Question

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?

Answer 1

Answer:

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.