A cylinder, with a weightless piston, has an internal diameter of $0.24 m$. The cylinder contains water and steam at $100 C$. It is situated in a constant temperature bath at $100 C$, Figure $2.1$. Atmospheric pressure is $1.01 × 10^5 Pa$. The steam in the cylinder occupies a length of $0.20 m$ and has a mass of $0.37 g$.(see diagram below).

I'd like to have a better understanding of the setup:

Why is the ideal gas equation, $pV=nRT$ not applicable to this system, does the fact that the steam is not isolated from the water below it (so that condensation occurs) violates the assumption that there are no molecular forces between ideal gas molecules?

The pressure also seems to remain constant at $p=p_{atm}$ as the piston is slowly pushed downwards/ pulled upwards, because the steam condenses/evaporates into water so there are less/more frequent collisions between the steam molecules and the cylinder, is my explanation satisfactory?

The diagram:

enter image description here


2 Answers 2


Saturated steam is NOT an ideal gas. It is water vapor that is in equilibrium with the liquid that it is in contact with. This means that any heat input to the system will vaporize some of the water at 100 deg C, and any heat removal from the system will condense some of the steam at 100 deg C. This type of system follows the Antoine equation rather than the ideal gas law. See https://en.wikipedia.org/wiki/Antoine_equation.

  • $\begingroup$ Thanks. Can I say that as the piston is being pushed downward, the pressure increases momentarily so does the temperature, which leads to evaporation of the water molecules near the surface, hence reduces the temperature then pressure, and in turn the pressure of the cylinder remains roughly constant throughout the process? $\endgroup$ Jun 25, 2020 at 17:25
  • 1
    $\begingroup$ You will not get condensation or evaporation without heat transfer with the environment. This means that if you push on the piston, the pressure will rise and the system will tend to condense some steam in order to lower the pressure back to 1 atm, but only if heat is transferred out of the cylinder. I note that your picture implies isothermal conditions, and if that is the case, such heat transfer will definitely take place. $\endgroup$ Jun 25, 2020 at 18:01

The water vapor does approximately follow the ideal gas law at 100 C. But the number of moles of water vapor has changed, and this needs to be taken into account.


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