### What is Charles Law?

Charles’ law is one of the gas laws which explains the relationship between volume and temperature of a gas. It states that when pressure is held constant, the volume of a fixed amount of dry gas is directly proportional to its absolute temperature. When two measurements are in direct proportion then any change made in one of them affects the other through direct variation. Charles’ Law is expressed by the equation:

**\(V\alpha T\)**

**Or**

** \(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}\)**

Where,

V1 and V2 are the Initial Volume and Final Volume respectively. T1 refers to the Initial Temperature and T2 refers to the Final Temperature. Both the temperatures are in the units of Kelvin.

Jacques Charles, a French scientist, in 1787, discovered that keeping the pressure constant, the volume of a gas varies on changing its temperature. Later, Joseph Gay-Lussac, in 1802, modified and generalized the concept as Charles’s law. At very high temperatures and low pressures, gases obey Charles’ law.

**Derivation:**

Charles’ Law states that at a constant pressure, the volume of a fixed mass of a dry gas is directly proportional to its absolute temperature. We can represent this using the following equation:

**\(V\alpha T\)**

Since V and T vary directly, we can equate them by making use of a constant k.

\(\frac{V}{T}=constant=k\)

The value of k depends on the pressure of the gas, the amount of the gas and also on the unit of the volume.

VT = k ———– (I)

Let V1 and T1 be the initial volume and temperature of an ideal gas. We can write equation I as:

\(\frac{V_{1}}{T_{1}}=k\) ———– (II)

Let’s change the temperature of the gas to T2. Consequently, its volume changes to V2. So we can write,

\(\frac{V_{2}}{T_{2}}=k\) ———– (III)

Equating equations (II) and (III),

\(\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}=k\)

Hence, we can generalize the formula and write it as:

\(\frac{(V_{1})}{(T_{1})}=\frac{(V_{2})}{(T_{2})}\)

Or

**\(V_{1}T_{2}=V_{2}T_{1}\)**

You know that on heating up a fixed mass of gas, that is, increasing the temperature, the volume also increases. Similarly, on cooling, the volume of the gas decreases. At 0 degree centigrade, the volume of the gas increases 1/273 of its original volume for a unit degree increase in temperature.

**It is to be noted here that the unit Kelvin is preferred for solving problems related to Charles’ Law, and not Celsius. **Kelvin (T) is also known as the Absolute temperature scale. For converting a temperature to Kelvin scale, you add 273 to the temperature in the centigrade/Celsius scale.

**Charles’ Law in Real Life:**

Charles’ law has a wide range of applications in our daily life. Some of the common examples are given below:

- In cold weather or environment, balls and helium balloons shrink.

**In bright sunlight, the inner tubes swell up.**

- In colder weather, the human lung capacity will also decrease. This makes it more difficult to do jogging or athletes to perform on a freezing winter day.

**Solved Examples:**

**Question 1: **** **A gas occupies a volume of 400cm3 at 0oC and 780 mm Hg. What volume (in litres) will it occupy at 80oC and 780 mm Hg?

**Solution: **Given,

V1= 400 cm³

V2 =?

T1= 0oC = 0+273 = 273 K

T2= 80oC = 80+273 = 353 K

Here the pressure is constant and only the temperature is changed.

Using Charles Law,

\(\frac{(V_{1})}{(T_{1})}=\frac{(V_{2})}{(T_{2})}\)

\(\frac{400}{273}=\frac{V_{2}}{353}\)

\(V_{2}=\frac{400\times 353}{273}\)

\(V_{2}=517.21cm^{3}\)

1 cubic centimeter = 0.001 litre =1 x 10-3 litre

∴ 517.21 cubic centimeter = 517.21 x 10-3 = 0.517 litres

**Question 2: **** **Find the initial volume of a gas at 150 K, if the final volume is 6 L at 100 K

**Solution: **Given,

V1=?

V2 =6 L

T1= 150 K

T2= 100 K

Using Charles Law,

\(\frac{(V_{1})}{(T_{1})}=\frac{(V_{2})}{(T_{2})}\)

\(\frac{(V_{1})}{(150)}=\frac{(6)}{(100)}\)

\(V_{1}=\frac{6\times 150}{100}\)

\(V_{1}=9L\)

The initial volume of a gas at 150 K is 9 litres.

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