# Phase change?

by doomed
Tags: phase
 P: 120 I'll give you some hints for the entropy: You have 2 thermodynamic processes there: 1) vaporization 2) isothermal expansion form 1 bar to 0.1 bar. Because the temperature is the same, the variation in entropy can easily be evaluated by $$\Delta S=\frac{Q_1+Q_2}{T}$$ where Q1+Q2 represents the total transfered heat. Now you have $$Q_1=m\cdot \lambda_v$$ for the vaporization at 100 C and $$Q_2=\nu R T \ln \frac{V_2}{V_1}$$ for the isothermal expansion and so on.......($$p_1 V_1=p_2 V_2$$ is the answer at your last question)