Question

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?

Answer 1

At constant temperature,By Boyle's law,

$P_1V_1=P_2V_2$

And in this case,$P_1=1.5\ atm,V_1=6L$

When $V_2=1L,P_2=\frac{P_1V_1}{V_2}=\frac{1.5\times6}{1}=9\ atm$

When $V_2=2500\ ml=2.5L,P_2=\frac{P_1V_1}{V_2}=\frac{1.5\times6}{2.5}=3.6\ atm$