If measurements if a gas are 75l and 300 full-scale and then the gas is measured a second time and found to be 50l, describe what had to happen to the pressure (if temperature remained constant).

The answer is a bit confusing, so I did some research and fount the original question. This is the complete question:

"If measurements of a gas are 75L and 300 kilopascals and then the gas is measured a second time and found to be 50L, describe what had to happen to the pressure (if temperature remained constant). Include which law supports this observation".

From that:

1) Data:

V1 = 75 liter
P1 = 300 kPa

V2 = 50 liter
P2 = x

T = constant.

2) Analysis (physical law)

Boyle's law states that the volume of a fixed amount of gas, at constant temperarature, is inversely related with the pressure.

In mathematical form that statement becomes into this equation:

PV = constant (at constant T)

=> P1V1 = P2V2.

Then, the decreasing of the volumen (compression) is accompanied by an increase of the pressure.

3) So, using Boyle's law with the data given:

300 kPa * 75 liter = x * 50 liter

=> x = 300 kPa * 75 liter / 50 liter = 450 kPa.

following Boyle's law, when the gas passed from 75 liter to 50 liter the pressure had to increase from 300 kPa to 450 kPa.

Milliseconds. . . . . .