If #8/3 L# of a gas at room temperature exerts a pressure of #30 kPa# on its container, what pressure will the gas exert if the container's volume changes to #19/12 L#?

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Answer 1 · 2017-02-20T08:12:46

The pressure is #=50.53kPa#

We apply Boyle's Law

#P_1V_1=P_2V_2#

#P_1=30kPa#

#V_1=8/3L#

#V_2=19/12L#

#P_2=V_1/V_2*P_1#

#=(8/3)/(19/12)*30#

#=50.53kPa#