Combined gas law formula

The information on this page is ✔ fact-checked.

The Combined gas law formula, expressed as P1V1/T1 = P2V2/T2, describes the relationship between pressure (P), volume (V), and temperature (T) of a given quantity of gas. In this formula, P1, V1, and T1 represent the initial pressure, volume, and temperature of the gas, while P2, V2, and T2 represent the final pressure, volume, and temperature, respectively. This formula is used to calculate the changes in pressure, volume, or temperature of a gas under different conditions while keeping the amount of gas constant.

Contents

Practice problems

Problem #1

A cylinder initially filled with 4 L of argon gas at 24 ℃ temperature has a pressure of 2 atm. What will be the final volume of the gas when the pressure increases to 8 atm and the temperature increases to 30 ℃?

Solution

Given data:

• Initial volume of the gas, V1 = 4 L
• Initial temperature of the gas, T1 = 24 ℃ = 297 K
• Initial pressure of the gas, P1 = 2 atm
• Final volume of the gas, V2 = ?
• Final pressure of the gas, P2 = 8 atm
• Final temperature of the gas, T2 = 30 ℃ = 303 K

Applying the formula:

• P1V1/T1 = P2V2/T2
• V2 = (P1 × V1 × T2) ÷ (T1 × P2)
• V2 = (2 × 4 × 303) ÷ (297 × 8)
• V2 = 303 ÷ 297
• V2 = 1.02 L

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

Problem #2

A balloon initially containing 12 L of helium gas at 20 ℃ and 5 atm pressure is heated to 40 ℃, causing the volume to increase to 60 L. What is the final pressure of the gas?

Solution

Given data:

• Initial volume of the gas, V1 = 12 L
• Initial temperature of the gas, T1 = 20 ℃ = 293 K
• Initial pressure of the gas, P1 = 5 atm
• Final temperature of the gas, T2 = 40 ℃ = 313 K
• Final volume of the gas, V2 = 60 L
• Final pressure of the gas, P2 = ?

Applying the formula:

P1V1/T1 = P2V2/T2
P2 = (P1 × V1 × T2) ÷ (T1 × V2)
P2 = (5 × 12 × 313) ÷ (293 × 60)
P2 = 313 ÷ 293
P2 = 1.06 atm

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

Problem #3

A rubber balloon is taken out of cold storage, and 8 L of helium gas is released into it at a temperature of 70 ℃ and pressure of 6 atm. If the initial temperature and pressure of the gas were 25 ℃ and 3 atm, respectively, what was the initial volume of the gas?

Solution

Given data:

• Final volume of the gas, V2 = 8 L
• Final temperature of the gas, T2 = 70 ℃ = 343 K
• Final pressure of the gas, P2 = 6 atm
• Initial temperature of the gas, T1 = 25 ℃ = 298 K
• Initial pressure of the gas, P1 = 3 atm
• Initial volume of the gas, V1 = ?

Applying the formula:

• P1V1/T1 = P2V2/T2
• V1 = (P2 × V2 × T1) ÷ (T2 × P1)
• V1 = (6 × 8 × 298) ÷ (343 × 3)
• V1 = 14304 ÷ 1029
• V1 = 13.90 L

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

Problem #4

At an initial temperature of 10 ℃, the container holds 6 L of neon gas, exerting a certain pressure on the container walls. When the temperature is raised to 80 ℃ and the volume of the gas is increased to 12 L, the pressure exerted by the gas on the container walls becomes 12 atm. What was the initial pressure of the gas in the container?

Solution

Given data:

• Initial temperature of the gas, T1 = 10 ℃ = 283 K
• Initial volume of the gas, V1 = 6 L
• Final temperature of the gas, T2 = 80 ℃ = 353 K
• Final volume of the gas, V2 = 12 L
• Final pressure of the gas, P2 = 12 atm
• Initial pressure of the gas, P1 = ?

Applying the formula:

• P1V1/T1 = P2V2/T2
• P1 = (P2 × V2 × T1) ÷ (T2 × V1)
• P1 = (12 × 12 × 283) ÷ (353 × 6)
• P1 = 40752 ÷ 2118
• P1 = 19.24 atm

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