At constant temperature,By Boyle's law,
P1V1=P2V2P_1V_1=P_2V_2P1V1=P2V2
And in this case,P1=1.5 atm,V1=6LP_1=1.5\ atm,V_1=6LP1=1.5 atm,V1=6L
When V2=1L,P2=P1V1V2=1.5×61=9 atmV_2=1L,P_2=\frac{P_1V_1}{V_2}=\frac{1.5\times6}{1}=9\ atmV2=1L,P2=V2P1V1=11.5×6=9 atm
When V2=2500 ml=2.5L,P2=P1V1V2=1.5×62.5=3.6 atmV_2=2500\ ml=2.5L,P_2=\frac{P_1V_1}{V_2}=\frac{1.5\times6}{2.5}=3.6\ atmV2=2500 ml=2.5L,P2=V2P1V1=2.51.5×6=3.6 atm
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