**Combined gas law formula** states that the product of **pressure **(P) and **volume** (V) of a given quantity of gas divided by the **temperature **(T) of that gas is constant. Here’s the formula of combined gas law: **P**_{1}**V**_{1}**/T**_{1}** = P**_{2}**V**_{2}**/T**_{2}

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

## Combined Gas Law Practice Problems

**Problem 1:** A 4 L of argon gas has a pressure of 2 atm at 24 °C temperature. Calculate the final volume of the gas, when the pressure of the gas increases to 8 atm and the temperature of the gas increases to 30 °C.

Solution:

Given data:

Initial volume of the gas, V_{1} = 4 L

Initial pressure of the gas, P_{1} = 2 atm

Initial temperature of the gas, T_{1} = 24 °C = 297 K

Final volume of the gas, V_{2} = ?

Final pressure of the gas, P_{2} = 8 atm

Final temperature of the gas, T_{2} = 30 °C = 303 K

Using the formula of combined gas law,

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

V_{2} = (P_{1} × V_{1} × T_{2}) ÷ (T_{1} × P_{2})

V_{2} = (2 × 4 × 303) ÷ (297 × 8)

V_{2} = 303 ÷ 297

V_{2} = 1.02 L

Therefore, the final volume of the gas is **1.02 L**.

**Problem 2:** A huge balloon contains 12 L of helium gas at 20 °C temperature and it is having a pressure of 5 atm. If the volume of the gas increases to 60 L at 40 °C temperature, then calculate the final pressure of the gas.

Solution:

Given data:

Initial volume of the gas, V_{1} = 12 L

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

Initial pressure of the gas, P_{1} = 5 atm

Final volume of the gas, V_{2} = 60 L

Final temperature of the gas, T_{2} = 40 °C = 313 K

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

Using the formula of combined gas law,

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

P_{2} = (P_{1} × V_{1} × T_{2}) ÷ (T_{1} × V_{2})

P_{2} = (5 × 12 × 313) ÷ (293 × 60)

P_{2} = 313 ÷ 293

P_{2} = 1.06 atm

Therefore, the final pressure of the gas is **1.06 atm**.

**Problem 3:** A rubber balloon is taken out from the cold storage and 8 L of helium gas is released in it at 70 °C temperature and 6 atm pressure. If the initial temperature of the gas is 25 °C and the initial pressure of the gas is 3 atm, then what is the initial volume of the gas?

Solution:

Given data:

Final volume of the gas, V_{2} = 8 L

Final temperature of the gas, T_{2} = 70 °C = 343 K

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

Initial temperature of the gas, T_{1} = 25 °C = 298 K

Initial pressure of the gas, P_{1} = 3 atm

Initial volume of the gas, V_{1} = ?

Using the formula of combined gas law,

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

V_{1} = (P_{2} × V_{2} × T_{1}) ÷ (T_{2} × P_{1})

V_{1} = (6 × 8 × 298) ÷ (343 × 3)

V_{1} = 14304 ÷ 1029

V_{1} = 13.90 L

Therefore, the initial volume of the gas is **13.90 L**.

**Problem 4:** A 6 L of helium gas filled in the container at 10 °C temperature exerts some amount of pressure on the walls of the container. The gas exerts 12 atm pressure on the walls of the container, when its volume is increased to 12 L and its temperature is increased to 80 °C. Calculate the initial pressure of the gas.

Solution:

Given data:

Initial volume of the gas, V_{1} = 6 L

Initial temperature of the gas, T_{1} = 10 °C = 283 K

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

Final volume of the gas, V_{2} = 12 L

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

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

Using the formula of combined gas law,

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

P_{1} = (P_{2} × V_{2} × T_{1}) ÷ (T_{2} × V_{1})

P_{1} = (12 × 12 × 283) ÷ (353 × 6)

P_{1} = 40752 ÷ 2118

P_{1} = 19.24 atm

Therefore, the initial pressure of the gas is **19.24 atm**.

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