Gay-Lussac’s Law

Gay-Lussac's Law, established by French scientist Joseph Louis Gay-Lussac in 1802, defines the relationship between the pressure and temperature of a gas when volume is held constant. The law states that the pressure of a fixed amount of gas is directly related to its absolute temperature (measured in Kelvin), as long as the volume does not change.

1.0Statement of Gay-Lussac’s Law

According to this law, for a fixed amount of an ideal gas at constant volume, the pressure is instantly related to its absolute temperature.

$P \propto T m$ and V are Constant

$\frac{P _{1}}{P _{2}} = \frac{T _{1}}{T _{2}}$

This law gives between pressure and temperature of a gas.It states that if the volume remains unchanged, the pressure of a given mass of a gas elevates or decreases by $\frac{1}{273.15}$ of its pressure at 0°C for every 1°C rise or fall of temperature.
If $P_{0}$ and $P_{t}$ are the pressure of a provided mass of gas at 0°C and t°C respectively,then according to Gay Lussac’s Law,

$P_{t} = P_{0} (1 + \frac{t}{273.15}) = P_{0} (\frac{273.15 + t}{273.15})$

$P_{t} = P_{0} \frac{T}{T _{0}}$

$T_{0} (K) = 273.15 and T (K) = 273.15 + t$

$\frac{P _{t}}{P _{0}} = \frac{T}{T _{0}}$

$\frac{P}{T} = Constant or P \propto T$

Volume remains unchanged, the pressure of a given mass of a gas is directly related to its absolute temperature.

2.0Derivation of Gay-Lussac’s Law

According to ideal gas equation,

PV=nRT

In Gay-Lussac's Law, both the volume (V)and the number of moles (n) are held constant.

$P = \frac{n RT}{v} \Rightarrow P \propto T$

This implies that pressure is directly proportional to temperature when volume and the number of moles are constant.

3.0Molecular Explanation

Gay-Lussac's Law states that, at unchanged volume, the pressure of a gas is directly related to its absolute temperature. This relationship is explained by the Kinetic Molecular Theory as follows:

Temperature and Kinetic Energy: Raising the temperature of a gas elevates the average kinetic energy of its molecules.
Molecular Motion: Higher kinetic energy leads to faster-moving molecules.
Collision Frequency and Force: Faster molecules collide with the walls of the container more often and with greater force.
Pressure Increase: These more frequent and forceful collisions result in an increase in pressure.
According to the Kinetic Molecular Theory, raising the temperature of a gas at unchanged volume leads to increased molecular motion, which in turn increases the pressure exerted by the gas.

By using Kinetic Theory of Gases

$P = \frac{1}{3} \frac{M}{V} \overline{v^{2}}$

For a given mass and constant volume V, $P \propto \overline{v^{2}}$

But $\overline{v^{2}} \propto T$ , Hence $P \propto T$ This proves Gay-Lussac’s Law and Regnault’s Law.

4.0Formula of Gay-Lussac’s Law

The relationship between pressure and temperature at unchanged volume can be written as:

$\frac{P _{1}}{T _{1}} = \frac{P _{2}}{T _{2}}$

$P_{1} an d P_{2}$ are the initial and final pressures,

$T_{1} an d T_{2}$ are the initial and final temperatures (in Kelvin).

5.0Experiment to Demonstrate Gay-Lussac’s Law

Experiment to Demonstrate Gay-Lussac’s Law

Take a gas in a sealed rigid container with a pressure gauge attached to it. The volume of the container should be constant.
Measure the initial temperature and pressure of the gas. Ensure the temperature is measured in Kelvin.
Gradually heat the gas and monitor the change in temperature and pressure.
Plot the data on a graph with pressure on the y-axis and temperature on the x-axis.
As the temperature of the gas increases, the pressure also increases. This confirms the direct proportionality between pressure and temperature.

6.0Graphs For Gay-Lussac’s Law

7.0Real-World Application of Gay-Lussac’s Law

Pressure cookers: The pressure inside a pressure cooker increases with temperature, speeding up cooking.
Tires of vehicles: The pressure inside the tires increases with temperature, which is why it’s important to check tire pressure when tires are cold.

8.0Solved Examples

A gas in a cylinder is initially at 27°C with a volume of 4 liters and a pressure of 100 N/m². The gas is then heated at unchanged volume, raising the temperature to 127°C. Calculate the resulting pressure.

Solution:

For Constant Volume

$\frac{P _{2}}{P _{1}} = \frac{T _{2}}{T _{1}} \Rightarrow P_{2} = \frac{T _{1}}{T _{2}} \times P_{1} = \frac{400 \times 150}{300} = 200 N / m^{2}$

A gas is contained in a closed container, and when the temperature is increased by 1°C, the pressure rises by 0.4%.Find the initial temperature of the gas?

Solution:

$P^{'} = P + \frac{0.4}{100} P, T^{'} = T + 1$

$\frac{P}{T} = \frac{P + \frac{0.4}{100} P}{T + 1} \Rightarrow T = 250, K$

Q-5.Find relation between $V_{1} an d V_{2}$ for a given curve.

Solution:

From Gay–Lussac's Law, $V_{1} < V_{2}$