If #9/4 L# of a gas at room temperature exerts a pressure of #42 kPa# on its container, what pressure will the gas exert if the container's volume changes to #12/7 L#?

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Answer 1 · 2016-11-19T13:34:14

The pressure is #=441/8kPa#

Here, we use Boyle's Law.

#P_1V_1=P_2V_2#

#P_1=42kPa#

#V_1=9/4l#

#V_2=12/7l#

#P_2=(P_1V_1)/V_2#

#P_2=42*(9/4)/(12/7)=(42*9*7)/(12*4)=441/8kPa#