Gay Lussac’s Law Formula | Problems

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Gay Lussac's Law Formula

Gay lussac’s law formula states that the pressure (P) of a given quantity of gas directly depends upon its temperature (T), which means that the pressure of the gas increases as its temperature increases. Here’s the formula of gay lussac’s law: P1/T1 = P2/T2

Let’s solve some problems based on this formula, so you’ll get a clear idea.

Gay Lussac’s Law Practice Problems

Problem 1: One gas cylinder is filled with a 2 atm pressure of hydrogen gas at 20 °C temperature. Calculate the final pressure of the gas, when the temperature of the gas cylinder is increased to 60 °C.

Solution:

Given data:
Initial pressure of the gas, P1 = 2 atm
Initial temperature of the gas, T1 = 20 °C = 293 K
Final pressure of the gas, P2 = ?
Final temperature of the gas, T2 = 60 °C = 333 K

Using the formula of gay lussac’s law,
P1/T1 = P2/T2
P2 = (P1 × T2) ÷ T1
P2 = (2 × 333) ÷ 293
P2 = 666 ÷ 293
P2 = 2.27 atm

Therefore, the final pressure of the gas is 2.27 atm.

Problem 2: A sample of 5 atm pressure of helium gas is filled inside a huge balloon. When a balloon is heated to 90 °C temperature, the pressure of the gas increases to 6 atm. Calculate the initial temperature of the gas.

Solution:

Given data:
Initial pressure of the gas, P1 = 5 atm
Final temperature of the gas, T2 = 90 °C = 363 K
Final pressure of the gas, P2 = 6 atm
Initial temperature of the gas, T1 = ?

Using the formula of gay lussac’s law,
P1/T1 = P2/T2
T1 = (P1 × T2) ÷ P2
T1 = (5 × 363) ÷ 6
T1 = 1815 ÷ 6
T1 = 302.5 K = 29.5 °C

Therefore, the initial temperature of the gas is 29.5 °C.

Problem 3: One cylinder filled with a lithium gas is heated at 25 °C temperature. When the temperature of the gas is increased to 80 °C, the pressure of the gas increases to 12 atm. What is the initial pressure of the gas?

Solution:

Given data:
Initial temperature of the gas, T1 = 25 °C = 298 K
Final temperature of the gas, T2 = 80 °C = 353 K
Final pressure of the gas, P2 = 12 atm
Initial pressure of the gas, P1 = ?

Using the formula of gay lussac’s law,
P1/T1 = P2/T2
P1 = (P2 × T1) ÷ T2
P1 = (12 × 298) ÷ 353
P1 = 3576 ÷ 353
P1 = 10.13 atm

Therefore, the initial pressure of the gas is 10.13 atm.

Problem 4: A 3 atm pressure of hydrogen gas is filled in a container at 30 °C temperature. If the pressure of the gas is increased to 4 atm, then calculate the final temperature of the gas.

Solution:

Given data:
Initial pressure of the gas, P1 = 3 atm
Initial temperature of the gas, T1 = 30 °C = 303 K
Final pressure of the gas, P2 = 4 atm
Final temperature of the gas, T2 = ?

Using the formula of gay lussac’s law,
P1/T1 = P2/T2
T2 = (P2 × T1) ÷ P1
T2 = (4 × 303) ÷ 3
T2 = 1212 ÷ 3
T2 = 404 K = 131 °C

Therefore, the final temperature of the gas is 131 °C.

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