All rights reserved 25

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro CONCEPT CHECK! Which best explains the final result that occurs once the gas sample has cooled? a) The pressure of the gas increases. b) The volume of the gas increases. c) The pressure of the gas decreases. d) The volume of the gas decreases. e) Both volume and pressure change. Copyright Cengage Learning. All rights reserved 26

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro CONCEPT CHECK! The gas sample is then cooled to a temperature of 15 C. Solve for the new condition. (Hint: A moveable piston keeps the pressure constant overall, so what condition will change?) Use Charles s Law to solve 6.00 L = V 2 (45+273) (15+273) 5.43 L Copyright Cengage Learning. All rights reserved 27

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Avogadro s Law Avogadro s hypothesis states that at a fixed temperature and pressure, the volume of a gas is directly proportional to the amount of gas (that is, to the number of moles of gas, n, or to the number of molecules of gas). Volume and number of moles are directly related (constant T and P). The formula: 1 2 1 2 Copyright Cengage Learning. All rights reserved 28 n V = n V

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro EXERCISE! If 2.45 mol of argon gas occupies a volume of 89.0 L, what volume will 2.10 mol of argon occupy under the same conditions of temperature and pressure? n V 1 2 = n V 1 2 2.45 mol = 2.10 mol 89.0 L V 2 76.3 L Copyright Cengage Learning. All rights reserved 29

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro Lussac s Law The pressure of a gas is directly proportional to its Kelvin temperature when the volume is held constant. P 1 = P 2 T 1 T 2

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro EXERCISE! A sample of nitrogen gas has a pressure of 6.58kPa at 539K. What will the pressure of the gas be when its temperature reaches 211K? P 1 = P 2 T 1 T 2 6.58kPa = P 2 539K 211K 2.58 kpa Copyright Cengage Learning. All rights reserved 31

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro The Combined Gas Law The combined gas law combines Boyle Law, Charles Law, and Lussac s Law. The formula: Formula combining n: P 1 V 1 = P 2 V 2 P 1 V 1 P 2 V 2 = n 1 T 1 n 2 T 2 T 1 T 2 You can cancel any term (P, V, n, T) that is the same (or held constant) on both sides.

Section 5.2 The Gas Laws of Boyle, Charles, and Avogadro EXERCISE! At what temperature (in ⁰C) does 121 ml of CO 2 at 27 ⁰ C and 1.05 atm occupy a volume of 293 ml at a pressure of 1.40 atm? P 1 V 1 = P 2 V 2 T 1 T 2 (1.05 atm)(121 ml) = (1.40 atm)(293 ml) (27+273) (T 2 + 273) 696 C Copyright Cengage Learning. All rights reserved 33

Chapter 5

Section 5.3 The Ideal Gas Law We can bring all of these laws together into one comprehensive law: PV = nrt Be careful of units! R = The universal gas constant = 0.08206 L atm/mol K =8.314 L kpa/mol K =62.36 L mmhg/mol K Copyright Cengage Learning. All rights reserved 35

Section 5.3 The Ideal Gas Law EXERCISE! An automobile tire at 23 ⁰C with an internal volume of 25.0 L is filled with air to a total pressure of 3.18 atm. Determine the number of moles of air in the tire. PV = nrt (3.18 atm)(25.0 L) = n(0.08206)(23+273k) 3.27 mol Copyright Cengage Learning. All rights reserved 37

Section 5.3 The Ideal Gas Law EXERCISE! What is the pressure in atm in a 304.0 L tank that contains 5.670 kg of helium at 25 ⁰C? First, convert kg of He into mol! 5.670 kg He 1000 g He 1mol He = 1416mol 1kg He 4.003 g He PV = nrt (P)(304.0L) = (1416 mol)(0.08206)(25+273k) = 114 atm Copyright Cengage Learning. All rights reserved 38

Section 5.3: 5.4 Gas Stoichiometry The Ideal Gas Law EXERCISE! A sample of oxygen gas has a volume of 2.50 L at STP. How many grams of O 2 are present? PV = nrt (1.00 atm)(2.50 L) = n(0.08206)(273) n = 0.112 mol O 2 32.00 g O 2 = 3.57 g O 2 1 mol O 2 Copyright Cengage Learning. All rights reserved 39

