Chapter 4 The Properties of Gases
Significant Figure Convention
At least one extra significant figure is displayed in all intermediate calculations.
The final answer is expressed with the correct number of significant figures.
The Nature of Gases (Sections 4.1–4.3)
Key Concepts
physical states (solid, liquid, gas), bulk matter, compressibility, pressure, gauge pressure,
barometer, pascal (Pa), manometer, mmHg (Torr), atmosphere (atm), bar
4.1 Observing Gases
Example 4.1
Relate the general properties of gas behavior to the molecular level.
Answer
The fact that gases are compressed with ease and expand rapidly to fill the available
space suggests that molecules of gases are widely separated and in ceaseless chaotic
motion.
4.2 Pressure
Example 4.2a
Suppose the height of the column of mercury in a barometer is exactly 760 mm.
Given that the density of mercury at 0°C is 13 595.1 kg⋅m–3, calculate the
atmospheric pressure.
Solution
P = dhg = (13 595.1 kg⋅m–3)(0.760 m)(9.806 65 m⋅s−2) = 101 325 kg⋅m−1⋅s−2 = 101 325
Pa = 101.325 kPa
Note: 1 Pa (pascal) = 1 kg⋅m−1⋅s−2 SI unit of pressure
Example 4.2b
At the same atmospheric pressure obtained in Example 4.2a, calculate the height of liquid
in a barometer that employs water at 20°C. [d (H2O) = 0.998 g⋅cm–3]
Answer
h = 10.4 m
Example 4.2c
The height of the system-side column in an open-tube mercury manometer was 15 mm
above that of the open side when the atmospheric pressure was 762 mmHg at 20°C.
Calculate the pressure in the system in units of mmHg and kPa.
Solution
In this case, the pressure in the system is equal to atmospheric pressure minus the
difference in height of the two columns. The system is at lower pressure than the
atmosphere (open end). The conversion to kPa requires the density of Hg at 20°C,
which is 13 546 kg⋅m−1⋅s−2.
P = 762 – 15 = 747 mmHg = 0.747 mHg
P = dhg = (13 546 kg⋅m−1⋅s−2) × (0.747 m) × (9.806 65 m⋅s−1) = 9.92 × 104 Pa = 99.2 kPa
Example 4.2d
Calculate the pressure inside a system when a closed-tube mercury
manometer shows a height difference of 12 cm (higher on the closed side) at 20°C.
Answer
P = 120 mmHg and 15.9 kPa
1
Chapter 4
4.3 Alternative Units of Pressure
Example 4.3
Convert an atmospheric pressure of 740 mmHg into each of the following pressure units: Torr,
atm, Pa, kPa, and bar.
Answer
P = 740 mmHg = 740 Torr = 0.974 atm = 9.87 × 104 Pa = 98.7 kPa = 0.987 bar
The Gas Laws (Sections 4.4–4.11)
Key Concepts
Boyle’s law, isotherm, hyberbola, Charles’s law, extrapolation, Kelvin scale,
molar volume, Avogadro’s principle, ideal gas law, gas constant, equation of state,
ideal gas, limiting law, combined gas law, standard temperature and pressure (STP),
standard ambient temperature and pressure (SATP), molar concentration, density,
amount-to-volume conversions in reacting gases, gas mixtures, partial pressure,
Dalton’s law of partial pressures, vapor pressure, mole fraction
4.7 The Ideal Gas Law
Example 4.7a
A 2.6-µL ampule of xenon has a pressure of 2.00 Torr at 15°C. How many
xenon atoms are present?
Solution
Convert the volume units into liters, the pressure units into atmospheres, and the
temperature units into kelvins. Solve for the number of moles of xenon, using the
ideal gas law, and then use Avogadro’s number to determine the number of atoms.
V = 2.6 × 10−6 L
P = (2.00/760) atm = 2.632 × 10−3 atm
T = 15 + 273.15 = 288.2 K
n=
PV
(2.632 ×10−3 atm)(2.6 × 10−6 L)
=
= 2.9 × 10−10 mol
RT
( 8.205 74 × 10−2 L ⋅ atm⋅K −1 ⋅mol−1 )(288.2 K)
n × N A = 1.7 × 1014 atoms
Example 4.7b
A 20-L flask at 200 K and 20 Torr contains nitrogen. What mass of nitrogen
is present (in grams)?
Answer
0.89 g
4.8 Applications of the Ideal Gas Law
2
Example 4.8a
A gas sample is compressed to half of its original volume and the absolute
temperature is increased by 15%. What is the pressure change?
Solution
Let indices 1 and 2 represent the initial and final states, respectively. Note the changes, if any,
in amount, volume, and temperature of the gas. Use the combined form of the ideal gas law
given above to solve for the pressure change.
The Properties of Gases
Change in the amount of gas:
50% decrease in the volume:
15% increase in the temperature:
n2 = n1
V2 = V1 – 0.50V1 = 0.50V1
T2 = T1 + 0.15T1 = 1.15T1
 V  n  T 
 V1   n1   1.15T1 
P2 = P1  1   2   2  = P1 
 
