Question

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?

Answer 1

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?