Question

A sample of water in equilibrium with its vapor in a closed 3-L container has a vapor pressure of 23.8 torr at 25°C. The container's volume is increased to 7 L at constant temperature and the liquid/vapor equilibrium is reestablished. What is the vapor pressure?

Answer 1

Answer:

The new vapor pressure is 10.2 torr or 0.01342 atm

Explanation:

Step 1: Data given

Volume of the container = 3.0 L

Vapor pressure = 23.8 torr = 23.8 / 760 = 0.03131579 atm

Temperature = 25 °C = 25 * 273 = 298 K

The volume is increased to 7.0 L

The temperature will stay constant

Step 2: Calculate the new pressure

P1*V1 = P2*V2

⇒with P1 = the initial pressure = 23.8 torr = 0.03131579 atm

⇒with V1 = the initial pvolume 3.0 L

⇒with P2 = the final pressure = TO BE DETERMINED

⇒with V2 = the increased volume = 7.0L

0.03131579 atm * 3.0 L = P2 * 7.0L

P2 = (0.03131579 atm * 3.0 L)/7.0L

P2 = 0.01342 atm

OR

23.8 torr * 3.0 L = P2 * 7.0 L

P2 = (23.8 * 3.0)/7.0

P2 = 10.2 torr

The new vapor pressure is 10.2 torr or 0.01342 atm

Answer 2

Answer:

23.8 torr

Explanation:

Vapor pressure of a liquid is simply the pressure of the vapor that arises when the liquid evaporates in a closed container above the liquid sample.

From the question given above, the vapor pressure is 23.8 torr at 25°C. Since the temperature is constant, the vapor will remain the same when equilibrium is reached in the new container as vapor pressure depends on the temperature of the liquid.