A simple of helium measuring 6L was kept at a pressure is of 1.5 atm, if pressure is doubled what would be its new volume?

3L

Explanation:

The following data were obtained from the question:

V1 (original volume) = 6L

P1 (original pressure) = 1.5 atm

P2 (new pressure) = 2 x P1 (since the new pressure is doubled)

P2 = 2 x 1.5

P2 = 3 atm

V2 (new volume) = ?

Using the Boyle's law equation P1V1 = P2V2, the new volume can be obtain as follow:

P1V1 = P2V2

1.5 x 6 = 3 x V2

Divide both side by 3

V2 = (1.5 x 6) / 3

V2 = 3L

From our calculations, the new volume will be half the original volume and this will be 3L