19. A sample of neon occupies a volume of 461 mL at STP.

What will be the

volume of the neon when the pressure is reduced to 93.3 kPa?

500.65mL

Explanation:

The following information were obtained from the question:

V1 (initial volume) = 461 mL

P1 (initial pressure) = stp = 101325Pa

P2 (final pressure) = 93.3 kPa

Recall: 1KPa = 1000Pa

Therefore, 93.3 kPa = 93.3x1000 = 93300Pa

V2 (final volume) = ?

Using the Boyle's law equation P1V1 = P2V2, the final volume of the gas can be obtained as follow:

P1V1 = P2V2

461 x 101325 = 93300 x V2

Divide both side by 93300

V2 = (461 x 101325) / 93300

V2 = 500.65mL

Therefore, the volume of Neon at 93.3 kPa is 500.65mL