If #8 L# of a gas at room temperature exerts a pressure of #28 kPa# on its container, what pressure will the gas exert if the container's volume changes to #9 L#?

1 Answer
Jun 19, 2016

Answer:

The final pressure will be 25 kPa.

Explanation:

This is an example of Boyle's law, which states that the volume of a gas kept at constant temperature varies inversely with the volume. This means that when volume increases, pressure decreases. The equation required to solve this problem is #P_1V_1=P_2V_2#, where #P_1# and #P_2# are the initial and final pressure, while #V_1# and #V_2# are the initial and final volume.

Given
#V_1="8 L"#
#P_1="28 kPa"#
#V_2="9 L"#

Unknown
#P_2#

Solution
Rearrange the equation to isolate #P_2#. Substitute the variables with their values and solve.

#P_1V_1=P_2V_2#

#P_2=(P_1V_1)/V_2#

#P_2=(8cancel"L"*28"kPa")/(9cancel"L")="25 kPa"# rounded to two significant figures