Charles’ Law: Relationship Between Temperature and Volume 

1.0What is Charles’ Law?

Charles’ Law states that at constant pressure, the volume of a given mass of a gas is directly proportional to its absolute temperature (measured in Kelvin).

In simpler words, if you heat a gas, its volume increases, provided the pressure remains constant. Conversely, if the temperature decreases, the volume of the gas decreases.

2.0Mathematical Expression of Charles’ Law

The law can be stated as:

“At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature (in Kelvin).”

Mathematically:

$V\propto T(atconstantpressure)or\frac{V}{T}=constant$

If:

• ​$V_1$ = Initial volume at temperature $T_1$
• $V_2$ = Final volume at temperature $T_2$

Then, Charles’ Law formula is:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

This formula helps calculate unknown volumes or temperatures in gas-related problems, making it important for JEE Chemistry numerical problems.

3.0Explanation of the Temperature-Volume Relationship

How does Charles’ Law work?

• When the temperature of a gas increases (at constant pressure), the kinetic energy of gas molecules increases.
• The molecules move faster and strike the walls of the container more frequently and forcefully, causing the volume to expand if the pressure is kept constant.
• Conversely, when the temperature decreases, the volume contracts.

Key Takeaway:
Direct Proportionality: If the temperature doubles (in Kelvin), the volume also doubles, provided the pressure remains unchanged.

Practical Example:
If a balloon is filled with air at room temperature and then warmed, it expands as the temperature rises, following Charles’s Law.

4.0Graphical Representation of Charles’ Law

The relationship between temperature and volume as per Charles’ Law, can be visualized using a graph:

• X-axis: Temperature (in Kelvin)
• Y-axis: Volume

The graph is a straight line passing through the origin, illustrating direct proportionality.

If the temperature is expressed in Celsius, the line intercepts the temperature axis at ($-273.15^{\circ }C$), which is absolute zero—theoretically, the temperature at which the volume of a gas becomes zero.

5.0Real-Life Applications of Charles’ Law

• Hot Air Balloons: As the air inside the balloon is heated, its volume increases, causing the balloon to rise.
• Automobile Tires: Tire volume changes with temperature; in winter, tires appear flatter due to lower temperature, reducing the gas volume.
• Breathing: Expansion and contraction of lung volume with temperature changes during respiration.
• Aerosol Cans: Gas inside expands on heating, sometimes causing cans to burst if overheated.

6.0Charles’ Law in Daily Life and Industry

• Refrigeration Systems: Expansion of gases absorbs heat, cooling the refrigerator.
• Meteorology: Understanding how air expands and contracts with temperature changes helps in weather prediction.
• Industries: Used in designing equipment where gases are stored or processed at varying temperatures.

7.0Limitations of Charles’ Law

• Non-Ideal Behavior: At high pressures and low temperatures, real gases deviate from ideal behavior, and Charles’ Law may not hold precisely.
• Condensation: If the temperature drops close to a gas’s condensation point, the gas may liquefy, making Charles’ Law inapplicable.
• Chemical Reactivity: If the gas reacts chemically at certain temperatures, the law may not apply.

8.0Relationship Between Temperature and Volume

Charles' Law relationship explains:

• If the temperature increases, gas particles gain kinetic energy and move faster, colliding with container walls with greater force, thereby expanding the volume.
• If the temperature decreases, the kinetic energy of the gas particles decreases, they slow down, and the gas occupies less volume.

9.0Practice Problems

Problem 1: Calculating Final Volume

A sample of gas occupies 3.0 L at 27°C. If the temperature increases to 87°C at constant pressure, what will be the new volume of the gas?

Solution:
Convert temperatures to Kelvin:
( T1 = 27°C + 273 = 300,K )
( T2 = 87°C + 273 = 360,K )

Apply Charles’ Law:

$\frac{V1}{T1}=\frac{V2}{T2}][\frac{3.0}{300}=\frac{V2}{360}][V2=\frac{3.0\times 360}{300}=3.6,L$

Problem 2: Determining Temperature for a Volume Change

A gas has a volume of 500 mL at 400 K. To what temperature must it be cooled for the volume to become 350 mL at constant pressure?

Solution:

$\frac{500}{400}=\frac{350}{T2}][T2=\frac{350\times 400}{500}=280,K$

Problem 3: Real-Life Scenario

A balloon has a volume of 2.5 L at 15°C. If taken outdoors at -5°C, what will be its new volume? (Assume constant pressure.)

Solution:
( T1 = 15°C + 273 = 288,K )
( T2 = -5°C + 273 = 268,K )

$\frac{2.5}{288}=\frac{V2}{268}][V2=\frac{2.5\times 268}{288}=2.33,L$

Problem 4: Find the Initial Temperature

A gas has a volume of 1.2 L at a certain temperature. When heated to 350 K, its volume becomes 1.5 L. What was the initial temperature?

Solution:

$\frac{1.2}{T1}=\frac{1.5}{350}][T1=\frac{1.2\times 350}{1.5}=280,K$