What Is The Formula For Gay-Lussac’s Law?Gay-Lussac's Law states that the pressure of a fixed amount of gas at fixed volume is directly proportional to its temperature in kelvins.

Gay-Lussac’s Law is a special case of the ideal gas law. This law only applies to ideal gases held at a constant volume allowing only the pressure and temperature to change.Answer: Gay-Lussac’s Law is expressed as:

Pi/Ti = Pf/Tf

wherePi = initial pressureTi = initial absolute temperaturePf = final pressureTf = final absolute temperature

Guy-Lussac's Gas Law ExampleIdeal Gas Law Example Problem

It is extremely important to remember the temperatures are absolute temperatures measured in Kelvin, NOT °C or °F.Guy-Lussac's gas law is a special case of the ideal gas law where the volume of the gas is held constant. When the volume is held constant, the pressure exerted by a gas is directly proportional to the absolute temperature of the gas. This example problem uses Guy-Lussac's law to find the pressure of a heated container.

Problem:

A 20 L cylinder containing 6 atm of gas at 27 °C. What would the pressure of the gas be if the gas was heated to 77 °C?

Solution

The cylinder's volume remains unchanged while the gas is heated so Gay-Lussac's gas law applies. Gay-Lussac's gas law can be expressed as

Pi/Ti = Pf/Tf

wherePi and Ti are the initial pressure and absolute temperaturesPf and Tf are the final pressure and absolute temperature

First, convert the temperatures to absolute temperatures.

Ti = 27 °C = 27 + 273 K = 300 KTf = 77 °C = 77 + 273 K = 350 K

Use these values in Gay-Lussac's equation and solve for Pf.

Pf = PiTf/TiPf = (6 atm)x(350K)/(300 K)Pf = 7 atm

Ideal Gas Example Problem - Constant VolumeWorked Chemistry Problems

QuestionThe temperature of a sample of an ideal gas confined in a 2.0 L container was raised from 27 °C to 77 °C. If the initial pressure of the gas was 1200 mm Hg, what was the final pressure of the gas?

Solution

Step 1

Convert temperatures from Celsius to Kelvin

K = °C + 273

Initial temperature (Ti): 27 °C

K = 27 + 273K = 300 KelvinTi = 300 K

Final temperature (Tf): 77 °C

K = 77 + 273K = 350 KelvinTf = 350 K

Step 2

Using the ideal gas relationship for constant volume, solve for the final pressure (Pf)

Pi / Ti = Pf / Tf

solve for Pf:

Pf = (Pi x Tf) / TiPf = (1200 mm Hg x 350 K) / 300 K

Pf = 420000 / 300Pf = 1400 mm Hg

AnswerThe final pressure of the gas is 1400 mm Hg.

Gay-Lussac's Law: Gay-Lussac's Law states that the pressure of a fixed amount of gas at fixed volume is directly proportional to its temperature in kelvins.

Gay-Lussac's Law Formula :Gas Equation: Pi/Ti = Pf / TfInitial Pressure(Pi) = PfTi / TfInitial Temperature(Ti) = PiTf / PfFinal Pressure(Pf) = PiTf / TiFinal Temperature(Tf) = PfTi / Piwhere,Pi = Initial Pressure, Ti = Initial Temperature, Pf = Final Pressure. Tf = Final Temperature,

Gay-Lussac's Law Example:Case 1: A cylinder contain a gas which has a pressure of 125kPa at a temperature of 200 K. Find the temperature of the gas which has a pressure of 100 kPa. Pi = 125 kPa, Ti = 200 K, Pf = 100 kPa Step 1: Substitute the values in the below final temperature equation: Final Temperature(Tf) = PfTi / Pi = (100 x 200) / 100 = 20000 / 125 Final Temperature(Tf) = 160 KThis example will guide you to calculate the final temperature manually.

Case 2: Find the final pressure of gas at 150 K, if the pressure of gas is 210 kPa at 120 K. Pi = 210 kPa, Ti = 120 K, Tf = 150 K Step 1: Substitute the values in the below pressure equation: Final Pressure(Pf) = PiTf / Ti = (210 x 150) / 120 = 31500 / 120 Final Pressure(Vf) = 262.5 kPaThis example will guide you to calculate the pressure manually.

P1               P2

------     =     ------T1               T2

P1               P2------     =     ------

T1               T2The above formula is Gay-Lussac's Law named after the French chemist and physicist Joseph Louis Gay-Lussac (1778 - 1850). The law states that the pressure of a fixed mass of gas at a constant volume is directly proportional to its absolute temperature.

In other words, when temperature increases, pressure increases.When pressure decreases, temperature decreases.