Answer to Question #107241 in General Chemistry for Breanna J McKnight

Question #107241
The air in a 6.00 L tank has a pressure of 1.50 atm . What is the final pressure, in atmospheres, when the air is placed in tanks that have the following volumes, if there is no change in temperature and amount of gas?
1
Expert's answer
2020-04-02T09:00:17-0400

At constant temperature,By Boyle's law,

P1V1=P2V2P_1V_1=P_2V_2

And in this case,P1=1.5 atm,V1=6LP_1=1.5\ atm,V_1=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\ 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\ atm



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