At constant temperature, the pressure on 8.0 L of a gas increases from 1.0 atm to 4.0 atm. What will be the new volume of the gas?

1 Answer
Jul 8, 2017

Answer:

Well, according to old Boyle's law #Pprop1/V#, and get #V_2=2*L#.........

Explanation:

And if #Pprop1/V# (#"temperature"#, and #"moles of gas"# CONSTANT), then #PV=k#, where #k# is some constant.....and so for the same amount of gas UNDER different conditions of #"pressure"# and #"volume"# but constant #"temperature"#, we may write............

#P_1V_1=P_2V_2#

.........when we solve for #k#.........

And we fill in the blanks, #V_1=8.0*L#, #P_1=1*atm#, and #P_2=4*atm#...........

#V_2=(P_1V_1)/P_2=(1*cancel(atm)xx8.0*L)/(4*cancel(atm))=2*L#.

Is it reasonable that volume should be so reduced under higher pressure?