If #5/7 L# of a gas at room temperature exerts a pressure of #9 kPa# on its container, what pressure will the gas exert if the container's volume changes to #8/9 L#?

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Answer 1 · 2017-03-07T16:08:34

The pressure is #=7.23kPa#

We apply Boyle's Law

#P_1V_1=P_2V_2#

#P_1=9kPa#

#V_1=5/7L#

#V_2=8/9L#

#P_2=V_1/V_2*P_1#

#=(5/7)/(8/9)*9#

#=5/7*9/8*9#

#=405/56#

#=7.23kPa#