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So, the first law in thermodynamics tells us that $\Delta U=Q+W$, so if we have 1 kg of water at 100 C and it turns to steam. The volume changes from $V_0$ to $V_1$. My first two considerations are:

  1. There is no change in temperature (it stays at 100 C), does that mean $Q=0$?
  2. How can I calculate the work done by the system , can we use the ideal gas law? (That would mean $dW=PdV$, right?)

Thanks a lot and I hope this question is allowed.

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Increasing the temperature of a system is not the only way that its internal energy can increase. It can also increase at constant temperature as the result of a change of phase. Liquid water turning to steam is an example of that. In this situation, there is both heat transferred and work done on the surroundings, at constant pressure. So, $$Q=\Delta U_{vap}+p(V_V-V_L)$$ where $\Delta U_{vap}$ is the change in internal energy in going from 1 mole of liquid at 100 C to 1 mole of vapor at 100 C, p is the pressure (1 atm), $V_L$ is the volume of one mole of vapor at 100 C and 1 atm., and $V_L$ is the volume of one mole of liquid.

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