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

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

Relating temperature, pressure, volume and amount of gases
Heat an aerosol can Blow up a balloon and place it in a freezer Adding 5 mols of nitrogen to a tank that already contains 10 mols of nitrogen Exploding tires on a hot summer day

Elements that exist as gases at 250C and 1 atmosphere

NO2 gas

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

Atmospheric Pressure Sea level 1 atm 4 miles 0.5 atm 10 miles 0.2 atm

(force = mass x acceleration)
Area Pressure = (force = mass x acceleration) Units of Pressure 1 pascal (Pa) = 1 N/m2 1 atm = 760 mmHg = 760 torr 1 atm = 101,325 Pa Barometer

1 atm = 760 mm Hg 1 atm 760 mmHg P = 749 mmHg x = 0.986 atm
The pressure outside of a jet plane flying at high altitude falls considerably below standard atmospheric pressure. Therefore the air inside of the cabin must be pressurized to protect the passengers. What is the pressure in the cabin if the barometer reading is 749 mmHg? 1 atm = 760 mm Hg 1 atm 760 mmHg P = 749 mmHg x = atm

Problem 5.13 Convert 562 mmHg to atm.

1 atm = 760 mm Hg = 101.325 kPa 101.325 kPa 744 mmHg x = 99.2 kPa P =
The atmospheric pressure in New York City on a given day is 744 mmHg. What was the pressure in kPa? 1 atm = 760 mm Hg = kPa kPa 760 mmHg 744 mmHg x P = = 99.2 kPa

Problem 5.14 The atmospheric pressure at the summit of Mt. McKinley is 606 mmHg on a certain day. What is the pressure in atm? In kPa?

Manometers Used to Measure Gas Pressures
closed-tube open-tube

Forces Affecting Gases
.

The Gas Laws Boyle’s Law Robert Boyle (1627-1691)
Relates pressure and volume at constant temperature For a given mass of a gas at constant temperature, the volume of the gas varied inversely with pressure.

Apparatus for Studying the Relationship Between
Pressure and Volume of a Gas As P (h) increases V decreases

Boyle’s Law P a 1/V Constant temperature P x V = k
Constant amount of gas P x V = k P1 x V1 = P2 x V2

P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL
A sample of chlorine gas occupies a volume of 946 mL at a pressure of 726 mmHg. What is the pressure of the gas (in mmHg) if the volume is reduced at constant temperature to 154 mL? P x V = constant P1 x V1 = P2 x V2 P1 = 726 mmHg P2 = ? V1 = 946 mL V2 = 154 mL P1 x V1 V2 726 mmHg x 946 mL 154 mL = P2 = = 4460 mmHg

The Gas Laws Charles's Law Jacques Charles (1746-1823)
Relates temperature and volume at constant pressure For a given mass of a gas at constant pressure, the volume of the gas varied directly with temperature. V = b x T V1 = V2 T1 T2

Variation in Gas Volume with Temperature at Constant Pressure
As T increases V increases

Variation of Gas Volume with Temperature
at Constant Pressure Charles’ Law V a T Temperature must be in Kelvin V = constant x T V1/T1 = V2 /T2 T (K) = t (0C)

A sample of carbon monoxide gas occupies 3. 20 L at 125 0C
A sample of carbon monoxide gas occupies 3.20 L at 125 0C. At what temperature will the gas occupy a volume of 1.54 L if the pressure remains constant? V1 /T1 = V2 /T2 V1 = 3.20 L V2 = 1.54 L T1 = K T2 = ? T1 = 125 (0C) (K) = K V2 x T1 V1 1.54 L x K 3.20 L = T2 = = 192 K

Problem 5.24 Under constant pressure conditions a sample of hydrogen gas initially at 88ºC and 9.6 L is cooled until its final volume is 3.4 L. What is the final temperature (in K)? V = b x T V1 = V2 T1 T2

The Gas Laws Gay-Lussac's Law Joseph Gay-Lussac (1778-1850)
Gay-Lussac studied the effect of temperature on the pressure of a gas at constant volume For a given mass of a gas at constant volume, the pressure of the gas varied directly with temperature. P1 = P2 T1 T2

