### Lecture 2 (Slides) September 6

```Charles’s Law – Gas Volume and
Temperature
• A fixed amount of a gas at a fixed pressure will
expand if the temperature increases. Under
these conditions the volume of gas is
proportional to the Celsius temperature (or,
any other temperature!). We have, eg.
• V(t) = mt + b where t is the Celsius
temperature. This graph does not pass
through the origin of a V vs t plot.
Charles’s Law – Kelvin Temperatures
• The graph of gas volume vs temperature can
be made to pass through the origin if we
employ an absolute temperature scale, such
as the Kelvin scale, where zero degrees is the
lowest temperature achievable. With the
Kelvin scale (for example) at fixed P the
volume of gas becomes directly proportional
to temperature. Leads to simpler calculations.
• V(T) = kT
Gas volume as
a function of
temperature
Vat
V = kT
General Chemistry: Chapter 6
Slide 3 of 19
Charles’s Law
• We can take two points from the last graph.
• V1 = kT1 and V2 = kT2
• Combined these two give
V1 V2
•
=
T1 T2
• This is obviously an equation used for “initial
state → final state” problems (similar to
Boyle’s Law). We will eventually consider
these empirical laws in the context of kinetic
theory.
Kelvin Temperature Scale
• Celsius and Kelvin degrees have the same size.
A temperature of zero degrees Kelvin, or 0 K,
corresponds to -273.15 oC or -491.67 oF
(sounds colder although it’s not!). We won’t
use Fahrenheit temperatures further. To
convert from Celsius to Kelvin temperatures
• T(K) =t(oC) + 273.15
Class Example
• At a pressure of 0.986 bar and -22.0 oC a
sample of ethane gas , C2H6(g) has a volume of
2.75 L. What volume would the gas occupy if
the temperature were raised to + 22.0 oC
without changing the pressure?
• Class demonstration: Effect of temperature on
gas volume at constant pressure.
Standard Temperature and
Pressure
•Gas properties depend on conditions.
•IUPAC defines standard conditions of temperature
and pressure (STP).
P = 1 Bar = 105 Pa
T = 0°C = 273.15 K
General Chemistry: Chapter 6
Slide 7 of 19
• Gay-Lussac 1808
– Small volumes of gases react in the ratio of small whole
numbers.
At a fixed temperature and pressure, the volume of a
gas is directly proportional to the amount of gas.
General Chemistry: Chapter 6
Slide 8 of 19
At fixed T and P
Vn
or
V=cn
At STP
1 mol gas = 22.711 L gas
Figure 6-9
Molar volume of a gas visualized
General Chemistry: Chapter 6
Slide 9 of 19
Figure 6-8
Formation of Water – actual observation and Avogadro’s
hypothesis
General Chemistry: Chapter 6
Slide 10 of 19
• On the previous slide it is assumed that all
reactants and products are gases. We would
write the chemical reaction describing the
change as
• 2 H2(g) + O2(g) → 2 H2O(g)
• At “low” temperatures we might expect to see
liquid water formed. This reaction produces
both heat and light and is featured in many
Hollywood movies – eg. The Hindenburgh.
6-3 Combining the Gas Laws:
The Ideal Gas Equation
and the General Gas Equation
• Boyle’s law
V  1/P
• Charles’s law
VT
• Avogadro’s law V  n
General Chemistry: Chapter 6
}
V 
nT
P
Slide 12 of 19
The Ideal Gas Equation
PV = nRT
PV
R=
nT
General Chemistry: Chapter 6
Slide 13 of 19
Applying the ideal gas equation
General Chemistry: Chapter 6
Slide 14 of 19
The General Gas Equation
R=
P1V1
P2V2
=
n1T1
n2T2
If we hold the amount and volume constant:
P1
T1
=
P2
T2
General Chemistry: Chapter 6
Slide 15 of 19
Using the Gas Laws
General Chemistry: Chapter 6
Slide 16 of 19
Class Example – Ideal Gas Law Eq.
• Find (a) the density of CO2(g) at 55.0 oC and a
pressure of 64.3 kPa and (b) the number of
gas molecules per cm3 at this T and P.
• Solution (partial): The problem could be
tackled using the Combined Gas Law eqtn. but
is better approached using the Ideal Gas Law.
Why? “Trick”. No amount of gas is specified.
Any amount of gas works since (at a given T
and P) density is an intensive quantity.