A closed vessel contains 2 g of water vapor under a pressure of 50 kPa and at a temperature of 100 ºС.

A closed vessel contains 2 g of water vapor under a pressure of 50 kPa and at a temperature of 100 ºС. Without changing the temperature, the volume of the vessel was reduced by 4 times. Find the mass of the resulting water.

P1 = 50 kPa = 50,000 Pa;

Т = 100о С = 373 К

n1 (H2O) = m1 (H2O) / M (H2O) = 2/18 = 0.111 mol;

Using the Mendeleev-Clapeyron law, we find the initial volume of the vessel:

V1 = n1 (H2O) * R * T / P1 = (0.111 * 8.314 * 373) / 50,000 = 0.00689 m3;

V2 = V1 / 4 = 0.00689 / 4 = 0.00172 m3;

After compression, the pressure in the vessel will be 1 atmosphere (101325 Pa), the excess water will condense. Let’s find the amount of water substance in the vapor phase:

n2 (H2O) = (P2 * V2) / (R * T) = (101325 * 0.00172) / (8.314 * 373) = 0.056 mol;

n (liquid water) = n1 (H2O) – n2 (H2O) = 0.111 – 0.056 = 0.055 mol;

m (liquid water) = 0.055 * 18 = 0.99 g.

Answer: m (liquid water) = 0.99 g.