2. Getting Informed
On June 24, 1983, Joseph Michel and John Etienne Montgolfier
used fire to inflate a spherical balloon that travelled more than a mile
– about 2 kilometers – and stayed aloft for 10 minutes. News of this
remarkable achievement spread throughout France, and Jacques
Alexandre Caesar Charles immediately tried to duplicate this
performance. As a result of his work with balloons, Charles noticed
that the volume of gas is directly proportional to its temperature
which we now call as Charles’ law.
3. His temperature on the relationship of the temperature and
volume provides an explanation of how hot-air balloons work.
Eversince, it has been known that an objects float when it weighs
less than fluid it displaces. A larger volume is occupied by gas
when it is heated and less is occupied when it is cooled. This is
because hot air is less dense than cold air. Once the air in a balloon
gets hot enough, the net weight of the balloon, plus this hot air is
less than the weight of an equivalent volume of cold air outside,
and the balloon starts to rise. When the gas in the balloon is
allowed to cool, the balloon shrinks and return to the ground.
4. This means that the volume is directly proportional to the
temperature and it is illustrated below:
5. Jacques Charles came up with a mathematical equation of the
dependence of volume to the temperature of gas at a constant
pressure. Thus,
V ∝ T
V = k2T
k2 =
𝑉
𝑇
The above equation is known as the Charles’ law where k2 is the
proportionality constant. Charles’ law states that the volume and
temperature are directly proportional to each other at a constant
pressure.
6. For a given sample of gas under two different sets of conditions,
then
𝑉1
𝑇1
= k2 =
𝑉2
𝑇2
Charles’ Law Equation 2
𝑉1
𝑇1
=
𝑉2
𝑇2
Where V1 and T1 are the initial volume and temperature,
respectively, and V2 and T2 are the final conditions.
7. Example 1.
A balloon has a volume of 2,500mL on a day when the
temperature is 303 K. If the temperature at night falls to 283 K,
will be the volume of the balloon if the pressure remains
Given: Solution:
V1 = 2, 500mL V2 = ?
T1 = 303 K T2 = 283 K
𝑉1
𝑇1
=
𝑉2
𝑇2
V1T2 = V2T1
V2 =
𝑉1
𝑇2
𝑇1
8. V2 =
𝑉1
𝑇2
𝑇1
V2 =
(2,500 𝑚𝐿)(283 𝐾)
(303 𝐾)
=
707,500 𝑚𝐿 .𝐾
303 𝐾
= 2,334.98 mL or 2335 L
Interpretation:
According to Charles’ law, a decrease in temperature from 303 K to
283 K could cause the volume to decrease from 2,500 mL to 2,334. 98 mL,
since they are directly proportional to each other.
9. Example 2.
A gas syringe contains 56.05 mL of gas at 315 K. Determine the
volume that the gas occupies if the temperature is increased to
K at the same pressure.
Given: Solution:
V1 = 56.05 mL T2 = 380 K
T1 = 315 K V2 = ?
V2 =
𝑉1
𝑇2
𝑇1
V2 =
(0.05605 𝐿)(360 𝐾)
(315 𝐾)
V2 =
(21.30 𝐿 .𝐾)
(315 𝐾)
= 0.068 L or 68.00
mL
10. Example 3.
If 15.0 L of Neon at 60°C is heated at constant pressure, what
temperature will it have if it occupies a volume of 4.50 L?
Given:
V1 = 15 L V2 = 4.50 L
T1 = 60°C T2 =
A. Converting 60°C to kelvin scale is
K = °C + 273
= 60 + 273 K
= 333 K
11. B. Finding the final temperature, T2.
T2 =
𝑉2𝑇1
𝑉1
T2 =
4.50 𝐿 (333𝐾)
15 𝐿
T2 = 99. 9 K
