Gases
5-1
Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
An Overview of the Physical States of Matter
The Distinction of Gases from Liquids and Solids
1. Gas volume changes greatly with pressure.
2. Gas volume changes greatly with temperature.
3. Gases have relatively low viscosity.
4. Most gases have relatively low densities under normal conditions.
5. Gases are miscible.
5-2
Figure 5.1
5-3
The three states of matter.
Effect of atmospheric pressure on objects
at Earth’s surface.
5-4
A mercury barometer.
5-5
5-6
Converting Units of Pressure
PROBLEM:
A geochemist heats a limestone (CaCO3) sample and collects
the CO2 released in an evacuated flask. The CO2 pressure is
291.4 mmHg. Calculate the CO2 pressure in torrs,
atmospheres, and kilopascals.
PLAN: Construct conversion factors to find the other units of pressure.
SOLUTION:
291.4 mmHg x
1 torr
= 291.4 torr
1 mmHg
291.4 torr x
1 atm
= 0.3834 atm
760 torr
0.3834 atm x 101.325 kPa
1 atm
5-7
= 38.85 kPa
Figure 5.4
The relationship between the volume
and pressure of a gas.
Boyle’s law
5-8
Figure 5.13
5-9
A molecular description of Boyle’s law.
Figure 5.5
The relationship between the volume and temperature of a gas.
Charles’s law
5-10
Figure 5.15
5-11
A molecular description of Charles’s law.
Boyle’s law At constant temperature, the volume occupied by a
fixed amount of gas is inversely proportional to the applied
(external) pressure.
n and T are fixed
V a
PV = constant
1
V = constant
P
P
P1V1 = P2V2
Charles’s law At constant pressure, the volume of a fixed amount
of gas is directly proportional to its absolute (Kelvin) temperature.
P and n are fixed
V
T
5-12
= constant
V a T
V = constant x T
V1
T1
=
V2
T2
Amontons’s law At constant volume, the pressure exerted by a fixed
amount of gas is directly proportional to its absolute temperature.
P
T
P1
= constant
Combined gas law
T1
V a
=
P2
T2
T
V = constant x
P
P1V1
T1
5-13
P = constant x T
P a T
V and n are fixed
=
P2V2
T2
T
PV
P
T
= constant
An experiment to study the relationship between the
volume and amount of a gas.
5-14
The volume of 1 mol of an ideal gas compared with some
familiar objects.
5-15
THE IDEAL GAS LAW
PV = nRT
3 significant figures
PV
R=
nT
=
R is the universal gas constant
Figure 5.9
5-16
1 atm x 22.414 L
1 mol x 273.15 K
=
0.0821 atm•L
mol•K
Applying the Volume-Pressure Relationship
PROBLEM:
PLAN:
5-17
Boyle’s apprentice finds that the air trapped in a J tube occupies
24.8 cm3 at 1.12 atm. By adding mercury to the tube, he increases
the pressure on the trapped air to 2.64 atm. Assuming constant
temperature, what is the new volume of air (in L)?
Applying the Volume-Pressure Relationship
n and T are constant and P1V1 = P2V2
SOLUTION:
P1 = 1.12 atm
P2 = 2.64 atm
V1 = 24.8 cm3 (convert to L)
V2 = unknown
1 mL
x
x
24.8 cm3
1 cm3
P1V1
=
n1T1
V2 = V1 x
= 0.0248 L
103 mL
P2V2
n2T2
P1
P2
5-18
L
=
1.12 atm x 0.0248 L
2.46 atm
= 0.0105 L
Applying the Pressure-Temperature Relationship
PROBLEM:
PLAN:
5-19
A steel tank used for fuel delivery is fitted with a safety valve that
opens if the internal pressure exceeds 1.00 x 103 torr. It is filled
with methane at 23oC and 0.991 atm and placed in boiling water
at exactly 100oC. Will the safety valve open?
Applying the Pressure-Temperature Relationship
SOLUTION:
V and n remain constant and
P1
T1
=
P1 = 0.991 atm (convert to torr) P2 = unknown
T1 = 23oC (convert to K)
P1V1
=
n1T1
P2V2
n2T2
0.991 atm x 760 torr
1 atm
P2 = P1 x
5-20
T2 = 100oC (convert to K)
T2
T1
= 753 torr
= 753 torr x
373 K
296 K
= 949 torr
P2
T2
Solving for an Unknown Gas Variable at Fixed
PROBLEM:
PLAN:
Conditions
A steel tank has a volume of 438 L and is filled with 0.885 kg of
O2. Calculate the pressure of O2 at 21oC.
V, T and mass, which can be converted to moles (n), are given. We
use the ideal gas law to find P.
SOLUTION:
0.885 kg x
V = 438 L
T = 21oC
n = 0.885 kg O2 (convert to mol)
P = unknown
103 g
kg
x
mol O2
32.00 g O2
= 27.7 mol O2
27.7 mol x 0.0821
P=
5-21
nRT
V
=
438 L
atm•L
mol•K
(convert to K)
21oC + 273.15 = 294 K
x 294 K
= 1.53 atm
Figure 5.11
Summary of the stoichiometric relationships among the
amount (mol, n) of gaseous reactant or product and the gas
variables pressure (P), volume (V), and temperature (T).
5-22
Using Gas Variables to Find Amounts of
PROBLEM:
Reactants or Products
Copper reacts with oxygen impurities in the ethylene used to
produce polyethylene. The copper is regenerated when hot H2
reduces the copper(II) oxide, forming the pure metal and H2O.
What volume of H2 at 765 torr and 225oC is needed to reduce
35.5 g of copper(II) oxide?
PLAN: Since this problem requires stoichiometry and the gas laws, we
have to write a balanced equation, use the moles of Cu to
calculate moles of H2 and then volume of H2 gas.
5-23
Using Gas Variables to Find Amounts of
Reactants or Products
SOLUTION:
35.5 g CuO x
CuO(s) + H2(g)
mol CuO
79.55 g CuO
0.0821
0.446 mol H2 x
x
Cu(s) + H2O(g)
1 mol H2
1 mol Cu
atm•L
mol•K
x
1.01 atm
5-24
= 0.446 mol H2
498 K
= 18.1 L
Using the Ideal Gas Law in a Limiting-Reactant Problem
PROBLEM: The alkali metals [Group 1A(1)] react with the halogens [Group 7A(17)]
to form ionic metal halides. What mass of potassium chloride forms
when 5.25 L of chlorine gas at 0.950 atm and 293 K reacts with 17.0 g
of potassium?
PLAN: After writing the balanced equation, we use the ideal gas law to find the
number of moles of reactants, the limiting reactant and moles of product.
SOLUTION:
2K(s) + Cl2(g)
P = 0.950 atm
T = 293K
2KCl(s)
V = 5.25 L
nCl = unknown
2
n Cl
2
PV = 0.950 atm x 5.25 L
=
0.0821 atm•L x 293 K
RT
mol•K
17.0 g K x
mol K
39.10 g K
5-25
= 0.207 mol Cl2
= 0.435 mol K
Using the Ideal Gas Law in a Limiting-Reactant Problem
SOLUTION:
0.207 mol Cl2 x
0.435 mol K x
2 mol KCl
1 mol Cl2
2 mol KCl
2 mol K
= 0.414 mol
KCl formed
= 0.435 mol
KCl formed
Cl2 is the limiting reactant.
0.414 mol KCl x
74.55 g KCl
mol KCl
5-26
= 30.9 g KCl
Sample Problems
5-27
A sample of chlorine gas is confined in a 5.0-L
container at 328 torr and 37°C. How many moles
of gas are in the sample?
5-28
Plan: Given the volume, pressure, and temperature of a gas, the number of moles of the
gas can be calculated using the ideal gas equation, PV = nRT.
The gas constant, R = 0.0821 L•atm/mol•K, gives pressure in atmospheres and
temperature in Kelvin. The given pressure in torr must be converted to atmospheres and
the temperature converted to Kelvin.
Solution: PV = nRT or n = PV/RT
P =  328 torr   1 atm  = 0.43158 atm;
V = 5.0 L;
T = 37°C + 273 = 310 K
 760 torr 
n=
5-29
PV
(0.43158 atm)(5.0 L)

