if a gas sample in a balloon occupies 1.5 L at atmospheric pressure, what would be the pressure (in mmHg) if the volume was reduced to 0.8 L?

1425 mmHg.

Explanation:

The following data were obtained from the question:

Initial volume (V1) = 1.5 L

Initial pressure (P1) = 1 atm

Final volume (V2) = 0.8 L

Final pressure (P2) =?

Next, we shall determine the final pressure of the gas by using the Boyle's law equation as follow:

P1V1 = P2V2

1 × 1.5 = P2 × 0.8

1.5 = P2 × 0.8

Divide both side by 0.8

P2 = 1.5/0.8

P2 = 1.875 atm

Finally, we shall convert 1.875 atm to mmHg.

This can be obtained as follow:

1 atm = 760 mmHg

Therefore,

1.875 atm = 1.875 × 760 = 1425 mmHg.

Therefore, the new pressure of the gas is 1425 mmHg.

it should be p1 v1 = p2 v2

a

explanation: