CH. 5 - GASES
I. Pressure
A. force per unit area
B. the gas molecules fill the container
C. the molecules move around and hit the sides of the container
D. collisions are the force
E. container is the area
F. measured with a barometer
1. Pressure of atmosphere at sea level holds column of mercury 760 mmHg
2. 1 atm = 760 mmHg
G. monometer uses column of mercury to measure pressure
H. units of measurement and their conversion factors
1 atm = 760 mmHg
1 mm Hg = 1 Torr
1 atm = 101,325 Pa = 101.325 kPa
What is 724 mmHg in kPa?
724 mmHg x atm x 101.325 kPa = 96.5 kPa
760 mmHg atm
II. Boyle’s Law
A. pressure and volume are inversely related at constant temperature
B. PV = k
C. as one value goes up, the other value goes down
D. P1V1 = P2V2 or P1 = P2
V2 V1
20.5 L of nitrogen at 25°C and 742 Torr are compressed to 9.8 atm at a constant temperature. What is the new volume?
P1V1 = P2V2 V2 = P1V1
P2
V2 = (242 Torr)(20.5 L) x atm x mmHg = 2.0 L nitrogen
9.8 atm 760 mmHg Torr
30.6 mL carbon dioxide at 740 Torr is expanded at constant temperature to 750 mL. What is the final pressure in kPa?
P1V1 = P2V2 P2 = P1V1
V2
P2 = (740 Torr)(30.6 mL) x mmHg x atm x 101.325 kPa = 4.0 kPa
750 mL Torr 760 mmHg atm
III. Charles’s Law
A. volume of a gas varies directly with absolute temperature at constant pressure
B. V = kT (where T is in Kelvins)
C. V1 = V2
T1 T2
What would the final volume be if 247 mL of gas at 22°C is heated to 98°C, if the pressure is held constant?
V1 = V2 V2 = V1T2
T1 T2 T1
(247 mL)(371 K) = 311 mL
295 K
At what temperature would 40.5 L of gas at 23.4°C have a volume of 81.0 L at constant pressure?
V1 = V2 T2 = V2T1
T1 T2 V1
(81.0 L)(296.4 K) = 593 K
40.5 L
IV. Avogadro’s Law
A. at constant temperature and pressure, the volume of a gas is directly related to the number of moles
B. V = kn (where n is the number of moles)
C. V1 = V2
n1 n2
V. Gay-Lussac’s Law
A. at constant volume, pressure and absolute temperature are directly related
B. P = kT
C. P1 = P2
T1 T2
VI. Combined Gas Law
A. if the moles of gas remain constant, use the formula and cancel out the things that don’t change P1V1 = P2V2
T1 T2
A deodorant can has a volume of 175 mL and a pressure of 3.8 atm at 22°C. what would the pressure be if the can is heated to 100°C?
P1V1 = P2V2 P2 = P1V1T2
T1 T2 T1V2
P2 = (3.8 atm)(175 mL)(373 K) = 4.8 atm
(295 K)(175 mL)
What volume of gas could the can release at 22°C and 743 Torr?
P1V1 = P2V2 V2 = P1V1T2
T1 T2 T1P2
V2 = (3.8 atm)(175 mL)(295 K) x Torr x 760 mmHg = 680 mL
(295 K)(743 Torr) mmHg atm
VII. Ideal Gas Law
STP = 22.41 L at 1 atm, 0°C, n = 1 mole
A. PV = nRT (where R is ideal gas constant)
B. R = 0.0821 L atm/mol K
62.364 L Torr/mol K
8.3145 L Pa/mol K
C. tells about a gas now, other laws tell about a gas changes when it changes
D. hypothetical substance - ideal gas
E. only approach ideal behavior at low pressure (<1 atm) and high temperature
A 47.3 L container containing 1.62 mol of He is heated until the pressure reaches 1.85 atm. What is the temperature?
