A gas at a pressure of #"203 kPa"# occupies #"40.0 L"#. If the pressure is decreased to #"35.0 kPa"#, what is the new volume? Assume the temperature is held constant.

1 Answer
Meave60
Jan 4, 2017

The final volume of helium gas will be #"232 L"#.

Explanation:

This is an example of Boyle's law, which states that the volume of a given amount of gas, held at a constant temperature, will vary inversely with its pressure. This means that when the pressure is increased, the volume will decrease and vice versa

The equation to use is #"P_1V_1=P_2V_2#, where #P# represents pressure, and #V# represents volume.

Known
#P_1="203 kPa"#
#V_1="40.0 L"#
#P_2="35.0 kPa"#

Unknown
#V_2#

Solution
Rearrange the equation to isolate #V_2#. Substitute the known values into the equation and solve.

#V_2=(P_1V_1)/(P_2)#

#V_2=(203cancel"kPa"xx40.0"L")/(35.0cancel"kPa")="232 L"#

The final volume of helium gas will be #"232 L"#.

Related questions
See all questions in Boyle's Law
Impact of this question
2759 views around the world
You can reuse this answer
Creative Commons License