Gay-Lassac's Law states that at **constant volume** the pressure of a gas is proportional to its absolute temperature specified in kelvins. For example, if you double the temperature of a gas, we double its pressure and vice versa.

**Gay Lussac's Law Formula** can be expressed as

Plot of P Vs T at constant volume is a straight line passing through origin. Such a plot is known as

Related Calculators | |

Gay Lussac's Law Calculator | Beer Lambert Law Calculator |

Boyle's Law Calculator | Charles Law Calculator |

Solved problems based on Gay-lussac's law are given below. ### Solved Examples

**Question 1: **The pressure of a container of helium is 650 torr. If the sealed container is cooled to 0^{o}C, what will the pressure be?

** Solution: **

Since the container is sealed, the volume does not change. Remember that temperature must be in kelvins.

P_{1} = 650 torr; T_{1} = 25^{o}C = 298K; T_{2} = 0^{o}C = 273K

Solving for P_{2}Since the problem is concerned only with temperature and pressure.

Gay-Lussac's law can be used

P_{2} = $\frac{[650\ torr][273K]}{298K}$ = 595 torr

**Question 2: **If we have a tank of gas at 1.520 torr pressure and a temperature of 300K and it is heated to 500K. What is the new pressure in atmospheres?

** Solution: **

Convert the initial pressure to atmospheres.

$\frac{1520\ torr}{1}$ $\times$ $\frac{1\ atm}{760\ torr}$ = 2.00atm

Solve for P_{2} in Gay-Lussac's law

P_{2} = $\frac{P_{1}T_{2}}{T_{1}}$

Substitute the values

P_{2} = $\frac{[2.00\ atm][500K]}{300K}$ = 3.33atm.

Since the container is sealed, the volume does not change. Remember that temperature must be in kelvins.

P

Solving for P

Gay-Lussac's law can be used

P

Convert the initial pressure to atmospheres.

$\frac{1520\ torr}{1}$ $\times$ $\frac{1\ atm}{760\ torr}$ = 2.00atm

Solve for P

P

Substitute the values

P