Charles Law is one of the fundamental laws used for the study of gases. Charles Law states that the volume of an ideal gas is directly proportional to the temperature (at absolute scale) at constant pressure. Famous French physicist Jacques Charles formulated this law in the year 1780.
Let's learn about Charles Law's derivation and others in detail in this article.
What is Charles Law?
Charles’ Law, also called the law of volumes, states that at constant pressure the volume of a gas varies directly with the temperature. i.e. increase in the temperature of the gas results in an increase in the volume and conversely decrease in the temperature of the gas results in a decrease in the volume of the gas.
Suppose the initial temperature of the gas is T1 when its volume is V1 also, its volume at temperature T2 is V2. Now according to Charles's Law.
V2/V1 = T2/T1
V1T2 = V2T1
This above equation establishes a proportional relation between the absolute temperature and the volume of the gas.
Defining Charles' Law
Charles's Law is defined as,
"The volume of a given mass of gas at constant pressure varies directly with its absolute temperature."
Jacques Charles
Jacques Charles is a French physicist that is famous for giving the Charles's Law of gases. He was born on 2nd November 1746 in Beaugency-sur-Loire. Charles along with Robert Brothers introduced hydrogen filled gas ballon. Using this balloon he ascended to a height of 1800 feet. This led to give him Charles's Law in 1787 which was later formulated by Gay-Lussac in 1802.
Charles Law Everyday Examples
Various examples where Charles Law is applicable in our everyday life are,
- In winters the tyres of vehicles shrinks and it is important to check their pressure regularly.
- If an air tube is placed in an pool outside in the sun then it can burst because of the air inside the tubes expands because of the Charles Law.
Charles Law equation states that, V α T. Now the Charles Law formula is,
V1/T1 = V2/T2
where,
- V1 is the Initial Volume
- T1 is the Initial Temperature in Kelvin
- V2 is the Final Volume
- T2 is the Final Temperature in Kelvin
This formula is used to find the volume or temperature of the gas if pressure of the gas is constant.
According to Charles Law, "Volume of a gas is Directly proportional to its temperature at constant pressure" The relation can be expressed as,
V α T …(At constant Pressure)
When we substitute constant k for the proportionality sign, we obtain
V = kT
Note initially when the temperature at absolute scale is T1 and its volume is V1
then,
V1 = kT1....(1)
Now, when the temperature at absolute scale is T2 and its volume is V2
then,
V2 = kT2....(2)
dividing eq(1) by eq(2)
V1/V2 = kT1/kT2
V1/V2 = T1/T2
Thus, the Charles law is derived.
Experiment Verification of Charles Law
Charles law can be experimentally verified with the help of the diagram discussed below,
Apparatus Required
The apparatus required for experimental verification of Charles law is,
- Conical Flask
- Beaker
- Thermometer
- Bunsen Burner
- Rubber Stopper
Experiment
The beaker is filled with water and the empty conical flask is submerged in the beaker the set-up is then heated above Bunsen Burner the air inside the conical flask is then heated. The temperature of the system is measured using a thermometer in this condition (say 1). The setup is then cooled and the air inside the flask gets contracted in this condition(say 2) also the temperature is measured using the thermometer.
Noting the value of temperature and volume in conditions 1 and 2 Charles law is verified.
Graphical Representation Of Charles Law
The graph of Charles law is an Isobaric graph. An Isobaric graph is a graph that gives a straight line if plotted between volume(V) and temperature(T). The image given below shows the pressure of the different gases.
The slope of this graph gives the pressure of the gas at a specific volume and temperature. The increase in pressure increases the slope of the V-T graph.
Charles Law Application In Real Life
Charles Law has a variety of applications some of the applications of Charles Law that we observed in our daily life are,
- Hot air Balloons use the concept of Charles law to levitate.
- Helium Balloons shrinking in colder environments is also an example of Charles law.
- The decrease in the capacity of the lungs in the winter is also explained by Charles law
- When the temperature inside the piston of a car engine increases the air expands and pushes the piston outwards which makes the car engine work this is also explained by Charles law.
Charles' Law Units
Charles's law defines the relation of volume of gas with the absolute temperature of the gas at constant pressure. The volume of the gas is measured in liter and 1 l = 10-3 m3. The temperature of the gas is measure in Kelvin which is the absolute scale of temperature.
