Charles Law Formula: Statement, Equation, Combined Gas Law, Numerical

Charles Law Formula: On increasing the temperature, the gas expands, and volume increases. For example, an inflated balloon expands in size on increasing in temperature, which may lead to the bursting of the balloon. Then, is the temperature and volume related to each other? What happens to the volume of the gas when we increase the temperature or temperature is kept constant?

You will quench the thirst for all these answers in this article, Charles law formula. In this article, you will deal with Charles law, its equation, derivation of equation, graph, practical application, numerical, etc. You will also derive combined gas law from Boyles and Charles law.

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What is Charles Law?

In \(1787,\) a French scientist, Jacques Charles, studied the effect of temperature on the volume of gases at constant pressure. The following generalisation about the relationship between temperature and volume of a gas was observed, which is known as Charles law. It states that the volume of a given mass of gas increases or decreases by \(\frac{1}{{273}}\) of its volume at \({\rm{0}}\,^\circ {\rm{C}}\) for each one-degree rise or fall in temperature provided pressure is kept constant.

Let \({{\rm{V}}_{\rm{0}}}\) be the volume of a given mass of a gas at \({\rm{0}}\,^\circ {\rm{C}}\) and V_t is its volume at any temperature \({\rm{t}}\,^\circ {\rm{C}},\) then the volume, \({{\rm{V}}_{\rm{t}}}\) may be written in terms of Charles law, at constant pressure as:

For \(1\) degree rise in temperature, volume increases \({\rm{ = \;}}{{\rm{V}}_{\rm{0}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{273}}}}\)

For t degree rise in temperature, volume increases \({\rm{ = \;}}{{\rm{V}}_{\rm{0}}}{\rm{ \times }}\frac{{\rm{1}}}{{{\rm{273}}}}{\rm{ \times t}}\)

Therefore, volume at \({{\rm{t}}^{\rm{o}}}{\rm{C,}}\) \({{\rm{V}}_{\rm{t}}}{\rm{ = }}\) Initial volume \( + \) Increase in volume

\({{\rm{V}}_{\rm{t}}}{\rm{ = }}{{\rm{V}}_{\rm{0}}}{\rm{ + }}\frac{{{{\rm{V}}_{\rm{0}}}{\rm{ \times t}}}}{{{\rm{273}}}}\)

\({{\rm{V}}_{\rm{t}}}{\rm{ = V}}{{\rm{\;}}_{\rm{0}}}\left[ {{\rm{1 + }}\frac{{\rm{t}}}{{{\rm{273}}}}} \right]\)

\({{\rm{V}}_{\rm{t}}}{\rm{ = }}{{\rm{V}}_{{\rm{0\;}}}}\left[ {\frac{{{\rm{273 + t}}}}{{{\rm{273}}}}} \right]\)

Which Law States \({\rm{V}} \propto {\rm{T}}?\)

According to the mathematical form of Charles Law:

\({{\rm{V}}_{\rm{t}}}{\rm{ = }}{{\rm{V}}_{{\rm{0\;}}}}\left[ {\frac{{{\rm{2\;73 + t}}}}{{{\rm{273}}}}} \right]\)

But \({\rm{273 + t  =  T}}\) \(\left( {\rm{K}} \right)\) on the Kelvin scale and \({\rm{273  =  }}{{\rm{T}}_{\rm{0}}}\)

Therefore, \({{\rm{V}}_{\rm{t}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{0}}}{\rm{\;T}}}}{{{{\rm{T}}_{\rm{0}}}}}\)

\(\frac{{{{\rm{V}}_{\rm{t}}}}}{{{{\rm{V}}_{\rm{0}}}}}{\rm{ = }}\frac{{\rm{T}}}{{{{\rm{T}}_{\rm{0}}}}}\)

Thus,

\({\rm{\;}}\frac{{\rm{V}}}{{\rm{T}}}{\rm{ = Constant\;}}\left( {{\rm{at\;constant\;pressure}}} \right)\)
\({\rm{V}} \propto {\rm{T\;}}\left( {{\rm{at\;constant\;pressure}}} \right)\)

Hence, Charles Law can also state that the volume of a given mass of gas is directly proportional to the absolute or Kelvin temperature at constant pressure.

