A liquid that has an equilibrium vapor pressure of 129 torr at 25°C is placed into a 1-L vessel like that shown below.
A few mL of the liquid is introduced into the vessel at 25°C while it has a pressure of 1.00 atm of air in it, without allowing any of the air to escape. After a few minutes a few drops of liquid remain in the vessel. What is the pressure difference shown on the manometer?
how do i start this?
I can't see the diagram and you can't draw it on this forum. If the diagram is what I think it is the pressure difference will be 129 torr.
The new pressure in the closed vessel is 760 mm + 129 mm. It was 760.
760 torr is wrong
760 is not what I said. Reread my answer.
To start solving this problem, you need to understand the concept of vapor pressure and how it affects the pressure inside a closed vessel. Here are the steps you can follow to solve the problem:
1. Determine the equilibrium vapor pressure of the liquid: In the problem, it is given that the liquid has an equilibrium vapor pressure of 129 torr at 25°C. This means that at that temperature, the liquid will exert a pressure of 129 torr when it is in equilibrium with its vapor.
2. Determine the initial pressure in the vessel: It is stated that the vessel initially contains 1.00 atm of air pressure. This air pressure does not change during the experiment since no air escapes.
3. Find the difference in pressure: The pressure difference is given by the difference between the equilibrium vapor pressure and the initial pressure in the vessel. This can be calculated by converting the equilibrium vapor pressure to the same units as the initial pressure (atm) and then subtracting the initial pressure from it.
4. Convert the pressure difference to the desired units: The problem asks for the pressure difference in torr, so make sure to convert the pressure difference to torr if necessary.
Follow these steps, and you should be able to determine the pressure difference shown on the manometer.