Section 5.3: 5.4 Gas Stoichiometry Calculating 22.42L Using The Ideal Gas Law Molar Volume of an Ideal Gas at STP For 1 mole of an ideal gas at 0 ⁰C and 1 atm, the volume of the gas is 22.42 L. 1.000 mol 0.08206 L atm/k mol 273.2 K nrt V = = = 22.42 L P 1.000 atm STP = standard temperature and pressure 0⁰C and 1 atm Therefore, the molar volume is 22.42 L at STP. Copyright Cengage Learning. All rights reserved 40

Section 5.4 Gas Stoichiometry Applications of the Ideal Gas Law: Molecular Mass Determination M = molar mass and m = mass in grams m (grams) m M (molar mass)= so n = n (moles) M The ideal gas equation rearranges to: PV n = RT Setting the equations equal to one another: m = PV M RT Rearranges to: mrt M = PV Alternative to equation: (A) find n using the ideal gas equation; (B) Divide m (grams) by n (moles) to get grams/mol.

Section 5.4 Gas Stoichiometry Applications of the Ideal Gas Law: Molecular Mass Determination Example If 0.550 g of a gas occupies 0.200 L at 0.968 atm and 289 K, what is the molecular mass of the gas? mrt M = PV (0.550g) (0.08206 L-atm/mol-K))(289K) M = (0.968atm)(0.200L) M = 67.4 g/mol

Section 5.4 Gas Stoichiometry Applications of the Ideal Gas Law: Molecular Mass / Density Determination mrt m MP M = rearranges to = PV V RT m MP and density = so d = V RT drt Rearranges to M = P Alternative: find volume of one mole (n = 1) or other fixed quantity of gas. Divide mass of that quantity by volume to find g/l.

Section 5.4 Gas Stoichiometry Applications of the Ideal Gas Law: Molecular Mass /Density Determination Molar Mass of a Gas Molar mass = drt P g L atm K g L mol K = = atm mol d = density of gas T = temperature in Kelvin P = pressure of gas R = universal gas constant Copyright Cengage Learning. All rights reserved 44

Section 5.4 Gas Stoichiometry Applications of the Ideal Gas Law: Molecular Mass / Density Determination EXERCISE! drt M = P What is the density of F 2 at STP (in g/l)? MP (19.00g/mol 2)(1.00 atm) d = = RT (0.08206 L-atm/mol-K)(273 K) OR, since it is at STP: d = mass = molar mass = 38.00g vol molar vol 22.42L OR, use PV=nRT: Copyright Cengage Learning. All rights reserved 45

Section 5.4 Gas Stoichiometry The Law of Combining Volumes When gases measured at the same temperature and pressure react, the volumes of gaseous reactants and products are in small wholenumber ratios. Example: At a given temperature and pressure, 2.00 L of H 2 will react with 1.00 L of O 2 Each flask contains the same number of molecules. Therefore, the ratio of volumes is the same as the mole ratio from the balanced equation: 2 H 2 + O 2 2 H 2 O

Section 5.4 Gas Stoichiometry We can use the law of combining volumes for stoichiometry only for gases and only if the gases are at the same temperature and pressure. Otherwise, we must use stoichiometric methods from Chapter 3 combined with the ideal gas equation.

Section 5.4 Gas Stoichiometry Example How many liters of O 2 (g) are consumed for every 10.0 L of CO 2 (g) produced in the combustion of liquid pentane, C 5 H 12, if all volumes are measured at the same temperature and pressure? C 5 H 12 (l) + 8O 2 (g) 5 CO 2 (g) + 6 H 2 O(l) 8 L O 2 10.0 L CO 2 x = 16.0 L O 2 5 L CO 2

Section 5.4 Gas Stoichiometry As in other stoichiometry calculations, the problem centers around the mole ratio: If A is a gas, we find moles of A first by using the ideal gas equation and P, V, and T. If B is a gas, we solve for moles of B (n), then use the ideal gas equation to find P, V, or T.