PV = nRT T = PV
nR
T = (1.85 atm)(47.3 L) = 658 K
(1.62 mol)(0.0821)
Kr gas in an 18.5 L cylinder exerts a pressure of 8.61 atm at 24.8°C. What is the mass of Kr?
PV = nRT n = PV
RT
n = (8.61 atm)(18.5 L) = 6.51 mol Kr
(0.0821)(297.8 K)
6.51 mol Kr x 83.8 g Kr = 546 g Kr
mol Kr
A sample of gas has a volume of 4.18 L at 29°C and 732 Torr. What would the volume be at 24.8°C and 756 Torr?
PV = nRT n = PV V = nRT
RT P
n = (732 Torr)(4.18 L) = 123 mol
(0.0821)(302 K)
V = (123 mol)(0.0821)(297.8 K) = 3.98 Torr
(756 Torr)
VIII. Gas Density and Molar Mass
A. D = m/V
B. let M stand for molar mass
M = m/n
n = PV/RT
M = m__
PV/RT
M = mRT = m RT = DRT
PV V P P
What is the density of ammonia at 23°C and 735 Torr?
M = DRT
P
17.01 g NH3 = D(62.364)(296 K)
735 Torr
D = 0.68 g/L
A compound has the empirical formula CHCl. A 256-mL flask at 100.°C and 750 Torr contains .80 g of the gaseous compound. What is the empirical formula?
PV = nRT
(750 Torr)(.256 L) = n(62.364)(373 K)
n = 0.0083 mol CHCl
0.80 g CHCl = 96 g/mol CHCl
0.0083 mol CHCl
96 g/mol CHCl = 1.98 mol CHCl » 2 mol CHCl
48.5 g CHCl
2(CHCl) = C2H2Cl2
IX. Gases and Stoichiometry
A. reactions happen in moles
B. at STP, I mole of gas occupies 22.41 L
C. if not at STP, use the ideal gas law to calculate the moles of reactant or volume of product
Mercury can be achieved by the following reaction: HgO à Hg + O2. What volume of oxygen gas can be produced from 4.10 g of mercury (II) oxide at STP?
2HgO à 2Hg + O2
4.10 g HgO x mol HgO x 1 mol O2 x 32.0 g O2 = 0.303 g O2
216.6 g HgO 2 mol HgO mol O2
0.303 g O2 x mol O2 = 0.00947 mol O2
32.0 g O2
0.00947 mol O2 x 22.42 L O2 = 0.212 L O2
mol O2
Using the following reaction: NaHCO3 + HCl à NaCl + CO2 + H2O calculate the mass of sodium hydrogen carbonate necessary to produce 2.87 L of carbon dioxide at 25°C and 2.00 atm.
NaHCO3 + HCl à NaCl + CO2 + H2O
PV = nRT PV = m RT PVM = m
M RT
(2.00 atm)(2.87 L CO2)(84.0 g/mol NaHCO3)(1 mol CO2) = 19.7 g NaHCO3
(0.0821)(298 K)(1 mol NaHCO3)
X. Dalton’s Law
A. the total pressure in a container is the sum of the pressure each gas would exert if it were alone in the container
B. in the same container, R, T, and V are the same
Ptotal = (n1 + n2 + n3 + …)RT à Ptotal = (ntotal)RT
V V
XI. Mole Fraction
A. the ratio of moles of substance to total moles
c1 = n1 = P1_
ntotal Ptotal
The partial pressure of nitrogen in the air is 592 Torr. If the air pressure is 752 Torr, what is the mole fraction of nitrogen?
c1 = n1_
ntotal
c1 = 592 Torr = 0.787
752 Torr
What is the partial pressure of nitrogen if the container holding the air is compressed to 5.25 atm?
5.25 atm x 760 Torr = 3990 Torr
atm
592 Torr = 0.148 Torr
3990 Torr
XII. Vapor Pressure
A. when water evaporates, vapor has a pressure
B. gases are often collected over water, so the vapor pressure of water must be subtracted from the total pressure