The vapor pressure of water at 25.0 C is 23.8 mm Hg. If 250. mL of water is placed in a chamber with a volume of 1.00 cubic meters, how many mL of water will evaporate
PV=nRT
YOu know P, V,R, T solve for n.
then, grams water=n*molmass water
and finally, ml of water=gramswater*1ml/1gram
To determine how many mL of water will evaporate, we need to consider the change in vapor pressure when the volume of the chamber is increased.
First, let's calculate the mole fraction of water in the chamber before evaporation:
Mole Fraction of Water = (moles of water) / (total moles of gas)
Since water is the only component in the chamber initially, the mole fraction of water is 1.
Next, we need to calculate the mole fraction of water in the chamber after evaporation. This can be found using Raoult's law, which states that the vapor pressure of a component in a mixture is proportional to its mole fraction:
Vapor Pressure = Mole Fraction of Water * Vapor Pressure of Pure Water
To apply Raoult's law, let's rearrange the formula:
Mole Fraction of Water = Vapor Pressure / Vapor Pressure of Pure Water
Now, we know the initial vapor pressure of water at 25.0°C is 23.8 mm Hg. However, to apply Raoult's law, we need to convert it to atm:
Vapor Pressure (atm) = 23.8 mm Hg / 760 mm Hg/atm
= 0.0313 atm
The vapor pressure of pure water at 25.0°C is the same as the initial vapor pressure, so it remains as 0.0313 atm.
Now, we can calculate the mole fraction of water in the chamber after evaporation:
Mole Fraction of Water = 0.0313 atm / 0.0313 atm
= 1
Since the mole fraction of water remains the same after evaporation and there are no other components in the chamber, the volume of water evaporated will depend on the change in the volume of the chamber.
The initial volume of water placed in the chamber is 250 mL, which is equivalent to 0.25 L.
The final volume of the chamber is 1.00 cubic meter, which is equivalent to 1000 L.
Therefore, the change in volume is:
Change in Volume = Final Volume - Initial Volume
= 1000 L - 0.25 L
= 999.75 L
Since the volume of the chamber increased, the water will evaporate until the mole fraction of water remains the same. Therefore, no water will evaporate.
Thus, there will be 0 mL of water evaporated.