Section 5.4 Gas Stoichiometry Example In the chemical reaction used in automotive air-bag safety systems, N 2 (g) is produced by the decomposition of sodium azide, NaN 3 (s), at a somewhat elevated temperature: 2 NaN 3 (s) 2 Na(l) + 3 N 2 (g) What volume of N 2 (g), measured at 25⁰C and 0.980 atm, is produced by the decomposition of 62.5 g NaN 3? 1 mol NaN 3 3 mol N 2 62.5 g NaN 3 x x = 1.44 mol N 2 65.01 g NaN 3 2 mol NaN 3 PV = nrt (0.980 atm)(v) = (1.44 mol)(0.08206 L-atm/mol-K) (273+25K) V = 35.9 L N 2

Section 5.4 Gas Stoichiometry For a mixture of gases in a container, P Total = P 1 + P 2 + P 3 +... The total pressure exerted is the sum of the partial pressures that each gas would exert if it were alone. n 1 RT P 1 = V n 2 RT P 2 = V n 3 RT P 3 = V 51

Section 5.4 Gas Stoichiometry EXERCISE! 27.4 L of oxygen gas at 25.0⁰C and 1.30 atm, and 8.50 L of helium gas at 25.0 ⁰ C and 2.00 atm were pumped into a tank with a volume of 5.81 L at 25 ⁰ C. Calculate the new partial pressure of oxygen. P 1 V 1 =P 2 V 2 (1.30 atm)(27.4 L) = (P 2 )(5.81 L) P 2 = 6.13 atm Calculate the new partial pressure of helium. P 1 V 1 =P 2 V 2 (2.00 atm)(8.50 L) = (P 2 )(5.81 L) P 2 = 2.93 atm Calculate the new total pressure of both gases. 6.13 + 2.93 = 9.06 atm. 52

Section 5.4 Gas Stoichiometry CONCEPT CHECK! Ne V Ne = 2V Ar Which of the following best represents the mass ratio of Ne:Ar in the balloons? 1:1 1:2 2:1 1:3 3:1 Ar 1 L Ne = 1/2 L Ar 1 mol Ne = 1/2 mol Ar 1 mol Ne = 20 g 1 mol Ar = 40 g; ½ mol Ar = 20g Therefore: 1:1 Mass ratio Copyright Cengage Learning. All rights reserved 53

Section 5.4 Gas Stoichiometry Example A 1.00-L sample of dry air at 25 ⁰C contains 0.0319 mol N 2, 0.00856 mol O 2, 0.000381 mol Ar, and 0.00002 mol CO 2. Calculate the partial pressure of N 2 (g) in the mixture. n N2 RT P N2 = V =0.0319 mol (0.08206 L- atm/mol-k) 298 K 1.00 L = 0.780 atm

Section 5.4 Gas Stoichiometry The mole fraction (x) is the ratio of the number of moles of a given component in a mixture to the total number of moles in the mixture n 1 x 1 = n total P 1 = P total P 1 = x 1 P total Partial Pressure = Mole fraction x Total Pressure

Section 5.4 Gas Stoichiometry Ex. The partial pressure of oxygen gas was observed to be 156 torr in air with an atmospheric pressure of 743 torr. Calculate the mole fraction of O 2 present. X 1 = P 1 = P total 156torr 743torr = 2.10x10-1 (no unit)

Section 5.4 Gas Stoichiometry Ex. The mole fraction of nitrogen in the air is 0.7808. Calculate the partial pressure (in torr) of N 2 at STP..7808 = P 1 760 torr P 1 = 593torr

Section 5.4 Gas Stoichiometry Collection of Gases over Water As (essentially insoluble) gas is bubbled into the container for collection, the water is displaced. The gas collected is usually saturated with water vapor. If O 2 is being generated Assuming the gas is saturated with water vapor, the partial pressure of the water vapor is the vapor pressure of the water. P gas = P total P H 2O(g) = P bar P H 2O(g) P bar = barometric pressure what TWO gases are present here?

