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?
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
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