Question

The vapor pressure of pure water at 296 K is 2778.5 Pa. The vapor forms an ideal gas. 1) In some oil, the equilibrium concentration of water molecules is only 1% as large as in pure water. Suppose that the pure water is covered with a layer of that oil. What's the equilibrium vapor pressure of water above the oil layer?

Answer 1

Explanation:

It is given that vapor pressure of pure water at 296 K is 2778.5 Pa.These vapors will result in the formation of an ideal gas.

Now, as water is covered with oil and contains only 1% molecules of water. Hence, the vapor pressure of this mixture will also be equal to the vapor pressure of pure water.

So, vapor pressure of mixture = 1% vapor pressure of pure water

Therefore, $\text{(Vapor pressure)}_{mixture}$ = $\frac{1}{100} \times 2778.5 Pa$

                                                 = 27.785 Pa

Thus, we can conclude that the equilibrium vapor pressure of water above the oil layer is 27.785 Pa.