# A closed vessel contains 6 g of water vapor at a pressure of 25 kPa and at a temperature of 100 ° C.

**A closed vessel contains 6 g of water vapor at a pressure of 25 kPa and at a temperature of 100 ° C. Without changing the temperature, the volume of the vessel was reduced by 8 times. Find the mass of steam left after that in the vessel.**

To find the mass of the water vapor remaining in a closed vessel, we use the equality (we take into account that after reaching a pressure of 1 * 10 ^ 5 Pa, water condensation will begin, and the pressure will remain constant): Pn * Vn / mn = R * T / M = Pk * Vk / mk, whence we express: mk = Pk * Vk * mn / (Pn * Vn).

Values of variables: Pк – pressure reached by steam (Pк = 10 ^ 5 Pa); Vк (final volume) = 1/8 Vн (initial volume of water vapor); mn is the initial mass of steam (mn = 6 g); Pн – initial pressure (Pн = 25 kPa = 25 * 10 ^ 3 Pa);

Let’s make a calculation: mk = Pk * Vk * mn / (Pn * Vn) = 10 ^ 5 * Vk * 6 / (25 * 10 ^ 3 * 8Vk) = 3 g.

Answer: After reducing the volume of the vessel, 3 g of water vapor will remain.