**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: **P**_{1}**/T**_{1}** = P**_{2}**/T**_{2}

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, P_{1} = 2 atm

Initial temperature of the gas, T_{1} = 20 °C = 293 K

Final pressure of the gas, P_{2} = ?

Final temperature of the gas, T_{2} = 60 °C = 333 K

Using the formula of gay lussac’s law,

P_{1}/T_{1} = P_{2}/T_{2}

P_{2} = (P_{1} × T_{2}) ÷ T_{1}

P_{2} = (2 × 333) ÷ 293

P_{2} = 666 ÷ 293

P_{2} = 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, P_{1} = 5 atm

Final temperature of the gas, T_{2} = 90 °C = 363 K

Final pressure of the gas, P_{2} = 6 atm

Initial temperature of the gas, T_{1} = ?

Using the formula of gay lussac’s law,

P_{1}/T_{1} = P_{2}/T_{2}

T_{1} = (P_{1} × T_{2}) ÷ P_{2}

T_{1} = (5 × 363) ÷ 6

T_{1} = 1815 ÷ 6

T_{1} = 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, T_{1} = 25 °C = 298 K

Final temperature of the gas, T_{2} = 80 °C = 353 K

Final pressure of the gas, P_{2} = 12 atm

Initial pressure of the gas, P_{1} = ?

Using the formula of gay lussac’s law,

P_{1}/T_{1} = P_{2}/T_{2}

P_{1} = (P_{2} × T_{1}) ÷ T_{2}

P_{1} = (12 × 298) ÷ 353

P_{1} = 3576 ÷ 353

P_{1} = 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, P_{1} = 3 atm

Initial temperature of the gas, T_{1} = 30 °C = 303 K

Final pressure of the gas, P_{2} = 4 atm

Final temperature of the gas, T_{2} = ?

Using the formula of gay lussac’s law,

P_{1}/T_{1} = P_{2}/T_{2}

T_{2} = (P_{2} × T_{1}) ÷ P_{1}

T_{2} = (4 × 303) ÷ 3

T_{2} = 1212 ÷ 3

T_{2} = 404 K = 131 °C

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

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**Related:**

- Boyle’s Law Formula
- Charles Law Formula
- Avogadro’s Law Formula
- Ideal Gas Law Formula
- Dalton’s Law Formula
- Henry’s Law Equation
- Combined Gas Law Formula

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