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 is decreased from 200 mL to 100 mL at constant temperature?
Did the vapor pressure of this liquid change when you squeezed the gas down?
I think not.
Therefore the pressure did not change.
You have half as many gas molecules above the fluid.
Half the gas molecules condensed back into liquid form.
Well, well, well, looks like we have ourselves a moving piston! How about that? Now, let me simplify this situation for you, my curious friend. You've got a liquid chilling in a cylinder, and the volume of space above it gets reduced. As a result, the equilibrium pressure above the liquid changes.
To figure out this pressure change, we need to remember that the pressure is determined by the vapor of the liquid. The higher the volume, the more molecules can escape from the liquid, resulting in a higher equilibrium vapor pressure. So, when you shrink the volume from 200 mL to 100 mL, imagine it's like trying to shove a bunch of clowns into a tiny car. They'll start spilling out all over the place, causing the pressure to increase.
Now, as for the numbers, when the volume of space above the liquid is halved, you can expect the pressure to, well, you guessed it, double! So, if the equilibrium vapor pressure above the liquid was initially 29.0 mm Hg, when you squeeze that space down to 100 mL, the equilibrium pressure above the liquid would become 58.0 mm Hg. It's like adding an extra high-pressure clown to the car!
Just remember, when you mess with volumes, these clowns... I mean, molecules, don't always like it, and they start clowning around by increasing the pressure.
To calculate the equilibrium pressure above the liquid when the volume of space is decreased from 200 mL to 100 mL at constant temperature, we can use Boyle's Law.
Boyle's Law states that the pressure of a gas is inversely proportional to its volume at constant temperature. Mathematically, it can be expressed as:
P1 * V1 = P2 * V2
Where:
P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume
In this case, the initial pressure (P1) is given as the equilibrium vapor pressure above the liquid, which is 29.0 mm Hg. The initial volume (V1) is 200 mL, and the final volume (V2) is 100 mL.
Let's substitute these values into the equation and solve for the final pressure (P2):
P1 * V1 = P2 * V2
29.0 mm Hg * 200 mL = P2 * 100 mL
5800 mm mL = P2 * 100 mL
Dividing both sides of the equation by 100 mL:
P2 = 5800 mm Hg / 100 mL
Simplifying:
P2 = 58.0 mm Hg
Therefore, the equilibrium pressure above the liquid when the volume is decreased to 100 mL is 58.0 mm Hg.
To find the equilibrium pressure above the liquid when the volume of space is decreased from 200 mL to 100 mL at constant temperature, we need to apply the ideal gas law.
The ideal gas law is expressed as:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
In this case, we are dealing with the vapor pressure of the liquid above it, which can be treated as an ideal gas. Therefore, we can use this equation to find the pressure.
Given that the volume of space above the liquid is decreased from 200 mL to 100 mL, we can assume that the volume occupied by the vapor is also reduced by the same amount. So initially, the volume of the vapor is 200 mL, and it becomes 100 mL.
Now, let's solve the problem step by step:
Step 1: Determine the initial pressure of the vapor
The initial pressure (P1) can be calculated using the equation:
P1 * 200 mL = n1 * R * T
We are given that the initial volume (V1) is 200 mL, and the pressure (P1) is 29.0 mm Hg. We also assume that the number of moles (n1) and temperature (T) remains constant throughout the process.
Step 2: Determine the final pressure of the vapor
Now we need to find the final pressure (P2) when the volume becomes 100 mL. We can use the equation:
P2 * 100 mL = n2 * R * T
Step 3: Relate the initial and final pressures
Since the number of moles and temperature remains constant in this process, we can equate the two equations to find the relationship between P1 and P2:
P1 * 200 mL = P2 * 100 mL
Step 4: Solve for P2
By rearranging the equation, we can solve for P2:
P2 = (P1 * 200 mL) / 100 mL
P2 = 2 * P1
Finally, substitute the value of P1, which is given as 29.0 mm Hg, into the equation:
P2 = 2 * 29.0 mm Hg
P2 = 58.0 mm Hg
Therefore, the equilibrium pressure above the liquid when the volume of space is decreased from 200 mL to 100 mL is 58.0 mm Hg.