Section 5.4 Gas Stoichiometry

Section 5.4 Gas Stoichiometry Vapor Pressure of Water as a Function of Temp Temp.(C) Pressure (mmhg) 15 12.8 16 13.6 17 14.5 18 15.5 19 16.5 20 17.5 21 18.7 22 19.8 23 21.1

Section 5.4 Gas Stoichiometry Example: Hydrogen produced in the following reaction is collected over water at 23⁰C when the barometric pressure is 742 Torr: 2 Al(s) + 6 HCl(aq) 2 AlCl 3 (aq) + 3 H 2 (g) What volume of hydrogen gas will be collected in the reaction of 1.50 g Al(s) with excess HCl(aq)? P total = P H2 + P H2O 742 Torr = P H2 + 21.1 Torr P H2 = 721 Torr =.949atm 1.50 g Al x 1 mol Al x 3 mol H 2 = 0.0834 mol H 2 26.98 g Al 2 mol Al V = nrt =0.0834 mol (0.08206 L-atm/mol-K)(273 + 23 K) P.949 atm = 2.14 L

Section 5.6 The Kinetic Molecular Theory of Gases Final units are in m/s. Copyright Cengage Learning. All rights reserved Root Mean Square Velocity Gas molecules do not all move at the same speed. The average velocity of the gas is known as the Root Mean Square Velocity. R = 8.315 L-Kpa/mol-K **Must use this R so units cancel! T = temperature of gas (in K) M = mass of a mole of gas in kg u rms = 3RT M

Section 5.6 The Kinetic Molecular Theory of Gases Example: Calculate the root mean square velocity for the atoms in a sample of oxygen gas at 32 C. * M must be in Kg 32.00 g 1 kg =.03200 kg 1000g * T must be in K T= 32 + 273 = 305 K u rms = 3RT M u rms 3(8.315)(305 K) 0.03200kg = 488m/s

Section 5.7 Effusion and Diffusion Diffusion: The mixing of gases. At a fixed temperature, molecules of higher mass (M) move more slowly than molecules of lower mass. Effusion Describes the passage of a gas through a tiny orifice into an evacuated chamber. Rate of effusion measures the speed at which the gas is transferred into the chamber. 64

Section 5.7 Effusion and Diffusion Gases with lower molecular masses will diffuse and/or effuse faster than gases with higher molecular masses.

Section 5.7 Effusion and Diffusion Concept Check: Arrange the following gases in order of increasing speed of the particles at the same temperature: SO 2, NH 4, Ne SO 2 = 64.1 g NH 4 = 18.0 g Ne = 20.2 g SO 2, Ne, NH 4

Section 5.7 Effusion and Diffusion Graham s Law of Effusion Rate Rate gas a Molarmass Molarmass gas b gas b gas a M 1 and M 2 represent the molar masses of the gases. Copyright Cengage Learning. All rights reserved 69

Section 5.7 Effusion and Diffusion Example: Which gas effuses faster, Helium or Nitrogen? How much faster? Rate Rate He N 2 28.0g 4.0g Helium is 2.6 times faster than N 2 ; Nitrogen is.38 times slower than He

Section 5.7 Effusion and Diffusion Example: Which gas effuses faster, Oxygen or Carbon Dioxide? How much faster? Rate Rate O 2 CO 2 44.0g 32.0g Oxygen effuses 1.17 times faster than CO 2 ; CO 2 effuses.855 times slower than O 2

Section 5.7 Effusion and Diffusion Since Rate is a unit per time (1/t), Graham s law can be rearranged to solve for time instead of rate assuming the same amount of each gas. Time Time a b molarmass( a) molarmass( b)

Section 5.7 Effusion and Diffusion Example: It takes 354 s for 1.00 ml of Xe to effuse through a small hole. Under the same conditions, how long will it take for 1.00 ml of nitrogen to effuse? Solution A: Time Time a b molarmass( a) molarmass() b x = 164 s TimeN 28.02 2 354s 131.3

