14/15 Semester 2
Physical Chemistry I
(TKK-2246)
Instructor: Rama Oktavian
Email: rama.oktavian86@gmail.com
Office Hr.: M-F 13-15
Outlines
1. Gas: Properties
2. Gas laws: Boyle and Charles law
3. Ideal gas law
4. Equation of state
Gas state
State of matter
There are 4 states of matter
Gas state
State of matter
Properties of solid
retains a fixed volume and shape
rigid - particles locked into place
not easily compressible
little free space between particles
Microscopic view of a solid
does not flow easily
rigid - particles cannot move/slide past one
another
Gas state
State of matter
Properties of liquid
assumes the shape of the part of the
container which it occupies
particles can move/slide past one another
not easily compressible
little free space between particles
Microscopic view of a solid
flows easily
particles can move/slide past one another
Gas state
State of matter
Properties of gas
Microscopic properties
assumes the shape and volume of its
container
particles can move past one another
compressible
lots of free space between particles
Microscopic view of a solid
flows easily
particles can move past one another
Gas state
State of matter
Properties of gas
Macroscopic properties
Properties that can be observed and
measured
• Properties of bulk gases
• Observable
Microscopic view of a solid
How to make relation between those
macroscopic properties of gas??
– Pressure, volume, mass, temperature..
The general form of an equation of state is
p=f(T,V,n)
Gas state
Gas properties
Gases Exert Pressure: What is Pressure?
Pressure is defined as the force exerted divided by the area it acts over
Pressure = Force/Area
The SI unit of pressure, the pascal(Pa), is
defined as 1 newton per metre-squared:
1 Pa =1 N m−2
1 Pa =1 kg m−1s−2
1 atm =1.013 25 ×105Pa exactly 1 bar =105Pa
Gas properties
Gases Exert Pressure: What is Pressure?
Pressure is defined as the force exerted divided by the area it acts over
Pressure = Force/Area
Self-test 1.1 Calculate the pressure (in pascals and atmospheres) exerted
by a mass of 1.0 kg pressing through the point of a pin of area 1.0 ×10−2
mm2 at the surface of the Earth.
Gas properties
Pressure measurement
Barometer – device that measures
atmospheric pressure
Invented by Evangelista Torricelli in 1643
the height of the mercury column is
proportional to the external pressure
Gas properties
Pressure measurement
Derive an equation for the pressure at the
base of a column of liquid of mass density
ρ(rho) and height h at the surface of the
Earth.
p=F/A
V = Ah
F = mg
m = ρAh
m = ρV
F = mg = ρAhg
the pressure is independent of the shape
and cross-sectional area of the column.
Gas properties
Pressure measurement
Gas properties
Pressure measurement
 A manometer
measures the
pressure of a gas in
a container
 Gas pressure is the
force exerted by the
collisions of gas
particles with a
surface
Gas properties
Pressure measurement
Atmospheric pressure
By definition the average pressure at sea level will support a column
of 760 mm of mercury. (760 torr)
g = 9.81 m.s-2 h = 0.76 m
ρ = 13.6 g.cm-3 = 13.6 kg.L-1 = 13.6x103kg.m-3
P = 9.81x0.76x13.6x103 = 1.013x105 Pa (N.m-2)
Gas properties
Pressure measurement
Atmospheric pressure problem
If we made a barometer out of water, what would be the
height of the water column if the pressure is 745 torr?
Gas properties
Pressure measurement
Variation in atmospheric pressure
– Changing weather conditions
Gas properties
Pressure measurement
Variation in atmospheric pressure
– Changing altitude
Gas laws
Boyle’s law
• Boyle’s Law is one of the laws in physics that concern the
behaviour of gases
• When a gas is under pressure it takes up less space:
• The higher the pressure, the smaller the volume
• Boyles Law tells us about the relationship between the volume of a
gas and its pressure at a constant temperature
• The law states that pressure is inversely proportional to the
volume
Gas laws
Boyle’s law
Pressure-volume relationship
pressure-volume behavior of gases were made by Robert Boyle in 1662
P a 1/V
P x V = constant
P1 x V1 = P2 x V2
Gas laws
Pressure and Volume: Boyle’s Law
•
Volume and pressure are inversely proportional.
– If one increases the other decreases.
Gas laws
Pressure and Volume: Boyle’s Law
•
Boyle’s Law and Breathing: Inhalation
During inhalation,
 the lungs expand
 the pressure in the lungs
decreases
 air flows towards the
lower pressure in the
lungs
Gas laws
Pressure and Volume: Boyle’s Law
•
Boyle’s Law and Breathing: Inhalation
During exhalation
• lung volume decreases
• pressure within the lungs
increases
• air flows from the higher
pressure in the lungs to
the outside
Gas laws
Pressure and Volume: Boyle’s Law
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?
P1 x V1 = P2 x V2
P1 = 726 mmHg
P2 =
V1 = 946 mL
P1 x V1
726 mmHg x 946 mL
=
V2
154 mL
P2 = ?
