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Phase Transformations

Problem 2.186

A vessel of volume V=6.0V=6.0 l contains water together with its saturated vapour under a pressure of 40 atm and at a temperature of 250∘C250^{\circ} \mathrm{C}. The specific volume of the vapour is equal to Vv′=50l/kgV_{v}^{\prime}=50 \mathrm{l} / \mathrm{kg} under these conditions. The total mass of the system water-vapour equals m=5.0 kg.m=5.0 \mathrm{~kg} . Find the mass and the volume of the vapour.

Reveal Answer
 2.186. mv=(V−mVl′)/(Vv′−Vl′)=20 \text { 2.186. } m_{v}=\left(V-m V_{l}^{\prime}\right) /\left(V_{v}^{\prime}-V_{l}^{\prime}\right)=20 g,V0=1.0l\mathrm{g}, \quad V_{0}=1.0 \mathrm{l}. Here Vi′V_{i}^{\prime} is the specific volume of water.
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