A gas at a pressure of 199 kPa and a temperature of 18.2 ℃ has a volume of 6.49 L. When the gas temperature rises to 62.5 ℃, the volume increases to 10.6 L. What is the new pressure of the gas (in kPa)?

 A gas at a pressure of 199 kPa and a temperature of 18.2 ℃ has a volume of 6.49 L. When the gas temperature rises to 62.5 ℃, the volume increases to 10.6 L. What is the new pressure of the gas (in kPa)?

Solution

Use the formula for the combined gas law.

P₁V₁/T₁ = P₂V₂/T₂

Known and Unknown

P₁ = initial pressure = 199 kPa

V₁ = initial volume = 6.49 L

T₁ = initial temperature = 18.2 °C + 273.15 = 291.35 K

P₂ = final pressure = ? kPa

V₂ = final volume = 10.6 K

T₂ = final temperature = 62.5 °C + 273.15 = 335.65 K

Solve for P₂.

P₂ = P₁V₁T₂/T₁V₂

P₂ = (199 kPa × 6.49 L × 335.65 K)/(291.35 K × 10.6 L) = 140. kPa (rounded to three significant figures)