V2 = 154 mL
= 4460 mmHg
Gas laws
Pressure and Volume: Boyle’s Law
Guide to Calculations with Gas Laws
Gas laws
Pressure and Volume: Boyle’s Law
Problem
A deep sea diver is working at a depth where the
pressure is 3.0 atmospheres. He is breathing out air
bubbles. The volume of each air bubble is 2 cm2. At
the surface the pressure is 1 atmosphere. What is
the volume of each bubble when it reaches the
surface?
Gas laws
Charles’s law
• French chemist Jacques Charles discovered that the volume of a gas at
constant pressure changes with temperature.
• As the temperature of the gas increases, so does its volume, and as its
temperature decreases, so does its volume.
• The law says that at constant pressure, the volume of a fixed number of
particles of gas is directly proportional to the absolute (Kelvin) temperature
Gas laws
Charles’s law
Volume-temperature relationship
Variation of gas volume with temperature
at constant pressure
VaT
V = constant x T
V1/T1 = V2/T2
Gas laws
Charles’s law
Volume-temperature relationship
 For two conditions, Charles’s law is written
V1 = V2
(P and n constant)
T1
T2
 Rearranging Charles’s law to solve for V2 gives
T2 x V1 = V2 x T2
T1
T2
V2
=
V1 x T2
T1
Gas laws
Charles’s law
Example problem
A balloon has a volume of 785 mL at 21 °C. If the
temperature drops to 0 °C, what is the new volume of
the balloon (P constant)?
STEP 1 Set up data table:
Conditions 1
Conditions 2
V1 = 785 mL
T1 = 21 °C
= 294 K
V2 = ?
T2 = 0 °C
= 273 K
Know
Predict
V decreases
T decreases
Be sure to use the Kelvin (K) temperature in gas
calculations.
Gas laws
Charles’s law
STEP 2 Solve Charles’s law for V2:
V1
= V2
T1
T2
V2
= V1 x T2
T1
Temperature factor
decreases T
STEP 3 Set up calculation with data:
V2 = 785 mL x 273 K = 729 mL
294 K
Gas laws
Charles’s law
A sample of oxygen gas has a volume of 420 mL at a
temperature of 18 °C. At what temperature (in °C) will
the volume of the oxygen be 640 mL (P and n
constant)?
1) 443 °C
2) 170 °C
3) –82 °C
Gas laws
Avogadro’s law
Avogadro’s law states that
 the volume of a gas is
directly related to the
number of moles (n) of
gas
 T and P are constant
V1 = V2
n1
n2
Ideal Gas law
The combination of those laws gives
Usually written as:
R is gas constant
Ideal Gas law
R is known as universal gas constant
Using STP conditions
PV
R
nT
(1atm)( 22.4 L)
R
(1mol)( 273.15K )
R  0.0821(atm.L)(mol.K )
1
Ideal gas and Real gas
Ideal gas
The ideal gas law is used to describe the
behavior of an ideal gas.
Ideal gas: hypothetical gas that obeys
kinetic molecular theory and the ideal
gas law
Ideal gas and Real gas
Ideal gas
pV  RT
The ideal gas law was useful in determining the properties
of a specific sample of gas at constant T, P, V, and n.
We often need to know how a change in one (or more)
properties impacts the other properties for a sample of a
gas
Equation of state
Equation of state
The general form of an equation of state is
p=f(T,V,n)
PV  nRT
Ideal gas equation is equation of state
Equation of state
Equation of state
PV  nRT
P, V, n, T are properties
Intensive properties – independent on the quantity of material
P, T
Extensive properties – dependent on the quantity of material
n, V
The ratio of any two extensive
variables is always an intensive
variable
Intensive properties
Equation of state
Equation of state
PV  nRT
Expressed in intensive properties gives
PV  RT
Ideal gas and Real gas
Real gas
pV  RT
deviations from the perfect gas law because molecules interact
with one another
Repulsive forces are significant only when molecules are
almost in contact
Attractive intermolecular forces have a relatively long range and are
effective over several molecular diameters
Ideal gas and Real gas
Real gas
Z
The compression factor
For ideal gas
Z 1
P  Z  1
P  Z  1
P moderate Z < 1
Ideal gas and Real gas
Real gas
The compression factor
PV
Z
RT
Z
Ideal gas and Real gas
Equation of state for Real gas
Virial equation of state
How to describe this P-V behavior?
Ideal gas and Real gas
Equation of state for Real gas
Virial equation of state
PV
Z
RT
Ideal gas and Real gas
Equation of state for Real gas
Virial equation of state
Ideal gas and Real gas
Equation of state for Real gas
Virial equation of state
The second virial coefficient B′ can be obtained from measurements of the
density ρ of a gas at a series of pressures. Show that the graph of p/ρ against
p should be a straight line with slope proportional to B′.