Charles' law describes how gas volume changes with temperature. It states that the volume of a gas is directly proportional to its temperature when pressure is kept constant. The document provides the formula for Charles' law and shows examples of using it to calculate unknown volumes or temperatures given other variables like initial and final volumes and temperatures. It also discusses the limitations of Charles' law and provides sample problems and solutions demonstrating how to apply the law to calculate unknown values.

1. Charles law is an experimental gas law. It explains how
gases tend to expand when heated. French physicist
Charles studied the effect of temperature on the quantity
of a gas at constant pressure.
This law describes how a gas expands because of the
temperature increases; conversely, a decrease in
temperature will cause a decrease in volume.

2. Definition of Charles Law Formula is, “When the
pressure on a sample of a dry gas is held constant,
the Kelvin temperature and therefore the volume is
going to be in direct proportion.”
The equation of the law is PV = k.
• k may be a constant.
• P= Pressure
• V= Volume

3. Therefore, V=kT . For comparing an equivalent substance under
two different sets of conditions,
the law is often written as
•V1 = First Volume
•V2= Second Volume
•T1= First Temperature
•T2 = Second Temperature
The equation shows that, as temperature increases, the quantity
of the gas also increases in proportion.

4. Charles’s law appears to imply that the quantity of gas will
descend to zero at a particular temperature (−266.66 °C
consistent with Gay-Lussac’s figures) or −273.15 °C. At
temperature, the gas is having zero energy and hence the
molecules restrict motion.
Charles from his experiments concluded that at constant
pressure, the quantity of a hard and fast amount of a gas
increases or decreases by 1⁄273 (now 1⁄273.15) times the
quantity at 0 °C for each 1 °C rise or fall in temperature.

5.  We can verify Charles’ law experimentally and
determine the value of temperature with regard
to the above figure. the first apparatus of the
experiment consists of a conical flask and a
beaker. The empty flask is submerged into the
water-filled beaker as shown within the above
diagram. When the warmth is supply to the
beaker by a burner, it also heats the air within
the flask.


6. As a consequence, the air inside the flask expands. this is
often our condition 1. The flask is later dipped during a
cistern at temperature. Now, the air within the flask
contracts since the temperature is decreased. this is often
condition 2. By knowing the temperature and volume of
both conditions, we will verify the law.
Limitations of Charles law
Charles’s law is applicable to only ideal gases. Charles law
holds good for real gases only at high temperatures and low
pressures. the connection between the quantity and
temperature isn’t linear in nature at high pressures.

7. A gas in a container has an unknown initial
volume with an initial temperature of 273 K. the
final volume is found to be 3.5 L at a
temperature of 325 K. What is the initial
volume?
Given: V1= ? V2= 6L T1= 150K T2= 100k
V₁ / T₁ = V₂ / T₂
(V1) (150K) = (6L) (100K) then cross multiply
(150) (6) = (100) (V1)
900/100 = (100) (V1)/ 100
V1= 9

8. A gas in a container has an initial volume of 5 L and an
initial temperature of 10°C. the final temperature increases
to 45°C. what is the final volume?
Given: V1= 5L V2= ? T1= 283K T2= 318k
- Covert Celsius into kelvin
10°C + 273 = 283K
45°C + 273 = 318K
V₁ / T₁ = V₂ / T₂
(5L) (283K) = (V2) (318K) then cross multiply
(5) (318) = (283) (V2)
1590/283 = (283) (V2)/ 283
V2= 5.6 L - the final volume is 5.6 L

9. A gas in a container has a initial volume of 7 L
and an unknown temperature. The final
temperature is found to be 250 K with a final
volume of 12 L. what is the initial temperature?
Given: V1= 7 L V2= 12 L T1= ? T2= 250k
V₁ / T₁ = V₂ / T₂
(7L) (T1) = (12L) (250K) then cross multiply
(7L) (250) = (12) (T1)
1750/12 = (12) (T1)/ 12
T1= 145K
- The initial temperature is 145K

10. A gas in a container has an initial volume of 2 L and
an initial temperature of 25°C. the final volume
decreases to 0.25 L. what is the final temperature?
Given: V1= 2 L V2= .25L T1= 298K T2=?
- Convert first Celsius to kelvin
25°C + 273 = 298 K
V₁ / T₁ = V₂ / T₂
(2L) (298K) = (.25L) (T2) then cross multiply
(298) (.25) = (2) (T2)
74.5/2 = (2) (T2)/ 2
T2= 37. 25K then subtract to 273 so,
- The Final temperature is 235.75

11.  Gay-Lussac’s law is a gas law which states that the pressure exerted
by a gas (of a given mass and kept at a constant volume) varies
directly with the absolute temperature of the gas. In other words, the
pressure exerted by a gas is proportional to the temperature of the gas
when the mass is fixed and the volume is constant.
 This law was formulated by the French chemist Joseph Gay-Lussac in
the year 1808. The mathematical expression of Gay-Lussac’s law can
be written as follows:

12.  P ∝ T ; P/T = k
 Where:
• P is the pressure exerted by the gas
• T is the absolute temperature of the gas
• k is a constant.
 The relationship between the pressure and absolute temperature of a given mass of
gas (at constant volume) can be illustrated graphically as follows.

13. From the graph, it can be understood that the
pressure of a gas (kept at constant volume) reduces
constantly as it is cooled until the gas eventually
undergoes condensation and becomes a liquid.