 = P1 ( 2 )(1)(1.15) = 2.30 P1
 V2   n1   T1 
 0.50V1   n1   T1 
Because P2 = P1 + 1.30P1, the pressure has increased by 130%.
Example 4.8b
A gas sample has its pressure quadrupled, absolute temperature doubled, and amount doubled.
Calculate the volume change.
Answer
V2 = V1 (no change in volume)
4.9 Gas Density
Example 4.9a
The density of an unknown hydrocarbon gas (molecules containing only carbon and hydrogen
atoms) is found to be 0.875 g⋅L−1 at 570 Torr and 20°C. Calculate the molar mass of the gas
and identify it.
Solution
Convert the pressure units into atmospheres and the temperature units into kelvin. Solve for
the unknown molar mass. Compare with molar masses of simple
hydrocarbons to identify the gas.
P = (570/760) atm = 0.7500 atm and T = 20 + 273.15 = 293.2 K
M =
dRT ( 0.875 g⋅L−1 )( 8.20574 ×10−2 L ⋅ atm⋅K −1 ⋅mol −1 )(293.2 K)
=
= 28.1 g⋅mol−1
(0.7500 atm)
P
If the gas contains 2 moles of carbon atoms (24.02 g⋅mol−1) and 4 moles of hydrogen atoms
(4.03 g⋅mol−1), then M = 28.1 g⋅mol−1. Gas is ethene C2H4.
Example 4.9b
Calculate the density of methane gas at STP.
Answer
d = 0.715 g⋅L−1
4.10 The Stoichiometry of Reacting Gases
Example 4.10a Calculate the volume of each gas produced when 225 g of NH4NO3 (ammonium nitrate)
decomposes completely to N2O and H2O gases at atmospheric pressure (1.00 atm) and 300°C
(573 K). [Note the convention for significant figures described at the beginning of this
chapter.]
Solution
Balance the chemical equation for the decomposition of ammonium nitrate solid. Calculate
the number of moles of solid and use mole-to-mole calculations to determine the number of
moles of each gas formed. Use the ideal gas law to calculate the volume of each gas formed
at 1.00 atm and 573 K.
NH4NO3 (s) → N2O (g) + 2 H2O (g)
(225 g NH4NO3)/(80.05 g⋅mol–1) = 2.811 mol
1 mol NH4NO3 → 1 mol N2O and 2.811 mol NH4NO3 → 2.811 mol N2O
1 mol NH4NO3 → 2 mol H2O and 2.811 mol NH4NO3 → 5.622 mol H2O
V N 2O = nRT/P = (2.811 mol)(0.082 057 4 L⋅atm⋅K–1⋅mol–1)(573 K)/(1.00 atm) = 132 L
V H2O = nRT/P = (5.622 mol)(0.082 057 4 L⋅atm⋅K–1⋅mol–1)(573 K)/(1.00 atm) = 264 L
3
Chapter 4
Example 4.10b An initial volume of 352 L of butane at 250°C and 3.42 atm is burned with excess
oxygen: 2 C4H10 (g) + 13 O2 (g) → 8 CO2 (g) + 10 H2O (g). Calculate the mass of
carbon dioxide released to the atmosphere.
Answer
m = 4.94 × 103 g
4.11 Mixtures of Gases
Example 4.11a A 1.00-L sample of 100% humid air from the atmosphere is taken at 15°C. An analysis
reveals the following composition by mass: 934 mg of nitrogen (N2), 285 mg of oxygen (O2),
12.8 mg of water vapor (H2O), 15.9 mg of argon (Ar), and 0.658 mg of carbon dioxide (CO2).
Calculate the mole fraction and partial pressure of each component of air in the sample.
Solution
Convert the mass of each component into amount (nA, nB, …) by using the appropriate molar
mass. Sum the amounts of each component to obtain the total amount n. Take the ratio of the
amount of each component to the total amount to obtain the mole fraction. The total pressure
P (atm) is obtained from the total number of moles, the volume of the sample, and the absolute
temperature by the ideal gas law. The partial pressures are then obtained from Dalton’s law
(PA = xAP, …).
 1 g   1 mol 
−2
nN 2 = 934 mg 