Let \({{\rm{V}}_{\rm{1}}}\) be the volume of a gas at temperature \({{\rm{T}}_{\rm{1}}}{\rm{.}}\) By keeping the pressure constant, if the temperature of the gas is increased to \({{\rm{T}}_{\rm{2}}}{\rm{,}}\) then the volume will also be increased to \({{\rm{V}}_{\rm{2}}}{\rm{.}}\) Therefore, according to law, \(\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{\;}}\left( {{\rm{at\;constant\;mass\;and\;pressure}}} \right)\)

Practical Importance of Charles’ Law

An interesting application of the Charles law is the use of hot air balloons in sports and also for meteorological observations. According to the law, the gases expand on heating. Since the mass of gas remains unchanged; therefore, the number of molecules per unit volume, that is, the density of the gas, decreases on heating. Hence, hot air is less dense than cold air. This enables the hot air balloons to rise up by displacing the cooler air of the atmosphere.

What is Boyle’s law?

Boyle’s law states that “The volume of a given mass of a gas is inversely proportional to its pressure provided the temperature remains constant”. Mathematically, Boyle’s law is expressed as:

\({\rm{V}} \propto \frac{{\rm{1}}}{{\rm{P}}}{\rm{,}}\) at constant temperature and for a given mass of the gas.

Or \({\rm{V = k\;}}\frac{{\rm{1}}}{{\rm{P}}}{\rm{,}}\) where \({\rm{k}}\) is the constant of proportionality, \({\rm{V}}\) is the volume and \({\rm{P}}\) is the pressure of the gas. The value of constant k depends upon the mass and the temperature of the gas.

By rearranging the equation, we get \({\rm{PV = k}}{\rm{.}}\) This implies that the product of the pressure and volume of a given mass of a substance is constant at a constant temperature.

What is the Formula of Combined Gas Law?

The effect of pressure on the volume of gas at a constant temperature can be calculated with the help of Boyle’s law. The effect of temperature on the volume of the gas at constant pressure can be calculated by Charles Law. The combined effect of temperature and pressure on the volume of the gas is given by the combined gas law. Let the volume of a certain mass of a gas change from \({{\rm{V}}_{\rm{1}}}\) to \({{\rm{V}}_{\rm{2}}}\) when the pressure is changed from \({{\rm{P}}_{\rm{1}}}\) to \({{\rm{P}}_{\rm{2}}}\) and temperature from \({{\rm{T}}_{\rm{1}}}\) to \({{\rm{T}}_{\rm{2}}}\) in the following two steps,

Step 1: Suppose the volume of a given mass of a gas changes from \({{\rm{V}}_{\rm{1}}}\) to \({\rm{v}}\) when the pressure is changed from \({{\rm{P}}_{\rm{1}}}\) to \({{\rm{P}}_{\rm{2}}}\) at a constant temperature \({{\rm{T}}_{\rm{1}}}{\rm{.}}\)

According to Boyle’s law

\({{\rm{P}}_{\rm{2}}}{\rm{\nu  = }}{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}\)

\({\rm{\nu  = }}\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{P}}_{\rm{2}}}}}{\rm{\;\;\;\;\;\; –  –  –  –  > }}\left( {\rm{1}} \right)\)

Step 2: Now, suppose the volume \({\rm{\nu }}\) changes to \({{\rm{V}}_{\rm{2}}}\) when the temperature is changed from \({{\rm{T}}_{\rm{1}}}\) to \({{\rm{T}}_2}\) at constant pressure \({{\rm{P}}_{\rm{2}}}{\rm{.}}\)

According to Charles law

\(\frac{{\rm{\nu }}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\)

