Assume that you have a liquid in a cylinder equipped with a movable piston. There is no air in the cylinder, the volume of space above the liquid is 200 mL, and the equilibrium vapor pressure above the liquid is 29.0 mm Hg.
What is the equilibrium pressure above the liquid when the volume of space goes from 200 mL to 100 mL at constant temperature?
To find the equilibrium pressure above the liquid when the volume of space goes from 200 mL to 100 mL, we can use Boyle's Law. Boyle's Law states that the pressure and volume of a gas are inversely proportional, as long as the temperature remains constant.
Boyle's Law can be expressed as:
P1V1 = P2V2
Where P1 and V1 refer to the initial pressure and volume, and P2 and V2 refer to the final pressure and volume.
In this case, the initial volume (V1) is 200 mL, and the final volume (V2) is 100 mL. The initial pressure (P1) is given as 29.0 mm Hg.
Substituting these values into the equation, we get:
(29.0 mm Hg) * (200 mL) = P2 * (100 mL)
Solving for P2:
P2 = (29.0 mm Hg) * (200 mL) / (100 mL)
Now, we can calculate the equilibrium pressure above the liquid:
P2 = 29.0 mm Hg * 2
Therefore, the equilibrium pressure above the liquid when the volume of space goes from 200 mL to 100 mL at constant temperature is 58.0 mm Hg.