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Charles Law describes how a gas expands due to the increase in temperature. Since the volume is directly proportional to the temperature hence a decrease in volume will cause a decrease in the temperature.
The effect of temperature on the volume of a gas at constant pressure was first carried out by a French scientist, in the year 1787 by Jacques Charles. The law is of volumes or Charles Law is an experimental gas law that explains how the gases lead to expanding upon heating.
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Keyterms: Gas, Heat, Kelvin, Flask, Charle’s Law, Boyle’s Law, Volume, Constant
Read Also: Specific Heat Capacity
Charles Law Formula
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Charles Law states that when the pressure on a dry gas is held constant, then the Kelvin temperature and therefore the volume will be directly proportional to each other.
As per the Charles Law equation,
V ∝ T,
The mathematical expression for the same is given as:
V =kT
=> V/T = K
Where,
V= Volume of gas
T= Temperature of gas (in Kelvin)
k= Constant (Non-Zero)
Charles Law illustrates that a rise in the temperature of a gas causes an increase in the volume, whereas a fall in temperature causes the volume to decrease. This law can be mathematically expressed under two different sets of conditions as follows:
In the equations above,
V1= Initial volume of the system
V2= Final volume of the system
T1= Initial temperature of the system (in Kelvin)
T2= Final temperature of the system (in Kelvin)
Hence, this law explains that as the absolute temperature of the gas increases, so does the volume of the gas.
Read Also: Boyle’s Law
Experimental Verification of Charles Law
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Charles Law can be verified experimentally to determine the temperature's value with regard to the figure below.
In the figure below, the first apparatus of the experiment contains a conical flask and a beaker. The empty flask is submerged into the beaker which is filled with water as shown in the figure. When the heat is supplied by a burner to the beaker filled with water, it also heats the air within the flask.
As a consequence, the air inside the flask begins to expand. This is often the first condition.
The flask inside the beaker is later submerged during a cistern at temperature. Now, the air inside the flask begins to contract since the temperature has decreased. This is often the second condition.
From both the conditions we can verify Charles Law that the temperature is directly proportional to volume.
Read More: Charles Law
Boyle’s Law
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According to Boyle's Law, the volume of an assigned mass of a gas is inversely proportional to its pressure given the temperature is held constant. The numerical representation for the law is as follows:
For an assigned mass of a gas, at a constant temperature,
V∝ 1/P or V = k (1/P)
Where,
k= Constant of proportionality. Its value depends on the temperature and mass of the gas.
P= Pressure of gas
V= Volume
On rearranging the above equation, we will get PV = k. This equation implies that at a constant temperature, the product of the pressure and volume of a given substance is constant.
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Things to Remember
- Charles Law states that when the pressure on a dry gas is held constant, then the Kelvin temperature and therefore the volume will be directly proportional to each other.
- The mass of gas remains unchanged.
- Upon heating, the number of molecules per unit volume decreases.
- Boyle's Law states that the volume of an assigned mass of a gas is inversely proportional to its pressure given the temperature is held constant
Previous Year Questions
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Sample Questions
Ques: Which law states that volume, V is directly proportional to temperature, T? (3 marks)
Ans: According to the Charles Law, Vt= (V0T) / T0
Vt/V0=T/T0
Thus, at constant pressure,
V/T =Constant
V ∝ T
Hence, Charles Law states that at a constant pressure the volume of a given mass of gas is directly proportional to the absolute/ Kelvin temperature.
Ques: Define the limitation or drawback of the Charles Law? (2 marks)
Ans: The drawback of Charles Law is that it is applicable only for ideal gas. It holds good for real gases only at high temperatures & low pressures but the relation between the volume and temperature is not linear at high pressures.
Ques: What will be the initial volume of gas at temperature 300 K, if the ultimate volume is 6L at 200 K? (3 marks)
Ans: From the question,
V2 = 6L, T1 = 300 K, T2= 200 K, V1 = ?
From Charles Law,
(V1)(T1) = (V2)(T2)
(V1) 300 = 6 (200)
V1 = 3L
Hence, the initial volume of gas at 300 K is 3 liter.
Ques: Why does the hot air balloon rise in the air? (2 marks)
Ans: This rise in hot air balloons is one of the applications of Charles's law. As per the law, the gases begin to expand upon heating but the mass of the gas remains the same. Therefore, the number of molecules per unit volume, i.e. density, begins to decrease upon heating. Since the hot air is less dense and cold; it enables the hot air balloon to rise by displacing the cooler air of the atmosphere.
Ques: List the applications of Charles's law. (3 marks)
Ans: The Charles law is an Ideal gas law that establishes a relation between volume and temperature at a constant pressure.
The below list contains a few of the applications of Charles law:
- Hot air balloon
- Bursting of a deodorant
- Turkey pop up timer
- Opening of a soda can
- Helium balloon on a cold day
- Bakery products.
Ques: At a temperature of 27.0 °C, a gas occupies 900.0 mL by volume. What will be the volume at 132.0 °C.? (3 marks)
Ans: Let's take V2= x,
Using the Charles law formula,
(V1) (T1) = (V2) (T2)
(900.0) / (300.0) = (x) / (405.0)
We get, x= 1215 mL
Ques: If 60 mL of gas is cooled from 33 °C to 5.00 °C then what will be the change in its volume? (3 marks)
Ans: From Charles law formula,
(60) / (306.0) = (x) / (278.0)
Solving the above equation we get,
306 (x) = 16680
x = 54.5 mL
The above value of x is the ending volume which is not the correct answer.
Hence, the volume is decreased by 5.5 mL.
Ques: At a standard temperature, a gas occupies 1.00 L by volume. Calculate the volume at 333.0 °C temperature. (3 marks)
Ans: Using the Charles law formula,
V1T2 = V2T1
V2 = (V1) [T2 / T1]
x = (1.00 L) [(606.0 K) / (273.0 K)]
x = 2.22 L
Hence, the volume will be, x = 2.22 L at a temperature of 333.0 °C.
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