P1 /T1 = P2 /T2 P1 = 2.50 kPa P2 = 1.54 kPa T1 = 298.15 K T2 = ?
A sample of nitrogen gas has a pressure of 2.50 kPa at 25 0C. At what temperature will the gas have a pressure of 1.54 kPa if the volume remains constant? P1 /T1 = P2 /T2 P1 = 2.50 kPa P2 = 1.54 kPa T1 = K T2 = ? T1 = 25 (0C) (K) = K P2 x T1 P1 1.54 kPa x K 2.50 kPa = T2 = = 184 K

Problem 5.20 At 46ºC a sample of ammonia gas exerts a pressure of 5.3 atm. What is the temperature when the pressure of the gas is reduced to one-tenth of the original pressure at the same volume?

Problem 5.22 A sample of air occupies 3.8 L when the pressure is 1.2 atm. (a) What volume (in L) does it occupy at 5010mmHg if the temperature remains constant? (b) At the original pressure and volume the temperature was 101ºC. What temperature is necessary to increase the pressure to 5010mmHg, but keep the volume constant?

Explain the following. State the gas law.
Helium balloon pops at a party on a summer day Predict how the size of exhaled bubbles change as a scuba diver approaches the surface of the water Aerosol cans have a warning not to put near a flame.

Describe the relationships of…
1. Boyle’s Law 2. Charles’s Law 3. Gay-Lussac’s Law

Avogadro’s Law V a number of moles (n) V = a x n V1 / n1 = V2 / n2
Constant temperature Constant pressure V a number of moles (n) V = a x n V1 / n1 = V2 / n2

4NH3 + 5O2 4NO + 6H2O 1 mole NH3 1 mole NO At constant T and P
Ammonia burns in oxygen to form nitric oxide (NO) and water vapor. How many volumes of NO are obtained from one volume of ammonia at the same temperature and pressure? 4NH3 + 5O NO + 6H2O 1 mole NH mole NO At constant T and P 1 volume NH volume NO

Summary of Gas Laws Boyle’s Law

Charles Law Gay-Lussac’s Law

Ideal Gas Equation 1 Boyle’s law: P a (at constant n and T) V
Charles’ law: V a T (at constant n and P) Avogadro’s law: V a n (at constant P and T) V a nT P V = constant x = R nT P R is the gas constant PV = nRT R = L • atm / (mol • K)

When the moles are constant,
The conditions 0 0C and 1 atm are called standard temperature and pressure (STP). Calculate the volume of 1 mole of an ideal gas at STP PV = nRT R = L • atm / (mol • K) When the moles are constant, PV = nR T P1V1 = P2V2 T T2 Therefore

PV = nRT nRT V = P 1.37 mol x 0.0821 x 273.15 K V = 1 atm V = 30.7 L
What is the volume (in liters) occupied by 49.8 g of HCl at STP? T = 0 0C = K P = 1 atm PV = nRT n = 49.8 g x 1 mol HCl 36.45 g HCl = 1.37 mol V = nRT P V = 1 atm 1.37 mol x x K L•atm mol•K V = 30.7 L

Problem 5.40 Calculate the volume (in L) of 88.4 g of CO2 at STP.

PV = nRT n, V and R are constant nR V = P T = constant P1 T1 P2 T2 =
Argon is an inert gas used in lightbulbs to retard the vaporization of the filament. A certain lightbulb containing argon at 1.20 atm and 18 0C is heated to 85 0C at constant volume. What is the final pressure of argon in the lightbulb (in atm)? PV = nRT n, V and R are constant nR V = P T = constant P1 = 1.20 atm T1 = 291 K P2 = ? T2 = 358 K P1 T1 P2 T2 = P2 = P1 x T2 T1 = 1.20 atm x 358 K 291 K = 1.48 atm

Extra Practice: Problem 5.36
The temperature of 2.5 L of a gas initially at STP is raised to 250ºC at constant volume. Calculate the final pressure of the gas in atm.

