If a gas is composed of molecules who attract eachother, then its pressure would increase were the interactions to suddenly disappear. This can be quantitatively and qualitatively discussed using Van der Waals' equation of state.

But the Van der Waals equation works only for a gas model. I was wondering what would happen to a liquid in the same situation. In other words:

What would happen to a liquid's pressure if all the molecular interactions suddenly ceased?

  • 3
    $\begingroup$ Wouldn't the liquid just turn into an ideal gas then? $\endgroup$ Dec 2 '18 at 5:01

The Van der Waals model can be defined as a model in which the number $n(E,V)$ of microstates as a function of total energy $E$ and total volume $V$ has this form, in units where Boltzmann's constant is $1$: $$ n(E,V)\propto\left(E+\frac{aN^2}{V}\right)^{DN/2}(V-bN)^N, \tag{1} $$ where

  • $N$ is the number of molecules,

  • $D$ is the number of spatial dimensions ($D=3$ in the real world),

  • $a$ is a constant that controls the strength of the attractive interaction between molecules,

  • $b$ is a constant that controls each molecule's "personal space" that cannot be penetrated by other molecules.

The corresponding entropy is \begin{align} S(E,V) &\equiv \log n(E,V)\\ &=\text{constant}+\frac{ND}{2}\log\left(E+\frac{aN^2}{V}\right) +N\,\log(V-bN). \tag{2} \end{align} To confirm that this is consistent with the Van der Waals equation of state, use the definitions of temperature and pressure $$ \frac{1}{T}=\frac{\partial S}{\partial E} \hskip2cm \frac{p}{T}=\frac{\partial S}{\partial V} \tag{3} $$ and use the first of these equations to eliminate the $E$ from the second of these equations. This gives the usual Van der Waals equation of state $$ \left(p+\frac{a}{(V/N)^2}\right)\left(\frac{V}{N}-b\right)=T. \tag{4} $$ The Van der Waals model defined by (1), when put in contact with a thermal reservoir, describes a fluid with a phase transition much like the liquid-gas phase transition (http://www.damtp.cam.ac.uk/user/tong/statphys/five.pdf). In the pressure-temperature plane, the phase transition line has an endpoint (the critical point), just like the phase diagram of carbon dioxide or water does, and you can move continuously from one "phase" to the other by going around this point instead of crossing the line. In other words, the liquid and gas phases are not really distinct phases. If you cross the line to get from one phase to the other, then there is a sudden change of density — much like the liquid/gas phase transition. Turning off the interactions makes this line disappear. The "liquid" and gas phases were never really distinct in the first place.

By the way, the reason I started with equation (1), instead of with equation (4), is that the preceding claims are based on an analysis (not shown here) that I did starting with equation (1). Equation (4) is presumably sufficient, but I haven't thought through that carefully enough to be sure.

What would happen to a liquid's pressure if all the molecular interactions suddenly ceased?

To make the question unambiguous, we need to specify what is held fixed when the molecular interactions are turned off. If the volume and temperature are both held fixed, and if the attractive interaction is turned off by setting $a=0$ in equation (4), then the pressure must increase, as stated in the question.

The point of this answer is that the conclusion is the same for a liquid, at least for the type of "liquid" described by the Van der Waals model. The liquid and gas phases are continously connected, and in this sense they are not really separate phases at all.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.