Chapter 5
Gases -1
Gas Laws
Dr. Sapna Gupta
Properties of Gases
1.
2.
3.
4.
5.
They are compressible.
They expand to fill the container.
Pressure, volume, temperature, and amount are related.
They have low density.
Form a homogeneous mixture.
Dr. Sapna Gupta/Gases
2
Kinetic Molecular Theory
A theory, developed by physicists, that is based on the assumption that a
gas consists of molecules in constant random motion.
Postulates of the Kinetic Theory
1.
2.
3.
4.
5.
Gases are composed of molecules whose sizes are negligible.
Molecules move randomly in straight lines in all directions and at
various speeds.
The forces of attraction or repulsion between two molecules
(intermolecular forces) in a gas are very weak or negligible, except
when the molecules collide.
When molecules collide with each other, the collisions are elastic.
The average kinetic energy of a molecule is proportional to the
absolute temperature.
Dr. Sapna Gupta/Gases
3
Pressure
Pressure is force exerted per unit area. P = Force/Area
The SI unit for pressure is the pascal, Pa.
Other Units
atmosphere, atm
mmHg
torr
bar
Dr. Sapna Gupta/Gases
4
Measurement of Pressure
A barometer is a device for
measuring the pressure of the
atmosphere.
A manometer is a device for
measuring the pressure of a gas
or liquid in a vessel.
Dr. Sapna Gupta/Gases
5
Gas Laws
• Gas laws – empirical relationships among gas parameters.
•
•
•
•
Volume (V)
Pressure (P)
Temperature (T)
Amount of a gas (n)
• By holding two of these physical properties constant, it becomes
possible to show a simple relationship between the other two
properties.
Dr. Sapna Gupta/Gases
6
Boyle’s Law
• The volume of a sample of gas is inversely proportional to pressure at
constant temperature. (one increases if the other decreases and vice
versa)
1
V
P
When a 1.00-g sample of O2 gas at 0oC
is placed in a container at a pressure of
0.50 atm, it occupies a volume of 1.40 L.
When the pressure on the O2 is doubled
to 1.0 atm, the volume is reduced to
0.70 L, half the original volume.
PV = constant
P1V1=P2V2
Dr. Sapna Gupta/Gases
7
Boyle’s Law Contd….
Graphical representation
Dr. Sapna Gupta/Gases
8
Solved Problem:
A volume of oxygen gas occupies 38.7 mL at 751 mmHg and 21°C. What
is the volume if the pressure changes to 359 mmHg while the
temperature remains constant?
Vi = 38.7 mL
Pi = 751 mmHg
Ti = 21°C
Vf = ?
Pf = 359 mmHg
Tf = 21°C
PiVi
Vf 
Pf
(38.7 mL)(751 mmHg)
Vf 
(359 mmHg)
= 81.0 mL
(3 significant figures)
Dr. Sapna Gupta/Gases
9
Charles’s Law
• The volume of a sample of gas at constant pressure is directly
proportional to the absolute temperature (K).
V T
At room Temp
At – 72oC
V
 constant
T
Vi Vf

Ti T f
Dr. Sapna Gupta/Gases
10
Charles’s Law contd….
Graphical Representation
The temperature -273.15°C is called absolute zero. It is the
temperature at which the volume of a gas is hypothetically zero.
This is the basis of the absolute temperature scale, the Kelvin scale (K).
Dr. Sapna Gupta/Gases
11
Solved Problem:
You prepared carbon dioxide by adding HCl(aq) to marble chips, CaCO3.
According to your calculations, you should obtain 79.4 mL of CO2 at 0°C
and 760 mmHg. How many milliliters of gas would you obtain at 27oC?
Vi = 79.4 mL
Pi = 760 mmHg
Ti = 0°C = 273 K
Vf = ?
Pf = 760 mmHg
Tf = 27°C = 300. K
TfVi
Vf 
Ti
Vf 
(300. K)(79.4 mL)
(273 K)
= 87.3 mL
(3 significant figures)
Dr. Sapna Gupta/Gases
12
Avogadro’s Law
The volume of a gas sample is directly proportional to the number of
moles in the sample at constant pressure and temperature.
V a n
Vi Vf
=
ni nf
𝑉
= 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝑛
Dr. Sapna Gupta/Gases
13
Combined Gas Law
The volume of a sample of gas at constant pressure is inversely
proportional to the pressure and directly proportional to the absolute
temperature.
The mathematical relationship:
In equation form:
V
T
P
PV
 constant
T
PiVi PfVf