= 0.08479 = 0.085 mol chlorine
L • atm 
RT 
 0.0821 mol • K  (310 K)


What is the effect of the following on the volume
of 1 mol of an ideal gas?
(a) The pressure is tripled (at constant T).
(b) The absolute temperature is increased by a
factor of 3.0 (at constant P).
5-30
Plan: Use the relationship PV
1 1
T1
=
P2V2
T2
Solution:
a) As the pressure on a gas increases, the molecules move closer together, decreasing
the volume. When the pressure is tripled, the volume decreases to one third of the
original volume at constant temperature (Boyle’s Law).
V2 =
V1 PT
(V )(P )(1)
1 2
= 1 1
P2T1
(3P1 )(1)
V2 = ⅓V1
b) As the temperature of a gas increases, the gas molecules gain kinetic energy. With
higher energy, the gas molecules collide with the walls of the container with greater
force, which increases the size (volume) of the container. If the temperature is
increased by a factor of 3.0 (at constant pressure) then the volume will increase by a
factor of 3.0 (Charles’s Law).
V1 PT
(V )(1)(3T1 )
1 2
= 1
P2T1
(1)(T1 )
V2 = 3V1
V2 =
5-31
How many grams of phosphorus react with 35.5
L of O2 at STP to form tetraphosphorus
decaoxide?
P4(s)  5O2(g) + P4O10(s)
5-32
Plan: We can find the moles of oxygen from the standard molar volume of gases (1 L
of gas occupies 22.4 L at STP) and use the stoichiometric ratio from the balanced
equation to determine the moles of phosphorus that will react with the oxygen.
Solution:
P4(s) + 5 O2(g)  P4O10(s)
 1 mol O2  1 mol P4  123.88 g P4 
Mass P4 =  35.5 L O2  



22.4
L
O
5
mol
O
1
mol
P
2 
2 
4 

= 39.2655 = 39.3 g P4
5-33
How many grams of phosphine (PH3) can form
when 37.5 g of phosphorus and 83.0 L of
hydrogen gas react at STP?
P4(s) + H2(g)  PH3(g) [unbalanced]
5-34
Plan: To find the mass of PH3, write the balanced equation and find the number of moles
of PH3 produced by each reactant. The smaller number of moles of product indicates the
limiting reagent. Solve for moles of H2 using the ideal gas equation.
Solution:
P4(s) + 6 H2(g)  4 PH3(g)
From the ideal gas equation:
Moles hydrogen = 3.705357 mol H2 (unrounded)



PH3 from P4 =  37.5 g P4   1 mol P4  4 mol PH3 
 123.88 g P4  1 mol P4 
= 1.21085 mol PH3
 4 mol PH3 

 6 mol H 2 
PH3 from H2 =  3.705357 mol H 2  
= 2.470238 mol PH3
P4 is the limiting reactant.
 1 mol P4  4 mol PH3   33.99 g PH3 



 123.88 g P4  1 mol P4  1 mol PH3 
Mass PH3 =  37.5 g P4  
= 41.15676 = 41.2 g PH3
5-35