If #9/4 L# of a gas at room temperature exerts a pressure of #15 kPa# on its container, what pressure will the gas exert if the container's volume changes to #7/3 L#?

1 Answer
Oct 12, 2017

Answer:

#14.5kPa#

Explanation:

Boyle's law states that for an ideal gas #P\propto1/V#, or #P=k/V# where #k# is a constant.

In this question we are given the pressure and volume, but not the constant, though it is easy to work out.

#PV=k#

#15kPa=1.5*10^4Pa#

#9/4L=2.25L=2.25*10^(-3)m^3#

Therefore, #(1.5*10^4)(2.25*10^(-3))=33.75#

Our new volume of #7/3L# gives us the equation of #P=33.75/(7/3000)=101250/7~~14,464Pa=14.464kPa~~14.5kPa#