Ti
Tf
Include the Avogadro’s law and it becomes (1=initial and 2= final)
P iV i
PfVf
=
n iT i
nfTf
OR
P 1 V1
P 2 V2
=
n1T1
n2T2
Dr. Sapna Gupta/Gases
14
Solved Problem:
Divers working from a North Sea drilling platform experience pressure of
5.0 × 101 atm at a depth of 5.0 × 102 m. If a balloon is inflated to a volume of
5.0 L (the volume of the lung) at that depth at a water temperature of 4°C,
what would the volume of the balloon be on the surface (1.0 atm pressure)
at a temperature of 11°C?
Vi = 5.0 L
Pi = 5.0 × 101 atm
Ti = 4°C = 277 K
Vf = ?
Pf = 1.0 atm
Tf = 11°C = 284 K
(284 K)(5.0  101atm)(5.0 L)
Vf 
(277 K)(1.0 atm)
Dr. Sapna Gupta/Gases
Vf 
Tf PiVi
Ti Pf
= 2.6  102 L
(2 significant figures)
15
Ideal Gas Law
Ideal Gas Law
• Combining the historic gas laws yields:
• Adding the proportionality constant, R
PV
R=
nT
Dr. Sapna Gupta/Gases
16
Standard Temperature and Pressure (STP)
• The reference condition for gases, chosen by convention to be exactly
0°C and 1 atm pressure.
• The ideal gas equation is not exact, but for most gases it is quite
accurate near STP (760 torr (1 atm) and 273 K)
• An “ideal gas” is one that “obeys” the ideal gas equation.
• At STP, 1 mol of an ideal gas occupies 22.41 L.
at STP
T = 0°C = 273 K
n = 1 mol
P = 1 atm
V = 22.41 L
Dr. Sapna Gupta/Gases
17
Solved Problem:
A steel cylinder with a volume of 68.0 L contains O2 at a pressure of
15,900 kPa at 23ºC. What is the volume of this gas at STP?
P1  15,900 kPa 
1 atm
 157 .0 atm
101.3 kPa
T1 = 23 + 273 = 296 K
V1 = 68.0 L
P2 = 1 atm
T2 = 273 K
V2 = ?
P1V1
P2V2
=
n1T1
n2T2
V2 
PV
1 1T2
T1 P2

157.0 atm68.0 L 273 K 
296 K 1 atm
Dr. Sapna Gupta/Gases
 9850 L
18
Solved Problem:
A 50.0-L cylinder of nitrogen, N2, has a pressure of 17.1 atm at 23°C.
What is the mass of nitrogen in the cylinder?
V = 50.0 L
P = 17.1 atm
T = 23oC = 296 K
n
n
(17.1 atm)(50.0 L)
L  atm 

 0.08206
(296 K)
mol  K 

mass  35.20 mol
PV
RT
= 35.20 mols
28.02 g
mol
mass = 986 g
(3 significant figures)
Dr. Sapna Gupta/Gases
19
Solved Problem:
For an ideal gas, calculate the pressure of the gas if 0.215 mol occupies
338 mL at 32.0ºC.
n = 0.215 mol
V = 338 mL = 0.338 L
T = 32.0 + 273.15 = 305.15 K
P = ?
PV = nRT 
nRT
P 
V
L×atm 

 0.215 mol   0.08206
  305.15 K 
mol×K 

P 
0.338 L
Dr. Sapna Gupta/Gases
= 15.928
= 15.9 atm
20
Gas Density and Molar Mass
• Using the ideal gas law, it is possible to calculate the moles in 1 L at
a given temperature and pressure. The number of moles can then
be converted to grams (per liter).
Relation to density
• Get n/V on one side (mol/vol)
• Multiply by molar mass M (g/mol)
• Units of density are g/L here.
To find molar mass, rearrange the above equation.
Dr. Sapna Gupta/Gases
21
Solved Problem:
What is the density of methane gas (natural gas), CH4, at 125°C and
3.50 atm?
Mm = 16.04 g/mol
P = 3.50 atm
T = 125°C = 398 K
MmP
d
RT
g
(16.04
)(3.50 atm)
mol
d
L  atm 

 0.08206
(398 K)
mol  K 

g
d  1.72
L
(3 significant figures)
Dr. Sapna Gupta/Gases
22
Key Points
• Properties of gases
• The kinetic molecular theory
• Gas pressure
• Units
• Calculation
• Measurement
• The gas laws
• Boyle’s law
• Charles’ law
• Avogadro’s law
• The ideal gas law
• Combined gas law
• Calculating density and molar mass
Dr. Sapna Gupta/Gases
23