If #7/5 L# of a gas at room temperature exerts a pressure of #3 kPa# on its container, what pressure will the gas exert if the container's volume changes to #2/3 L#?

1 Answer
Nov 28, 2016

Answer:

The final pressure will be #63/10"kPa"##=##"6.3 kPa"#

Explanation:

This is a question involving Boyle's law, which states that the volume of a gas varies inversely with its pressure, as long as the amount and temperature are held constant.

http://www.one-school.net/Malaysia/UniversityandCollege/SPM/revisioncard/physics/heat/gaslaw.html
#P_1="3 kPa"#
#V_1="7/5 L"#
#V_2="2/3 L"#
#P_2="???"#

Rearrange the equation to isolate #P_2#. Substitute the known values into the equation and solve.

#P_2=(P_1V_1)/V_2#

#P_2=(3"kPa"·7/5cancel"L")/(2/3 cancel"L")#

Simplify.

#P_2=(21/5 "kPa")/(2/3)#

Multiply by the reciprocal of #2/3#.

#P_2=(21/5"kPa")·(3/2)#

Simplify.

#P_2=63/10 "kPa"##=##"6.3 kPa"#