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Gay-Lussac’s Law states that the pressure of a given mass of gas fluctuates clearly with the temperature of the gas when the volume is kept constant. In Gay-Lussac’s law, the volume remains constant, whereas the pressure remains directly proportional to the temperature. The usual equations for Gay-Lussac’s law are P/T = constant or

PiTi = PfTf

where

Let Pi be the initial pressure  and Ti be the absolute temperatures

Let Pf be the final pressure and Tf be the initial temperature

Gay-Lussac Law

Gay-Lussac’s law says that a gas’s pressure fluctuates directly with temperature whenever the volume is constant. So if the temperature rises, the pressure rises with it. As the temperature rises, the kinetic energy of the gas particles also increases. Because of the increased energy, the molecules hit the container’s walls with more force, resulting in higher pressure.

Edmonton’s Law is another name for Gay Lussac’s Law. Edmonton demonstrated the same logic by inventing the thermometer with a reading for the current temperature.

History 

In 1802 a French scientist and physicist named Joseph Louis Gay-Lussac observed that as you hold an overall gas consistent and apply heat, the pressure of the gas will rise. This is due to the gases’ higher kinetic energy, which causes them to collide more forcefully with the container’s walls (resulting in greater pressure).

Tips – 

When addressing a Gay-law Lussac’s problem, keep the following considerations in mind:

  • The only constant is the volume of the gas.
  • Pressure is directly proportional to the rise in the temperature of the gas.
  • If the temperature drops, so do the pressure of the gas

The K.E. of gas particles is estimated by temperature. Particles move even more relaxed at a lower temperature and will undoubtedly hit the container’s wall—the speed of the particles increases, considering how the temperature rises. 

Gay-Lussac Law related Problem –

  • Deduce the pressure change when a steady volume of gas is at 1.00 atm and the temperature increases from 20.0 °C to 30.0 °C.

Solution:

Before solving, convert temperature into kelvin scale:

T1= 273+20=293 K

T2= 273+30=303 K

P1 / T1 = P2 / T2

1.00 atm / 293 = x / 303

x = 1.03 atm

  • A 30.0 L sample of nitrogen inside a metal compartment at 20.0 °C is put inside a broiler whose temperature is 50.0 °C. The pressure inside the compartment at 20.0 °C was 3.00 atm. What is the pressure of the nitrogen after its temperature is increased?

Step by step Solution of the given problem:

P1 / T1 = P2 / T2

3.00 atm / 293.0 K = x / 323.0 K

x = 3.31 atm

  • A certain gas at 3.00 x 103 mmHg inside a steel tank is cooled from 500.0 °C to 0.00 °C. What is the final pressure of the gas in the steel tank?

Step by step Solution of the given problem:

P1 / T1 = P2 / T2

P1T2 = P2T1

P2 = (P1T2) / T1

P2 = [(3000 mmHg) (273 K)] / 773 K

  • Calculate the last pressure inside a scuba tank after it cools from 1.00 x 103 °C to 25.0 °C. The underlying pressure in the tank is 130.0 atm.

Step by step Solution of the given problem:

P1 / T1 = P2 / T2

P1T2 = P2T1

P2 = (P1T2) / T1

P2 = [(130.0 atm) (298 K)] / 1273 K

P2=30.43 atm

  • Think about an optimal gas with a flat temperature of T1. To what outright temperature could the gas be warmed to twofold its tension? 

Step by step Solution of the given problem:

P1 / T1 = P2 / T2

We need the pressure to be twofold, so set P1 = 1 and P2 = 2. The units don’t make any difference since they are very similar: atm, kPa, mmHg, torr. It doesn’t make any difference which one you use.

We need to see what the temperature does, so set T1 to 1 K and T2 to x. The temperature matters; it should be in Kelvins.

Therefore:

1 / 1 = 2 / x

x = 2 K

  • The temperature of a certain gas in a steel compartment at 30.0 kPa is expanded from −100.0 °C to 1.00 x 103 °C. What is the final pressure inside the tank?

Solution:

P2 = (P1T2) / T1

P2 = [(30.0 kPa) (1273 K)] / 173 K

P2 = 220.75 atm

Conclusion

When mass and volume are kept consistent, Gay- Lussac’s law states that a gas’ pressure fluctuates with temperature. As the temperature increases, the pressure increases with it. 

The Gay-Lussac’s Law has various significances. In this article, we learned how to find the temperature and pressures of a gas system using this law.