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Gay Lussac's Law Formula

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

The mathematical expression at constant volume and fixed number of moles is given by

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

Gay Lussac's Law Formula of pressure and temperature is equal to

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Gay-Lussacs Law Problems

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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 0oC, what will the pressure be?
Since the container is sealed, the volume does not change. Remember that temperature must be in kelvins.

P1 = 650 torr; T1 = 25oC = 298K; T2 = 0oC = 273K

Solving for P2
Since the problem is concerned only with temperature and pressure.
Gay-Lussac's law can be used
P2 = $\frac{[650\ torr][273K]}{298K}$ = 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?
Convert the initial pressure to atmospheres.

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

Solve for P2 in Gay-Lussac's law

P2 = $\frac{P_{1}T_{2}}{T_{1}}$

Substitute the values

P2 = $\frac{[2.00\ atm][500K]}{300K}$ = 3.33atm.

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