# Pressure equilibrium of liquid and ideal gas

Hello there I was wondering how pressure equilibrium of ideal gases and liquids work.

Example: Suppose we have an ideal gas at 1 bar and another one at the same pressure, seperated by a movable piston. The one gas' volume is halved, so the pressure doubled. The piston moves, and both gases equilibrate at 1.5 bar.

But what if one of the gases is replaced by a liquid. The volume of the gas is halved; but due to incompressibility of the liquid, will the gas' pressure stay at 2 bar? Or at which pressure does the system equilibriate?

• The gases don't equilibrate at 1.5 bar in the first example; please see my answer. Commented Aug 21, 2023 at 20:36

It's a fun exercise to derive the conditions of equilibrium here, as this allows us to consider various general combinations of compressible and incompressible materials.

At equilibrium, the total entropy $$S$$ is maximized: $$dS=0$$ (and $$d^2S<0$$).

From the fundamental relation, for container $$i$$,

$$dS_i=\frac{dU_i}{T_i}+\frac{P_i}{T_i}dV_i.$$

Assume a constant temperature $$T$$; then,

$$dS=dS_1+dS_2=\frac{dU_1+dU_2}{T}+\frac{P_1-P_2}{T}dV=0$$

because $$dV_1=-dV_2$$. Since this relation must hold for independent changes in $$dV$$, we have $$P_1=P_2$$ at equilibrium. That's familiar enough; pressures generally equalize on either side of a movable partition.

So what is this final pressure? Consider the isothermal bulk modulus $$K\equiv -V\left(\frac{\partial P}{\partial V}\right)_T$$, which describes the pressure–volume interplay.

From the exchange in volume $$dV_1=-dV_2$$, we can use the bulk modulus definition to write

$$\frac{V_1}{K_1}dP_1=-\frac{V_2}{K_2}dP_2,$$

which can be integrated to give

$$\int_{P_{1,\text{initial}}}^{P_{\text{final}}}\frac{V_1}{K_1}dP_1=\int_{P_{\text{final}}}^{P_{2,\text{initial}}}\frac{V_2}{K_2}dP_2.$$

Let's first assume both containers contain equal amounts of an ideal gas ($$PV=nRT$$, $$K=P$$). Then via integration and algebraic manipulation, we obtain

$$\left[-\frac{nRT}{P_1}\right]_{P_{1,\text{initial}}}^{P_{\text{final}}}=\left[-\frac{nRT}{P_2}\right]_{P_{\text{final}}}^{P_{2,\text{initial}}};$$

$$nRT\left(\frac{1}{P_{1,\text{initial}}}-\frac{1}{P_{\text{final}}}\right)=nRT\left(\frac{1}{P_{\text{final}}}-\frac{1}{P_{2,\text{initial}}}\right);$$

$$P_\text{final}=\frac{2P_{1,\text{initial}}P_{2,\text{initial}}}{P_{1,\text{initial}}+P_{2,\text{initial}}},$$

which is notably not the arithmetic average; the final pressure in your example is not $$\frac{3}{2}\,\text{atm}$$ but $$\frac{4}{3}\,\text{atm}$$. (We could have also obtained this from the ideal gas law applied to the final configuration: $$P=2nRT/\left(\frac{3}{2}V\right)$$.)

Now let's assume container 2 contains a liquid ($$V_2$$ approximately constant, equivalent to an assumption of $$K\gg P$$) to obtain

$$nRT\left(\frac{1}{P_{1,\text{initial}}}-\frac{1}{P_{\text{final}}}\right)=\frac{V_2}{K_2}(P_{2,\text{initial}}-P_{\text{final}}),$$

which can be solved for $$P_\text{final}$$ as

$$P_\text{final}=\frac{\sqrt{(nRT)^2+2nRTP_{1}V_2(2P_{1}-P_{2})/K+(P_{1}P_{2}V_2/K)^2}-nRT+P_{1}P_{2}V_2/K}{2P_{1}V_2/K},$$

where $$P_1$$ and $$P_2$$ refer to the initial pressures. For large $$K$$, a Taylor series expansion of the square-root term to third order gives

$$\sqrt{(nRT)^2+2nRTP_{1}V_2(2P_{1}-P_{2})/K+(P_{1}P_{2}V_2/K)^2}\approx nRT+\frac{P_1V_2(2P_1-P_2)}{K}-\frac{[P_1V_2(2P_1-P_2)]^2}{2nRTK^2},$$

yielding

$$P_\text{final}\approx P_{1}-\frac{P_1V_2(2P_1-P_2)^2}{4nRTK};$$

$$\frac{\Delta V}{V_2}= \frac{P_\text{final}-P_2}{K}.$$

For $$K\to\infty$$, $$P_{\text{final}}\to P_{1,\text{initial}}$$, as you suspected.

(A nuance that surprised me was the temperature $$T$$ appearing in the denominator: Shouldn't a higher gas temperature produce more energetic compression of the liquid? Yes, and this is explained by the cubic $$P_1$$ term appearing in the numerator. For constant $$n$$ and $$V_1$$, a greater $$T$$ must lead to a greater $$P_1$$.)

• What about forming a 2nd phase on the liquid side of the partition? Commented Aug 22, 2023 at 11:25
• The question describes a container completely filled with liquid that experiences only pressure increases. (The gas on the other side is pressurized from its volume being halved by a separate mechanism.) I don’t address pressure-induced solidification in my answer; if it occurred, it would proceed normally. Commented Aug 22, 2023 at 15:43

If the liquid fills its half, it just will be on a pressure of about 2bar, since it is almost incompressible.

If you move the partition to compress the ideal gas, part of the liquid on the other side of the partition will vaporize to form a vapor phase; both the liquid and vapor will be at the equilibrium vapor pressure.

• True, but not really what I meant; I meant we had two containers which each have a mole of gas at 1 bar, and the one volume container of one ideal gas is halved, so that the compressed gas moves the piston in the direction of the uncompressed gas
– 冰淇淋
Commented Aug 22, 2023 at 12:03