\({{\rm{V}}_{\rm{2}}}{\rm{ = \nu  \times }}\frac{{{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{\;\;\;\;\;\;\;\;\; –  –  –  –  > }}\left( {\rm{2}} \right)\)

On substituting the value of v from Eq. \((1)\) to Eq. \((2),\) we get

\({{\rm{V}}_{\rm{2}}}{\rm{ = }}\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{P}}_{\rm{2}}}}}{\rm{ \times }}\frac{{{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}}}\)

\(\frac{{{{\rm{P}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{P}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{\;\;\;\;\; –  –  –  –  –  > }}\left( {\rm{3}} \right)\)

Equation \((3)\) is the most convenient form of the ideal gas equation for the purpose of calculation when the five variables are given, and the \({{\rm{6}}^{{\rm{th}}}}\) is to be calculated. This equation is also known as the combined gas law.

Numerical Problems On Charles Law Formula

1. A sample of gas occupies \({\rm{1}}{\rm{.50 L}}\) at \({\rm{25}}\,^\circ {\rm{C}}.\) If the temperature is raised to \({\rm{6}}{{\rm{0}}^{\rm{o}}}{\rm{C,}}\) what is the new volume of the gas if the pressure remains constant?
Ans: \({{\rm{V}}_{\rm{1}}}{\rm{ =  1}}{\rm{.50\, L}}\)

\({{\rm{T}}_{\rm{1}}}{\rm{ =  273  +  25  = 298\, K}}\)

\({{\rm{T}}_{\rm{2}}}{\rm{ =  273  +  60  =  333\, K}}\)

According to Charles Law

\(\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{\;}}\left( {{\rm{at\;constant\;pressure}}} \right)\)

\({{\rm{V}}_{\rm{2}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{\;}}\left( {{\rm{at\;constant\;pressure}}} \right)\)

\({{\rm{V}}_{\rm{2}}}{\rm{ = \;}}\frac{{{\rm{1}}{\rm{.50\; \times \;333}}}}{{{\rm{298}}}}\) \({{\rm{V}}_{\rm{2}}}{\rm{ = 1}}{\rm{.68\;L}}\)

2. A sample of helium gas has a volume of \({\rm{520 mL}}\) at \({\rm{100}}\,^\circ {\rm{C}}.\) Calculate the temperature at which the volume will become \({\rm{260 mL}}{\rm{.}}\) Assume that pressure is constant.
Ans: \({{\rm{V}}_{\rm{1}}}{\rm{ =  520\, mL}}\)

\({{\rm{V}}_{\rm{2}}}{\rm{ =  260\, mL}}\)

\({{\rm{T}}_{\rm{1}}}{\rm{ =  273  +  100  = 373\, K}}\)

\({{\rm{T}}_{\rm{2}}}{\rm{ =  ?}}\)

According to Charles Law

\(\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{\;}}\left( {{\rm{at\;constant\;pressure}}} \right)\)

\({{\rm{T}}_{\rm{2}}}{\rm{ = }}\frac{{{{\rm{V}}_{{\rm{2\;}}}}{{\rm{T}}_{\rm{1}}}}}{{{{\rm{V}}_{\rm{1}}}}}{\rm{\;}}\left( {{\rm{at\;constant\;pressure}}} \right)\)

\({{\rm{T}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{260\;mL\; \times \;373\;K}}}}{{{\rm{520\;mL}}}}\)

\({{\rm{T}}_{\rm{2}}}{\rm{ = 186}}{\rm{.5\;K}}\) Or \({\rm{t = 186}}.{\rm{5 – 273 = }}\;{\rm{ – 86}}.{\rm{5}}\,^\circ {\rm{C}}\)

3. At \({\rm{25}}\;^\circ {\rm{C}}\) and \({\rm{760\, mm}}\) of \({\rm{Hg}}\) pressure, a gas occupies \({\rm{600 mL}}\) volume. What will be its pressure at a height where the temperature is \({\rm{10}}{\,^{\rm{o}}}{\rm{C}}\) and volume of the gas is \({\rm{640\, mL}}{\rm{.}}\)
Ans: \({{\rm{p}}_{\rm{1}}}{\rm{ =  760\, mm}}\) of \({\rm{Hg}}\)

