Convert 77.0 L at 18.0 mm of Hg to its new volume at standard pressure.

1 Answer
Stefan V. · Media Owl
Dec 15, 2014

The answer is #1.8L#.

Standard Temperature and Pressure stipulates a pressure of #100.0kPa#. In order to solve this problem, we'll use Boyle's law, #P_1V_1 = P_2V_2#, which states that the pressure and volume of a gas are proportional to one another.

Let's calculate the volume by converting #mmHg# to #kPa# first.

#18 mmHg * (1.0 atm)/(760 mmHg) * (101.325 kPa)/(1.0 atm) = 2.4kPa#

So, #V_2 = P_1/P_2 * V_1 = (2.4 kPa)/(100.0 kPa) * 77L = 1.8L#

Now calculate the volume by converting #kPa# to #mmHg#.

#100.0 kPa * (1 atm)/(101.325 kPa) * (760 mmHg)/(1 atm) = 750 mmHg#

#V_2 = P_1/P_2 * V_1 = (18 mmHg)/(750 mmHg) * 77L = 1.8L#

Related questions
See all questions in Boyle's Law
Impact of this question
10826 views around the world
You can reuse this answer
Creative Commons License