Limitations of Charles law
The limitation of Charles Law is that it only applies to perfect gas. For a real gas, it is only valid at high temperatures and low pressures also, at high pressures, the relationship between volume and temperature is not linear.
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Charles Law Solved Examples
Example 1: At 19°C, a sample of gas takes up 2.55 L. What is the new volume of the gas if the pressure stays the same and the temperature is increased to 30°C?
Solution:
T1 = 273 + 19 = 292 K
T2 = 273 + 30 = 303 K
Given,
- Initial Temperature = T1 = 292 K
- Initial Volume = V1 = 2.55 l
- Final Temperature = 303 K
- Final Volume = V2 = ?
Now, according to Charles's Law
V1/T1 = V2/T2
2.55 / 292 = V2 /303
V2 = 2.55 × 303 / 292
V2 = 772.65 / 292
V2 = 2.64 L
Final Volume = V2 = 2.64 l
Example 2: At 123 degrees Celsius, helium gas has a volume of 1690 ml. Determine the temperature at which the capacity expands to 472 ml. Count on pressure remaining steady.
Solution:
T1 = 273 + 123 = 396 K
Given,
- Initial Temperature = T1 = 396 K
- Initial Volume = V1 = 1690 ml
- Final Temperature = ?
- Final Volume = V2 = 472 ml
Now, according to Charles's Law
V1/T1 = V2/T2
V1/T1V2 = 1/T2
T2 = T1V2 / V1
T2 = 396 × 1690 / 472
T2 = 669240 / 472
T2 = 1417.9 K
T2 = 1417.9 - 273 = 1144.9°C
Final Temperature = T2 = 1144.9°C
Example 3: What will the initial gas volume be at 400 K if the final volume is 13 L at 270 K?
Solution:
Given,
- Initial Temperature = T1 = 400 K
- Initial Volume = V1 = ?
- Final Temperature = 270 K
- Final Volume = V2 = 13 l
Now, according to Charles's Law
V1/T1 = V2/T2
V1 = V2T1 / T2
V1 = 13 × 400 / 270
V1 = 5200 / 270
V1 = 19.26 L
Initial Volume = V1 = 19.26 L
Example 4: What will happen to the gas's volume if 55 mL of it is cooled from 67°C to 30°C?
Solution:
T1 = 273 + 67 = 340 K
T2 = 273 + 30 = 303 K
Given,
- Initial Temperature = T1 = 340 k
- Initial Volume = V1 = 55 ml
- Final Temperature = 303 K
- Final Volume = V2 = ?
Now, according to Charles's Law
V1/T1 = V2/T2
V2 = V1T2 / T1
V2 = 55 × 303 / 340
V2 = 16665 /340
V2 = 49.015 ml
Final Volume = V2 = 49.015 ml
Example 5: A gas has a volume of 1.5 L at normal temperature. At a temperature of 333 K, determine the volume.
Solution:
Given,
- Initial Temperature = T1 = 273 k
- Initial Volume = V1 = 1.5 l
- Final Temperature = T2 = 333k
- Final Volume = V2 = ?
Now, according to Charles's Law
V1/T1 = V2/T2
V2 = V1T2 / T1
V2 = 1.5 × 333 / 273
V2 = 499.5 / 273
V2 = 1.8 L
Final Volume = V2 = 1.8 L
Example 7: At standard temperature the volume is 30 L, find the final temperature has a volume of 50 L.
Solution:
Given,
- Initial Temperature = T1 = 273 k
- Initial Volume = V1 = 30 l
- Final Temperature = ?
- Final Volume = V2 = 50 l
Now, according to Charles's Law
V1/T1 = V2/T2
T2 = T1V2 / V1
T2 = 273 × 30 / 50
T2 = 8190 / 50
T2 = 163.8 K
Final Temperature = T2 = 163.8 K
Related Article,
Charles’s Law Practice Problems
P1. At standard temperature the volume is 32 L, find the final temperature has a volume of 86 L.
P2. A gas has a volume of 9 L at STP. At a temperature of 450 K, determine the volume.
P3. At standard temperature the volume is 87 L, find the final temperature has a volume of 5 L.
P4. A gas has a volume of 17 L at normal temperature. At a temperature of 789 K, determine the volume.
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