\({{\rm{p}}_{\rm{2}}}{\rm{ =  ?}}\)

\({{\rm{V}}_{\rm{1}}}{\rm{ =  600\, mL}}\)

\({{\rm{V}}_{\rm{2}}}{\rm{ =  640\, mL}}\)

\({{\rm{T}}_{\rm{1}}}{\rm{ =  273 + 25  =  298\, K}}\)

\({{\rm{T}}_{\rm{2}}}{\rm{ =  273 + 10  =  283\, K}}\)

According to combined gas law equation

\({\rm{\;}}\frac{{{{\rm{p}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{p}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}\)

\({{\rm{p}}_{\rm{2}}}{\rm{ = }}\frac{{{{\rm{p}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}{{\rm{T}}_{\rm{2}}}}}{{{{\rm{V}}_{\rm{2}}}{{\rm{T}}_{\rm{1}}}}}\)

\({{\rm{p}}_{\rm{2}}}{\rm{ = }}\frac{{{\rm{760}}\;{\rm{mm}}\;{\rm{of}}\;{\rm{Hg}}\;\; \times \;\;{\rm{600}}\;{\rm{mL}}\;\; \times \;\;{\rm{283}}\;{\rm{K}}}}{{{\rm{640}}\;{\rm{mL}}\;\; \times \;\;{\rm{298}}\;{\rm{K}}}}\,\,{{\rm{p}}_{\rm{2}}}{\rm{ = 676}}.{\rm{6}}\;{\rm{mm}}\;{\rm{of}}\;{\rm{Hg}}\)

Summary

In the article, Charles law formula you have understood Boyle’s law, Charles law and combined gas law. You can derive mathematical form of Charles law, i.e., \({\rm{\;}}\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{\;}}\) and combined gas law, i.e., \({\rm{\;}}\frac{{{{\rm{p}}_{\rm{1}}}{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{p}}_{\rm{2}}}{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{.}}\) You can explain the practical application of Charles law. Using Charles law and combined gas equation, you can solve for unknown volume, temperature or pressure.

FAQs

1. How to use the Charles law formula?
Ans: Charles law is used to calculate the unknown volume or temperature of the reaction or process at constant pressure using the following equation, \(\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{\;}}\left( {{\rm{at\;constant\;pressure}}} \right)\)

2. How to solve the Charles law formula?
Ans: Charles Law states that the volume of a given mass of gas is directly proportional to the absolute or Kelvin temperature at constant pressure.
Let \({{\rm{V}}_{\rm{1}}}\) be the volume of a gas at temperature \({{\rm{T}}_{\rm{1}}}{\rm{.}}\) By keeping the pressure constant, if the temperature of the gas is increased to \({{\rm{T}}_{\rm{2}}}{\rm{,}}\) then the volume will also increase to \({{\rm{V}}_{\rm{2}}}{\rm{.}}\) Therefore, according to law,              
\(\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{\;}}\left( {{\rm{at\;constant\;mass\;and\;pressure}}} \right)\)

3. How to derive Charles law formula?
Ans: According to the mathematical form of Charles Law:
\({{\rm{V}}_{{\rm{t\;}}}}{\rm{ = }}{{\rm{V}}_{{\rm{0\;}}}}\left[ {\frac{{{\rm{273 + t}}}}{{{\rm{273}}}}} \right]\)
But \({\rm{273 + t  =  T}}\) \(\left( {\rm{K}} \right)\) on the Kelvin scale and \({\rm{273  =  T,}}\)Therefore, \({{\rm{V}}_{\rm{t}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{0}}}{\rm{T}}}}{{{{\rm{T}}_{\rm{0}}}}}\)
\(\frac{{{{\rm{V}}_{\rm{t}}}}}{{{{\rm{V}}_{\rm{0}}}}}{\rm{ = }}\frac{{\rm{T}}}{{{{\rm{T}}_{\rm{0}}}}}\)
Thus, \({\rm{\;}}\frac{{\rm{V}}}{{\rm{T}}}{\rm{ = Constant\;}}\left( {{\rm{at\;constant\;pressure}}} \right)\) \({\rm{V}} \propto {\rm{T\;}}\left( {{\rm{at\;constant\;pressure}}} \right)\)

