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Suppose we have a mass of water at room pressure and temperature, then we started heating the water.

What will happen to the temperature if we decreased the pressure while the sample of water is undergoing phase change from liquid to vapor(keeping in mind that heat is being added all the time)?

My book ( thermodynamics by Yunus A. Cengel) says that the temperature would drop but the book does not provide explanation.

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Use a water phase diagram to work it out.

enter image description here

Figure 1. Water phase diagram. Source: SSC Chemistry.

If we choose the 100°C and 1.0 atm intersection that represents where you're at when boiling a kettle. Water is undergoing a phase change from liquid to gas as you specified.

If you now drop the pressure (going vertically down the dashed line) it should be clear that the liquid will tend to change to gas. You are aware, however, that the liquid has a latent heat of evaporation so energy is required to convert from liquid to gas and, in the short term, this will come from the water. As a result the temperature will drop - probably running down along the green-yellow border until the situation stabilises and boiling begins again.

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  • $\begingroup$ If the pressure were reduced "quickly", I believe that would usually be approximated as an adiabatic (no heat flow across system boundary) expansion. But the question makes it clear heat is added during the transition. So I think quasi static process with heat maintained such as to follow the boiling line, as you say. $\endgroup$ – Pete W Apr 1 at 22:49
  • $\begingroup$ This is true, but some teachers might take exception to using a phase diagram as the "solution". I would back this up with the rule that boiling point always decreases with decreasing air pressure (to some limit as you near vacuuum). $\endgroup$ – Carl Witthoft Apr 5 at 14:21
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You are adding heat at a constant rate $\dot{q}_o$ (W) to vaporize an amount $n$ (moles) water at a constant temperature. When you drop the external pressure, you also drop the partial pressure of the water vapor in the surrounding gas. This increases the thermodynamic driving force for the water to evaporate (Le Chatelier's principle should come to mind here). Decreasing the partial pressure of the water vapor above the liquid water also increases the driving force for convective mass transport of water from the liquid to the gas.

Decreasing the pressure increases both the thermodynamic and the kinetic driving forces for the water to evaporate. The evaporation rate will increase.

We can write an instantaneous energy balance to predict the direction of the temperature change with time $dT/dt$ based on the molar specific heat $\bar{C}_p$ (J/mol $^o$C).

$$ \dot{q}_o = n \bar{C}_p \frac{dT}{dt} + \Delta_{vap}\bar{H}\ r_{vap} $$

Vaporization rate $r_{vap}$ (mols/s) increases to carry away the molar vaporization enthalpy $\Delta_{vap}\bar{H}$ (J/mol). Note that $r_{vap} = k\ dn/dt$ (rate constant times moles evaporated per time), so the solution is not to take an integration of temperature with time.

With constant $\dot{q}_o$, an instantaneous increase in $r_{vap}$ must be accompanied by a commensurate decrease in $T$.

In practice, the water will cool to its new equilibrium point of $p,T$ on the vapor pressure line. The process is irreversible and water is certainly not an ideal gas anywhere near its condensation $p,T$. The path that the water takes to get to the new $p,T$ cannot be predicted. You could carry out the process in a quasi-static "slow" decrease in pressure or a rapid quench in pressure. The former may allow the water to remain as somewhat like an ideal gas with homogeneous temperature and pressure throughout during the process. The latter may lead to localized cooling. Presuming that enough liquid water is still present, the end result will be the same, simply taking different times to reach the new equilibrium.

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  • $\begingroup$ Water stratifies so it will not change temperature evenly or all at one time. Even seen hot water go through a cold tank of water and come out with little change in temperature - we all expected it to mix... So we had to reverse the flow for the desired output. Water behaves weirdly... $\endgroup$ – Solar Mike Apr 2 at 21:18
  • $\begingroup$ @SolarMike Noted accordingly. Thanks. $\endgroup$ – Jeffrey J Weimer Apr 3 at 0:16
  • $\begingroup$ @SolarMike ok so we'll put a blender into the water tank :-) $\endgroup$ – Carl Witthoft Apr 5 at 14:21

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