Extra Practice: Problem 5.32
Given that 6.9 moles of carbon monoxide are present in a container of volume 30.4 L, what is the pressure of the gas (in atm) if the temperature is 62ºC?

d is the density of the gas in g/L
Density (d) Calculations m is the mass of the gas in g d = m V = PM RT M is the molar mass of the gas n = m M Molar Mass (M ) of a Gaseous Substance dRT P M = d is the density of the gas in g/L

dRT P M = d = m V = = 2.21 1 atm x 0.0821 x 300.15 K M = M =
A 2.10-L vessel contains 4.65 g of a gas at 1.00 atm and C. What is the molar mass of the gas? dRT P M = d = m V 4.65 g 2.10 L = = 2.21 g L 2.21 g L 1 atm x x K L•atm mol•K M = M = 54.5 g/mol

Problem 5.44 At 741 torr and 44ºC 7.10 g of a gas occupy a volume of 5.40 L. What is the molar mass of the gas (in g/mol)?

C6H12O6 (s) + 6O2 (g) 6CO2 (g) + 6H2O (l)
Gas Stoichiometry What is the volume of CO2 produced at 37 0C and 1.00 atm when 5.60 g of glucose are used up in the reaction: C6H12O6 (s) + 6O2 (g) CO2 (g) + 6H2O (l) g C6H12O mol C6H12O mol CO V CO2 1 mol C6H12O6 g C6H12O6 x 6 mol CO2 1 mol C6H12O6 x 5.60 g C6H12O6 = mol CO2 0.187 mol x x K L•atm mol•K 1.00 atm = nRT P V = = 4.76 L

Problem 5.60 Ethanol (C2H5OH) burns in air: C2H5OH(l) + O2(g)  CO2(g) + H2O(l) Determine the volume of air in liters at 35.0ºC and 790 mmHg required to burn 227 g of ethanol. Assume that air is 21.0 percent O2 by volume.

Dalton’s Law of Partial Pressures
V and T are constant P1 P2 Ptotal = P1 + P2 Studying gaseous mixtures shows that each gas behaves independantly of the others

PA = nART V PB = nBRT V XA = nA nA + nB XB = nB nA + nB PT = PA + PB
Consider a case in which two gases, A and B, are in a container of volume V. PA = nART V nA is the number of moles of A PB = nBRT V nB is the number of moles of B XA = nA nA + nB XB = nB nA + nB PT = PA + PB PA = XA PT PB = XB PT Pi = Xi PT mole fraction (Xi ) = ni nT

Pi = Xi PT PT = 1.37 atm 0.116 8.24 + 0.421 + 0.116 Xpropane =
A sample of natural gas contains 8.24 moles of CH4, moles of C2H6, and moles of C3H8. If the total pressure of the gases is 1.37 atm, what is the partial pressure of propane (C3H8)? Pi = Xi PT PT = 1.37 atm 0.116 Xpropane = = Ppropane = x 1.37 atm = atm

Problem 5.63 A mixture of gases contains 0.31 mol CH4, 0.25 mol C2H6, and 0.29 mol C3H8. The total pressure is 1.50 atm. Calculate the partial pressure of each gas. = _____ = ______ = ______

Collecting a Gas over Water
2KClO3 (s) KCl (s) + 3O2 (g) PT = PO + PH O 2

Vapor of Water and Temperature

Problem 5.66 A mixture of helium and neon gasses is collected over water at 28.0ºC and 745 mmHg. If the partial pressure of helium is 368 mmHg, what is the partial pressure of neon (in mmHg)? (vpH20 = 28.3 mmHg at 28.0ºC)

Problem 5.68 A sample of zinc metal reacts completely with an excess of hydrochloric acid: Zn(s) + 2HCl(aq)  The hydrogen gas produced is collected over water at 25.0ºC. The volume of the as is 7.80L, and the pressure is atm. Calculate the amount of zinc metal (in grams) consumed in the reaction. (vpH20 = 28.3 mmHg at 28.0ºC)