4. What is Charles Law and Boyle’s law?
Ans: Charles Law states that “the volume of a given mass of gas is directly proportional to the absolute or Kelvin temperature at constant pressure”.
\({\rm{V}} \propto {\rm{T\;at\;constant\;pressure}}\)
Boyle’s law states that “The volume of a given mass of a gas is inversely proportional to its pressure provided the temperature remains constant”. \({\rm{V}} \propto \frac{{\rm{1}}}{{\rm{P}}}{\rm{,}}\) at constant temperature and for a given mass of the gas.

5. What is Charles Law? Derive the equation?
Ans: Charles Law states that “the volume of a given mass of gas is directly proportional to the absolute or Kelvin temperature at constant pressure”.
According to the mathematical form of Charles’s Law:
\({{\rm{V}}_{\rm{t}}}{\rm{ = }}{{\rm{V}}_{\rm{0}}}\left[ {\frac{{{\rm{273 + t}}}}{{{\rm{273}}}}} \right]\)
But \({\rm{273 + t  =  T }}\left( {\rm{K}} \right)\) on the Kelvin scale and \({\rm{273  =  T,}}\) Therefore, \({{\rm{V}}_{\rm{t}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{0}}}{\rm{T}}}}{{{{\rm{T}}_{\rm{0}}}}}\)
\(\frac{{{{\rm{V}}_{\rm{t}}}}}{{{{\rm{V}}_{\rm{0}}}}}{\rm{ = }}\frac{{\rm{T}}}{{{{\rm{T}}_{\rm{0}}}}}\)
Thus, \({\rm{\;}}\frac{{\rm{V}}}{{\rm{T}}}{\rm{ = Constant\;}}\left( {{\rm{at\;constant\;pressure}}} \right)\)              
\({\rm{V}} \propto {\rm{T\;}}\left( {{\rm{at\;constant\;pressure}}} \right)\)

6. How do you calculate Charles Law?
Ans: Charles Law can be calculated as follows,
Let \({{\rm{V}}_{\rm{1}}}\) be the volume of a gas at temperature \({{\rm{T}}_{\rm{1}}}{\rm{.}}\) By keeping the pressure constant, if the temperature of the gas is increased to \({{\rm{T}}_{\rm{2}}}{\rm{,}}\) then the volume will also increase to \({{\rm{V}}_{\rm{2}}}{\rm{.}}\) Therefore, according to law,              
\(\frac{{{{\rm{V}}_{\rm{1}}}}}{{{{\rm{T}}_{\rm{1}}}}}{\rm{ = }}\frac{{{{\rm{V}}_{\rm{2}}}}}{{{{\rm{T}}_{\rm{2}}}}}{\rm{\;}}\left( {{\rm{at\;constant\;mass\;and\;pressure}}} \right)\)

7. What is Charles’ law state?
Ans: Charles Law states that “the volume of a given mass of gas is directly proportional to the absolute or Kelvin temperature at constant pressure”.              
\({\rm{V}} \propto {\rm{T\;\;\;\;}}\left( {{\rm{at\;constant\;pressure}}} \right)\)

8. What is Boyle’s law?
Ans: Boyle’s law states that “The volume of a given mass of a gas is inversely proportional to its pressure provided the temperature remains constant”.
\({\rm{V}} \propto \frac{{\rm{1}}}{{\rm{P}}},\) at constant temperature and for a given mass of the gas.

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