 = 3.333 × 10 mol
1000
mg
28.02
g



 1 g   1 mol 
−3
nO2 = 285 mg 

 = 8.906 × 10 mol
1000
mg
32.00
g



 1 g   1 mol 
−4
nH 2O = 12.8 mg 

 = 7.105 × 10 mol
 1000 mg   18.016 g 
 1 g   1 mol 
−4
nAr = 15.9 mg 

 = 3.980 × 10 mol
1000
mg
39.95
g



 1 g   1 mol 
−5
nCO2 = 0.658 mg 

 = 1.495 × 10 mol
1000
mg
44.01
g



n = nN 2 + nO2 + nH 2O + nAr + nCO2 = 4.336 × 10−2 mol
xN 2 = nN 2 / n = 0.7687 ;
xAr = nAr / n = 0.009 179 ;
xO2 = nO2 / n = 0.2054 ;
xH 2O = nH 2O/ n = 0.016 39;
xCO2 = nCO2 / n = 0.000 344 8
P = nRT/V = (0.043 33 mol)(0.082 057 4 L⋅atm⋅K–1⋅mol–1)(288.2 K)/(1.00 L) = 1.025 atm
PN 2 = x N 2 P = 0.788 atm; PO2 = xO2 P = 0.211 atm; PH 2O = xH 2O P = 0.0168 atm;
PAr = xAr P = 0.009 41 atm; PCO2 = xCO2 P = 0.000 353 atm
Example 4.11b A sample of damp air in a 1.00-L container exerts a pressure of 762.0 Torr at 20°C; but when it
is cooled to −10.0°C, the pressure falls to 607.1 Torr as the water condenses. What mass of
water was present?
Assume the vapor pressure of ice at −10°C is negligible.
Answer
4
m = 8.44 × 10−2 g
The Properties of Gases
Molecular Motion (Section 4.12–4.14)
Key Concepts
diffusion, effusion, Graham’s law of effusion, kinetic model of gases, pressure-volume
product proportional to square of speed, momentum, root mean square speed, Maxwell
distribution of speeds
4.12 Diffusion and Effusion
Example 4.12a It takes a certain amount of hydrogen gas 45 s to effuse through a porous barrier. How long, in
minutes, does it take the same amount of uranium hexafluoride (UF6) to effuse under the same
conditions?
Solution
First determine the molar mass of H2 and UF6. The time for effusion is proportional to the
square root of the molar mass. Because UF6 has a higher molar mass, it takes longer to effuse.
M H 2 = 2(1.007 9) = 2.0158 g⋅mol −1; M UF6 = 238.03 + 6(19.00) = 352.03 g⋅mol −1
(time for UF6 to effuse) = (time for H 2 to effuse)
M UF6
M H2
= (45 s)
352.03
2.0158
= (45 s) 174.635 = (45 s) (13.215) = 5.9 × 102 s = 9.9 min
Example 4.12b It takes 50.0 mL of nitrogen 55.0 s to effuse through a porous barrier. An unknown compound
takes 164 s to effuse through the same barrier under the same conditions. What is the molar
mass of this compound?
Answer
M of unknown compound = 248 g⋅mol–1
Example 4.12c How much does the rate of effusion (and average speed) in H2 change when the temperature is
decreased from 25°C to −120°C?
Solution
Convert temperature to the absolute scale before using the above relationship.
T1 = 25 + 273.15 = 298.2 K and T2 = −120 + 273.15 = 153.2 K
Rate of effusion (average speed) at 153 K
153.2 K
=
= 0.5137 = 0.717
Rate of effusion (average speed) at 298 K
298.2 K
Example 4.12d The average speed of molecules in a gas increases by 50% when the temperature is increased
from an initial value of 320 K. What is the final temperature of the gas?
Answer
T2 = 720 K
Example 4.12e Which has the higher average speed, N2O molecules at 50°C or HCN molecules at 25°C?
Solution
Calculate the molar mass of N2O and HCN. Convert Celsius temperature to the absolute scale.
Determine which gas has the larger value of the square root of the temperature over the molar
mass.
M N 2O = 2(14.01) + 16.00 = 44.02 g⋅mol −1
M HCN = 1.007 9 + 12.01 + 14.01 = 27.027 9 g⋅mol −1
TN 2O = 50 + 273.2 = 323.2 K and THCN = 25 + 273.2 = 298.2 K
Average speed of N2O ∝ (323.2/44.02)1/2 = 2.71
Average speed of HCN ∝ (298.2/27.027 9)1/2 = 3.32
