Gay-Lussac’s Law

Propane tanks are extensively used in the kitchen. It’s not enjoyable, however, to discover you’ve run out of gas halfway through a meal. On a hot day, gauges are used to measure the pressure inside gas tanks that read greater than on a cool day. When deciding whether or not to replace the tank before your next cookout, keep the air temperature in mind. In this article, we’ll go over Gay Lussac’s Law in detail, including its formula and derivation.

Table of Contents

• What is Gay-Lussac’s Law?
• Recommended Videos
• Formula and Derivation
• Examples of Gay-Lussac’s Law
• Solved Exercises on Gay-Lussac’s Law
• Frequently Asked Questions – FAQs

What is Gay-Lussac’s Law?

Gay-Lussac’s law is a gas law which states that the pressure exerted by a gas (of a given mass and kept at a constant volume) varies directly with the absolute temperature of the gas. In other words, the pressure exerted by a gas is proportional to the temperature of the gas when the mass is fixed and the volume is constant.

This law was formulated by the French chemist Joseph Gay-Lussac in the year 1808. The mathematical expression of Gay-Lussac’s law can be written as follows:

P ∝ T ; P/T = k

Where:

• P is the pressure exerted by the gas
• T is the absolute temperature of the gas
• k is a constant.

The relationship between the pressure and absolute temperature of a given mass of gas (at constant volume) can be illustrated graphically as follows.

From the graph, it can be understood that the pressure of a gas (kept at constant volume) reduces constantly as it is cooled until the gas eventually undergoes condensation and becomes a liquid.

Recommended Videos

Formula and Derivation

Gay-Lussac’s law implies that the ratio of the initial pressure and temperature is equal to the ratio of the final pressure and temperature for a gas of a fixed mass kept at a constant volume. This formula can be expressed as follows:

(P₁/T₁) = (P₂/T₂)

Where:

• P₁ is the initial pressure
• T₁ is the initial temperature
• P₂ is the final pressure
• T₂ is the final temperature

This expression can be derived from the pressure-temperature proportionality for gas. Since P ∝ T for gases of fixed mass kept at constant volume:

P₁/T₁ = k (initial pressure/ initial temperature = constant)

P₂/T₂ = k (final pressure/ final temperature = constant)

Therefore, P₁/T₁ = P₂/T₂ = k

Or, P₁T₂ = P₂T₁

Examples of Gay-Lussac’s Law

When a pressurized aerosol can (such as a deodorant can or a spray-paint can) is heated, the resulting increase in the pressure exerted by the gases on the container (owing to Gay-Lussac’s law) can result in an explosion. This is the reason why many pressurized containers have warning labels stating that the container must be kept away from fire and stored in a cool environment.

An illustration describing the increase in pressure which accompanies an increase in the absolute temperature of a gas kept at a constant volume is provided above. Another example of Gay-Lussac’s law can be observed in pressure cookers. When the cooker is heated, the pressure exerted by the steam inside the container increases. The high temperature and pressure inside the container cause the food to cook faster.

Solved Exercises on Gay-Lussac’s Law

Exercise 1

The pressure of a gas in a cylinder when it is heated to a temperature of 250K is 1.5 atm. What was the initial temperature of the gas if its initial pressure was 1 atm?

Given,

Initial pressure, P₁ = 1 atm

Final pressure, P₂ = 1.5 atm

Final temperature, T₂ = 250 K

As per Gay-Lussac’s Law, P₁T₂ = P₂T₁

Therefore, T₁ = (P₁T₂)/P₂ = (1*250)/(1.5) = 166.66 Kelvin.

Exercise 2

At a temperature of 300 K, the pressure of the gas in a deodorant can is 3 atm. Calculate the pressure of the gas when it is heated to 900 K.

Initial pressure, P₁ = 3 atm

Initial temperature, T₁ = 300K

Final temperature, T₂ = 900 K

Therefore, final pressure (P₂) = (P₁T₂)/T₁ = (3 atm*900K)/300K = 9 atm.

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