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.

$P\propto Tm$ 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,

$P_t=P_0\left(1+\frac{t}{273.15}\right)=P_0\left(\frac{273.15+t}{273.15}\right)$

$P_t=P_0\frac{T}{T_0}$

$T_0(K)=273.15\text{and}T(K)=273.15+t$

$\frac{P_t}{P_0}=\frac{T}{T_0}$

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

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.

$P=\frac{nRT}{v}\Rightarrow P\propto T$

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:

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

$P_{1}and{P}_2$ are the initial and final pressures,

$T_{1}and{T}_2$ are the initial and final temperatures (in Kelvin).

5.0Experiment 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

1. 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

$\frac{P_2}{P_1}=\frac{T_2}{T_1}\Rightarrow P_2=\frac{T_1}{T_2}\times P_1=\frac{400\times 150}{300}=200N/m^2$

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:

$P^{\prime }=P+\frac{0.4}{100}P,T^{\prime }=T+1$

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

Q-5.Find relation between $V_{1}and{V}_2$ for a given curve.

Solution:

From Gay–Lussac's Law, $V_1<V_2$