Therefore, HCN has the greater average speed under these conditions.
5
Chapter 4
Example 4.12f
Answer
What ratio of temperatures is required for methane (CH4 ) and helium (He) to have the same
average speed?
TCH 4
THe
=
M CH 4
M He
=
16.0416
= 4.01
4.00
4.13 The Kinetic Model of Gases
Example 4.13a Calculate the root mean square speed of helium and radon gases at 300 K and 1200 K.
Solution
Look up the molar masses of helium and radon and convert them to kg⋅mol−1. Use the above
expression for root mean square speed to calculate c at the two temperatures. Be careful to use
the gas constant in the proper SI units.
MHe = (4.00 g⋅mol−1) × (10−3 kg⋅g−1) = 4.00 × 10−3 kg⋅mol−1
MRn = (222 g⋅mol−1) × (10−3 kg⋅g−1) = 2.22 × 10−1 kg⋅mol−1
vr m s =
3RT
M
=
3 (8.314 47 J⋅K −1 ⋅mol−1 )(300 K)
4.00 × 10
−3
kg⋅mol
−1
=
1.871× 106 m 2⋅s −2 = 1.37 × 103 m⋅s −1
The ratio of temperatures is (1200/300) = 4. Because vr ms ∝ T , the speed at 1200 K will be
4 = 2 times that at 300 K The final speed for helium is therefore vr m s = 2.74 × 103 m⋅s−1 at
1200 K.
For radon, the same procedure yields
vr m s = 184 m⋅s−1 at T = 300 K and vr m s = 368 m⋅s−1 at T = 1200 K
Example 4.13b Calculate the root mean square speed for ethene (ethylene, C2H4) gas at 298 K. At what
temperature will ethyne (acetylene, C2H2) gas have the same root mean square speed?
Answer
vr m s = 515 m⋅s–1 and T = 277 K for ethyne gas
The Impact on Materials: Real Gases
(Sections 4.15–4.17)
Key Concepts
real gases, compression factor ( Z ), intermolecular forces, gas liquefaction,
Joule-Thomson effect, equation of state, virial equation, virial coefficients,
van der Waals equation, van der Waals parameters
6
The Properties of Gases
4.17 Equations of State of Real Gases
Example 4.17a Use both the ideal gas law and van der Waals equation to calculate the pressure of CO2 gas at
298 K and 200 K (5 K above the sublimation temperature). Assume the molar volume is
exactly 1 L⋅mol−1.
Note: The van der Waals parameters are a = 3.640 L2⋅atm⋅mol−1 and b = 4.267 × 10−2 L⋅mol−1.
Solution
Use the ideal gas law and the rearranged form of the van der Waals equation.
T = 298 K ⇒ P =
P=
nRT
( 0.082 057 L⋅atm⋅K −1⋅mol−1 ) (298 K)
=
= 24.45 atm
V
(1 L⋅mol−1 )
(0.082 057 L ⋅ atm ⋅ K −1 ⋅ mol−1 )(298 K) (3.640 L2 ⋅ atm ⋅ mol−1 )
nRT
n2
−a
=
−
V − nb
(1 − 4.267 × 10−2 )(L ⋅ mol−1 )
(1 L ⋅ mol−1 )2
V2
= ( 25.543 − 3.640 ) atm = 21.90 atm
Change from ideal = (24.45 – 21.90)/(24.45) = (2.55/24.45) = 0.104 or 10.4%
T = 200 K ⇒ P =
P=
nRT
(0.082 057 L ⋅ atm ⋅ K −1 ⋅ mol−1 )(200 K)
=
= 16.41 atm
V
(1 L ⋅ mol−1 )
nRT
n2
(0.082 057 L ⋅ atm ⋅ K −1 ⋅ mol−1 )(200 K) (3.640 L2 ⋅ atm ⋅ mol−1 )
−a
=
−
V − nb
V2
(1 − 4.267 × 10−2 )(L ⋅ mol−1 )
(1 L ⋅ mol−1 ) 2
= (17.143 − 3.640 ) atm = 13.50 atm
Change from ideal = (16.41 – 13.50)/(16.41) = (2.91/16.41) = 0.177 or 17.7%
Example 4.17b Use both the ideal gas law and van der Waals equation to calculate the pressure of H2O gas at
298 K and 373 K (boiling temperature of water). Use the molar volumes of 782.3 L⋅mol−1 at
298 K and 30.61 L⋅mol−1 at 373 K, which correspond to the vapor pressure of liquid water.
Note: The van der Waals parameters are a = 5.536 L2⋅atm⋅mol−1 and b = 3.049 × 10−2 L⋅mol−1.
Answer
P = 0.031 26 atm (ideal gas) and P = 0.031 25 atm (van der Waals) at 298 K
P = 0.999 9 atm (ideal gas) and P = 0.995 0 atm (van der Waals) at 373 K
Surprisingly, water vapor behaves as an ideal gas up to its boiling temperature.
7