14. Gay-Lussac’s law implies that the ratio of the initial pressure and temperature is equal
to the ratio of the final pressure and temperature for a gas of a fixed mass kept at a
constant volume. This formula can be expressed as follows:
(P1/T1) = (P2/T2)
Where:
• P1 is the initial pressure
• T1 is the initial temperature
• P2 is the final pressure
• T2 is the final temperature

15. This expression can be derived from the pressure-temperature
proportionality for gas. Since P ∝ T for gases of fixed mass kept
at constant volume:
P1/T1 = k (initial pressure/ initial temperature = constant)
P2/T2 = k (final pressure/ final temperature = constant)
Therefore, P1/T1 = P2/T2 = k
Or, P1T2 = P2T1

16.  When a pressurized aerosol can (such as a deodorant can or a spray-paint can) is heated, the resulting
increase in the pressure exerted by the gases on the container (owing to Gay-Lussac’s law) can result in
an explosion. This is the reason why many pressurized containers have warning labels stating that the
container must be kept away from fire and stored in a cool environment.

17.  An illustration describing the increase in pressure which accompanies an increase in the absolute
temperature of a gas kept at a constant volume is provided above. Another example of Gay-Lussac’s law
can be observed in pressure cookers. When the cooker is heated, the pressure exerted by the steam
inside the container increases. The high temperature and pressure inside the container cause the food to
cook faster.

18. The pressure of a gas in a cylinder when it is heated to a temperature of 250K is 1.5 atm. What was the
initial temperature of the gas if its initial pressure was 1 atm.
 Given,
 Initial pressure, P1 = 1 atm
 Final pressure, P2 = 1.5 atm
 Final temperature, T2 = 250 K
 As per Gay-Lussac’s Law, P1T2 = P2T1
 Therefore, T1 = (P1T2)/P2 = (1*250)/(1.5) = 166.66 Kelvin.

19. At a temperature of 300 K, the pressure of the gas in a deodorant can is 3 atm.
Calculate the pressure of the gas when it is heated to 900 K.
Initial pressure, P1 = 3 atm
Initial temperature, T1 = 300K
Final temperature, T2 = 900 K
Therefore, final pressure (P2) = (P1T2)/T1 = (3 atm*900K)/300K = 9 atm.

20. The combined gas law combines the three gas laws: Boyle's
Law, Charles' Law, and Gay-Lussac's Law. It states that the ratio of the
product of pressure and volume and the absolute temperature of a gas
is equal to a constant. When Avogadro's law is added to the combined
gas law, the ideal gas law results. Unlike the named gas laws, the
combined gas law doesn't have an official discoverer. It is simply a
combination of the other gas laws that works when everything except
temperature, pressure, and volume are held constant.

21.  There are a couple of common equations for
writing the combined gas law.
 The classic law relates Boyle's law and Charles' law
to state:
 PV/T = k
 where P = pressure, V = volume, T = absolute
temperature (Kelvin), and k = constant.
 The constant k is a true constant if the number of
moles of the gas doesn't change. Otherwise, it varies.

22.  Another common formula for
the combined gas law relates
"before and after" conditions of
a gas:
P1V1 / T1 = P2V2 / T2

23.  Find the volume of a gas at STP when 2.00 liters is collected at 745.0
mm Hg and 25.0 degrees Celsius.
 To solve the problem, you first need to identify which formula to use. In
this case, the question asks about conditions at STP, so you know you're
dealing with a "before and after" problem. Next, you need to
understand STP. If you haven't memorized this already (and you
probably should, since it appears a lot), STP refers to "standard
temperature and pressure," which is 273 Kelvin and 760.0 mm Hg.

24.  Because the law works using absolute temperature, you
need to convert 25.0 degrees Celsius to the Kelvin scale.
This gives you 298 Kelvin.
 At this point, you can plug the values into the formula and
solve for the unknown. A common mistake some people
make when they're new to this kind of problem is confusing
which numbers go together. It's good practice to identify
the variables. In this problem they are:

25. P1 = 745.0 mm Hg
V1 = 2.00 L
T1 = 298 K
P2 = 760.0 mm Hg
V2 = x (the unknown you're solving for)
T2 = 273 K
Next, take the formula and set it up to solve for the unknown "x," which in this
problem is V2:
P1V1 / T1 = P2V2 / T2
Cross-multiply to clear the fractions:
P1V1T2 = P2V2T1
Divide to isolate V2:
V2 = (P1V1T2) / (P2T1)
Plug in the numbers and solve for V2:
V2 = (745.0 mm Hg · 2.00 L · 273 K) / (760 mm Hg · 298 K)
V2 = 1.796 L
Report the result using the correct number of significant figures:
V2 = 1.80 L

26. The combined gas law has practical applications when dealing
with gases at ordinary temperatures and pressures. Like other
gas laws based on ideal behavior, it becomes less accurate at
high temperatures and pressures. The law is used in
thermodynamics and fluid mechanics. For example, it can be
used to calculate pressure, volume, or temperature for the gas in
clouds to forecast weather.

28. FIND THE INITIAL VOLUME OF A GAS
AT 150 K, IF THE FINAL VOLUME IS 6 L
AT 100 K
Explanation:
Given,
V1=?
V2 =6 L
T1= 150 K
T2= 100 K
Using Charles Law,
(V1)(T1)=(V2)(T2)
(V1)(150)=(6)(100)
V1=6×150100
V1=9L
The initial volume of a gas at 150 K is 9 liters.