Gay-Lussac's Law

03rd Apr 2019 @ 8 min read

Gay-Lussac's law is also known as pressure law or Amontons's law. The law correlates how the pressure of a gas increases with an increase in temperature. This law is named after French chemist Joseph Louis Gay-Lussac. He formulated this relationship in 1808. Gay-Lussac's law is similar to Charles's law, the only difference is the “volume” term in Charles's law is interchanged by the “pressure” term in Gay-Lussac's law.

Statement

For a fixed amount of an enclosed ideal gas, its pressure is directly proportional to its absolute temperature at a constant volume.

$P=\text{constant} \times T$

Explanation

As the law states: the pressure and temperature of an ideal gas are directly proportional to each other at a constant volume and for a given mass of a gas. The statement can mathematically be expressed as:

$P \propto T$

By removing the proportionality, we get,

$P=k \times T$

where k is a constant of proportionality.

The above expression can be rearranged as:

$\frac{P}{T}=k$

The above expression is valid for a given mass of a gas and at a constant volume. Hence as the temperature increases, the pressure of the gas also increases and as the temperature decreases, the pressure decreases. It is clear from the above equation that the ratio of the pressure to temperature is independent of the pressure or temperature at a constant volume.

Note: In the above expression temperature should be in kelvin.

For a given amount ideal gas, P₁, T₁ and P₂, T₂ are the pressures and temperatures of condition 1 and condition 2 at a constant volume. From Gay-Lussac's law,

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$

If the pressure of a gas obeying Gay-Lussac's law is double, then the temperature also gets double which is explained below.

$P_2=2 P_1$ $\frac{P_1}{T_1} = \frac{P_2}{T_2} = \frac{2 P_1}{T_2}$ $T_2 =2 T_1$

Gay-Lussac's law experiment — Experimental apparatus of Gay-Lussac's law

The above diagram is a typical experimental setup required for verification of Gay-Lussac's law. When the air inside the chamber is heated with a heating source (electric heater is used in the diagram) via water, the temperature of the air increases i.e. the kinetic energy associated with the molecules of the air increases. Thus, molecules of the air exert more force in the outwards direction on all the walls of the container. Since the boundaries are rigid, the entrapped air inside the chamber is unable to expand. The number of molecular collisions increases and the pressure of the air also increases. By the same analogy, when the air is cooled, the pressure of the air decreases.

Note: in Charles's law, at least one of the boundaries of the vessel is movable. Thus, with the rise of temperature, the air inside the vessel expands. This expansion is what makes the pressure constant in Charles' law. While in Gay-Lussac's law the volume remains constant and the pressure varies.

Graphical Representation

The graphical representation of Gay-Lussac's law is demonstrated in the graphs below.

Gay-Lussac's law graph — Pressure is linearly proportional to temperature. The graph is a straight line with positive slope and passes through the origin.

From the above graph, the pressure increases linearly with an increase in the temperature. In the above figure, the temperature is taken in absolute scale. It can also be observed that as the temperature approaches zero, the pressure also approaches zero. All the above four lines are plotted at a constant volume (isochore).

As mentioned earlier, the ratio of the pressure to temperature is independent of temperature (or pressure) at a constant volume. This can be observed from the above graph.

Limitations of Gay-Lussac's law

The limitations are as follows:

The law is only applicable to ideal gases.
Gay-Lussac's law holds good for real gases at high temperatures and/or low pressure.
The ratio of the pressure to temperature deviated at high pressures. The ratio decreases with increasing the pressure. This decrease is due to an increase in volume at high pressures which is explained by an increase in repulsive forces among the molecules at high pressures.

Real World Applications of Gay-Lussac's Law

There are numerous applications of Gay-Lussac's law can be observed in day to day life. Below are some of the mentioned:

Pressure cooker

When a pressure cooker is kept on a heating source (stove). As per Gay-Lussac's law, the pressure of the fluid in the cooker increases with the rising of the temperature.

Aerosol can

An aerosol can on exposure to the sunlight or high temperatures, might bursts. The reason for bursting is increased in the pressure inside the can beyond the threshold limit. The pressure increases on the account of the rise of the temperature inside the can.

Gun bullet

When the bullet from a gun is ignited, the chemical energy stored in the shell of the bullet is converted into heat by chemical reactions. This heat increases the temperature which as per Gay-Lussac's law increases the pressure. Because of the high pressure, the bullet is fired from the gun.

Automobile tyres

The rupture of automobile tyres on subjection to high temperature is a classic example of Gay-Lussac's law. The high temperature pressurizes the air inside the tyres and beyond a point, they explode.

Gay-Lussac's law along with Boyles' law, Charles's law and Avogadro's law form ideal gas law.

Examples

Example 1

Consider a fixed amount of butane gas at temperature 65 °C at pressure 1.2 × 10⁴ N m⁻² undergoes a change from stage 1 to stage 2 such that the initial and final volumes remain the same. The new temperature of the gas is 200 °C. Calculate the new pressure?

First, convert both the temperatures in the problem statement from celsius to kelvin.

$T_1 =65+273.15 =338.15 \, \text{K}$ $T_2 =200+273.15 =473.15 \, \text{K}$

As from Gay-Lussac's law at constant volume and for a given mass of gas,

$\frac{P_1}{T_1} = \frac{P_2}{T_2}$ $P_2 &= \frac{P_1}{T_1} \times T_2\\&= \frac{1.4 \times 10^4}{338.15} \times 473.15\\&= 1.9 \times 10^4 \, \text{N}\,\text{m}^{-2}$

Therefore, the new pressure is 1.9 × 10⁴ N m⁻².

Example 2

The pressure of water vapour in a pressure cooker is 112 700 N m⁻². Estimate the boiling point of water in the pressure cooker. Assume the vapour behave ideally and obeys Gay-Lussac's law.

We all know that the boiling point of water at 1 atmospheric pressure is 100 °C. And 1 atmosphere is equivalent to 101 325 N m⁻². Thus, applying Gay-Lussac's law,

$\frac{P_1}{T_1}= \frac{P_2}{T_2}$ $T_2 &= \frac{P_2}{P_1} \times T_1 = \frac{112\,700}{101\,325} \times (100+273.15) \\&=\frac{112\,700}{101\,325} \times 373.15= 415.04 \, \text{K}\approx 142 \, \degree \text{C}$

The boiling point of the water inside the pressure cooker is 142 °C.

Example 3

A soda bottle at the room the temperature of 25 °C and the pressure of 2 atm is heated to the temperature of 330 °C at which it bursts. Calculate the pressure of the heated soda bottle. Converting the initial and final temperatures are from degree celsius to kelvin.

$T_1 =25+273.15 =298.15 \, \text{K}$ $T_1 =330+273.15 =603.15 \, \text{K}$

From the Gay-Lussac's law,

$\frac{P_1}{T_1} =\frac{P_2}{T_2}$ $P_2 = \frac{P_1}{T_1} \times T_2&= \frac{2}{298.15} \times 603.15 \\&\approx 4.05 \, \text{atm}$

The soda bottle bursts at the pressure of 4.05 atm.

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