Section 5.7 Effusion and Diffusion Example: It takes 354 s for 1.00 ml of Xe to effuse through a small hole. Under the same conditions, how long will it take for 1.00 ml of nitrogen to effuse? Solution B: Rate Rate gas a Molarmass Molarmass gas b gas b gas a x = 164 s 1.00mL X 131.3g 1.00mL 28.02g 354 s

Section 5.8 Real Gases Real Gases and van der Waals Equation An ideal gas is a hypothetical concept. No gas exactly follows the ideal gas law. Real gas molecules do have volume, and the molecules of real gases do experience attractive forces. Gases behave most like ideal gases at low pressure, high temperature. Copyright Cengage Learning. All rights reserved 75

Real Gases (van der Waals Equation) n 2 a (P + ) (V-nb) = nrt V 2 Corrected pressure Corrected volume a and b are constants derived from experimental results a has units of L 2 atm/mol 2 b has units of L/mol. Copyright Cengage Learning. All rights reserved 76

Two corrective Factors: 1. n 2 a/v 2 alters the pressure It accounts for the intermolecular attractive forces between gas molecules A low value for a reflects weak IM forces; a high value reflects stronger IM forces. 2. nb alters the volume n 2 a (P + )(V-nb) = nrt V 2 It accounts for the volume occupied by the gas molecules The value of b is generally much lower than a The value of b generally increases with the size of the molecule 77

Example: 1.00 mol of carbon dioxide gas at 373 K occupies 536 ml. What is the calculated value of the pressure in atm using: (a) Ideal gas equation (b) Van der Waals equation? Van der Waals constants for carbon dioxide: a = 3.61 L 2 atm mol -2 ; b = 0.0428 L mol -1. (a) Use the ideal gas law equation! PV=nRT (P) (.536L) = (1.00mol)(0.08206)(373K) P = 57.1atm

Example: 1.00 mol of carbon dioxide gas at 373 K occupies 536 ml. What is the calculated value of the pressure in atm using: (a) Ideal gas equation (b) Van der Waals equation? Van der Waals constants for carbon dioxide: a = 3.61 L 2 atm mol -2 ; b = 0.0428 L mol -1. n (a) Use Van der Waals equation! 2 a (P + )(V-nb) = nrt V 2 (1.00) 2 (3.61) [P + -----------------][0.536 (1.00)(0.0428)]=(1.00)(0.08206)(373) (.536) 2 (P + 12.6)(0.493) = 30.6 P + 12.6 = 62.1 P = 49.5 atm

Predict which substance has the largest "b" constant: NH 3, N 2, CH 2 Cl 2, Cl 2, CCl 4 The value of b constant is merely the actual volume of the gas. A larger molecular volume results in a larger b value. From the compound in the list, CCl 4 is the largest so it will have the greatest b constant.

1.Using Van der Waals equation, calculate the temperature of 20.0 mole of helium in a 10.0 liter cylinder at 120 atm. Van der Waals constants for helium: a = 0.0341 L 2 at mol -2 ; b = 0.0237 L mol -1 2. Compare this value with the temperature calculated from the ideal gas equation.

(i) Using the Van der Waals equation: Substituting in (P + an 2 /V 2 )(V - nb) = nrt, P = 120 atm n = 20.0 mol V = 10.0 L (120 + 0.0341 x (20.0/10.0)2)(10.0-20.0 x 0.0237) = 20.0 x 0.0821 x T [Note the value of R = 0.0821 because P is in atm and V in L] (120 + 0.1364)(10.0-0.5) = 1.64 x T T = 696 K [Note that the correction to P is not significant as T is well above the temperature at which helium gas will liquefy while the volume correction is significant due to the high pressure.] (ii) Using the Ideal Gas Equation: Substituting in PV = nrt, 120 x 10.0 = 20.0 x 0.0821 x T T = 1200/1.642 = 731 K

Section 5.10 Chemistry in the Atmosphere

Section 5.10 Chemistry in the Atmosphere EXERCISE! Consider the following apparatus containing helium in both sides at 45 C. Initially the valve is closed. After the valve is opened, what is the pressure of the helium gas? 2.00 atm 9.00 L 3.00 atm 3.00 L Copyright Cengage Learning. All rights reserved 85