A 0.0138-m3 container is initially evacuated. Then, 4.73 g of water is placed in the container, and, after some time, all of the water evaporates. If the temperature of the water vapor is 358 K, what is its pressure?

56.7kpa

Explanation:

V = 0.0138-m³

M = 4.73g

T = 358k

R = ideal gas constant = 8.314

Number of moles of water n = 4.73/18 = 0.2628 (Remember that the molecular mass of water is 18).

Pv = nRT

P = nRT/V

P = (0.2628x8.314x358)/0.0138

P = 56681.23

P = 56.7Kpa

56.7Kpa is therefore the pressure of the water vapor.