increase which could be potentially fatal.
Example:
A sample of helium gas has a pressure of 2.50 atm and a volume
of 125 mL.
What will the volume be in mL if the pressure
increases to 3.50 atm?Assume temperature and amount of gas
remain constant.
Summarize the problem:

##### We have textbook solutions for you!

**The document you are viewing contains questions related to this textbook.**

**The document you are viewing contains questions related to this textbook.**

Expert Verified

Sarasua CHM 2045 Lecture Notes Tro Chapter 5 Sp2015
12
J.A.C. Charles 1787, a French mathematician and physicist
observed the variation of volume and temperature when amount
of gas was fixed and pressure was constant.
Figure 5.10 p. 202 As T increases, gas expands (V increases).
As T decreases, gas contracts (V decreases). You can see that all
lines (data is linear) can be extrapolated to absolute zero of
temperature 0.00K = -273.15ºC.
There is a direct relationship between volume & temperature.
V
∝ T
or V=k
∙
T or
𝑉
𝑇
= k
or
𝑉
𝑇
= constant
For Charles’ Law, V and T vary while P,T and n are constant.
Rearrange the Ideal gas Equation to get what changes (V and T)
on left and what stays constant on right (P,n,R):
PV=nRT
?𝑉
𝑇
=nR

Sarasua CHM 2045 Lecture Notes Tro Chapter 5 Sp2015
13
𝑉
𝑇
=
?𝑅
?
= constant
A working relationship can be derived
V
1
T
1
=
V
2
T
2
T must be Kelvin; 1 and 2 correspond to initial and final
conditions.
(KMT) On a molecular level:
heated gas molecules move faster,
strike the container walls more frequently and with greater force
so to keep the P constant, V must increase (density decreases) so
that this motion is distributed over larger space. At lower T,
molecules move more slowly so V decreases (density increases)
to keep P constant. See Figure 5.11 p. 204.
Example:
A balloon has a volume of 5.0 L at room temperature (25.0
⁰
C).
When placed in a freezer at -20.0
⁰
C, what will its new volume
be in L?
Assume pressure and amount of gas are constant.
V
1
= 5.0L
T
1
=25.0
⁰
C = 298.2K
V
2
=?
T
2
= -20.0
⁰
C = 253.2K
V
1
T
1
=
V
2
T
2

Sarasua CHM 2045 Lecture Notes Tro Chapter 5 Sp2015
So V
2
=
V
1
T
1
x T
2
=
V1 x T2
T
1
=
5.0? 𝑥 253.2?
298.2K
= 4.2 L
3.
Amonton’s Law
or
Gay-
Lussac’s Law
P,T vary at constant V, n
The relationship between pressure and temperature is known as
Amonton’s Law
(end of 1600s) or
Gay-
Lussac’s Law
(see p.
191). Pressure is directly related to kelvin T for a fixed amount
of gas at constant V: P
∝
T
or P=kT so
?
𝑇
= k or
?
𝑇
= constant
PV = nRT so
?
𝑇
=
?𝑅
𝑉
= constant
A working relationship can be derived
P
1
T
1
=
P
2
T
2
(KMT) From a molecular standpoint, at higher T, molecules
strike the container walls more often and with greater force
therefore pressure increases.
Example:
A sample of air in a closed 250.00 mL flask is heated from
20.0
⁰
C to 75.0
⁰
C.
The pressure of the gas at 20.0
⁰
C = 0.98
atm.
Calculate the pressure at 75.0
⁰
C.

14

Sarasua CHM 2045 Lecture Notes Tro Chapter 5 Sp2015
15
4.
Avogadro’s Law
: Volume and Amount (in Moles)
Avogadro’s law assumes constant pressure and temperature.
n,V vary at constant P,T
It applies to all gases regardless of identity. See Figure 5.12 p.
189. It can be seen that
𝑉 ∝ n
with a linear relationship. If moles