Question

A gas occupies a volume of 845.4 mL of at standard temperature. If the temperature increases to 20.1oC, what is the new volume?

Answer 1

According to the ideal gas law of the isobaric process:

V₁/T₁ = V₂/T₂

where V₁ - initial volume, V2 - final volume, T₁ - initial temperature, T₂ - final temperature (20 °C = 293.25 K).

V₂ = V₁T₂ / T₁ = 845.4 mL × 293.25 K / 273.15 K = 907.6 mL

Answer: 907.6 mL