# Moles of gas in a VLE mixture

I have a system in vapor-liquid equilibrium which also has some gas inside (let's say that the VLE system is Water and the gas is Air). The system looks like this:

There is some heat applied to it and therefore more water will vaporize and the Saturation Pressure of the vapor will increase. Since it is a closed system, I the amount of gas stays the same, while the amount of vapor increases. I would like to calculate the pressure of this system and I'm not sure 100% that my approach is correct.

I was thinking that the total pressure of the system is:

$P_{tot} = P_{sat} + P_{gas}$

So the Saturation Pressure can be calculated using the Antoine equation or taken from tables. I'm fine with that. Here Question 1 arises: how can one check if the VLE condition still applies once heat is applied? Compare Antoine with Ideal Gas?

Now the main question is... what to do with the gas? If I assume that it is ideal gas, it can be written as:

$P_{gas} = \frac{n_{gas} R T}{V_{gas}}$

And here is where I get confused. Question 2: Is $V_{gas}$ the volume of the gas phase only or it is considered together with the vapor like in the figure ($V_{gas}=V_{vap}$)? And if it is considered together with the vapor, then obviously it should be a variable, i.e. $V_{gas} = m_{vap} \rho_{vap}$. Another question arises with this: Question 3: If I don't know the amount of gas, how do I find it? (I know that one can find the amount at the initial state of the system or STP conditions $n_{gas} = \frac{(P_{init}-P_{sat})V}{RT}$, but there are two unknowns: $n_{gas}$ and $V$, unless the V is known). So basically the volume is the one that confuses me and I am thinking of using Raoult's law to get the fractions between the gas and the vapor, but I'm not sure this is the right approach. Can anyone advice me on this?

Thank you!

Paul

If you assume that the gas is insoluable in the liquid, then you could analyze it as follows (neglecting the effect of temperature on the density of the liquid):

Let $n_g$ equal the number of moles of the "gas" in the container

Let $n_v$ equal the total number of moles (vapor plus liquid) of the "vapor" species in the container

Let V be the total volume of the container

Let x be the fraction of the "vapor species" that is liquid

Then the volume of liquid in the container is:$$V_L=n_vxv_L\tag{1}$$where $v_L$ is the molar volume of the liquid. The remaining volume of the container is occupied by gas and vapor. Since the vapor is saturated, the partial pressure of the "vapor" is equal to the equilibrium vapor pressure:$$p_v=P_{sat}(T)$$This must be consistent with the remaining volume and the ideal gas law:$$P_{sat}(T)(V-V_L)=n_v(1-x)RT\tag{2}$$ Eqns. 1 and 2 can be used to solve for x, the split between the number of moles in the liquid and the number of moles in the vapor.

As far as the gas is concerned, its partial pressure can be determined from the ideal gas law (now that the volume of liquid is known):$$p_g(V-V_L)=n_gRT$$

• Thank you! I have some questions. So is $n_v$ the moles of vapor or of the entire mixture? And $n_g$ is a constant, right? Since we have no gas generation of release. And how would you define $x$ mathematically? $x = n_v/n_{mix}$ or? I also need to make sure that $P_{sat} + P_g = P_{amb}$ at the beginning of the system. – Physther Oct 1 '16 at 9:37
• "So is nvnv the moles of vapor or of the entire mixture?" The entire mixture. "And ng is a constant, right?" Yes. "And how would you define x mathematically? $x=n_{vapor}/(n_{vapor}+n_{liquid})$ "I also need to make sure that Psat+Pg=Pamb at the beginning of the system. " Yes, if that actually is the initial pressure in the container. – Chet Miller Oct 1 '16 at 12:09
• Okay. So I followed the right path. But do you think that it is normal that the pressure will decrease with temperature? The volume of gas (V-V_L) increases and for some reason, my total pressure decreases. It's somehow intuitive that the pressure should increase, but of course, if one increases the volume, the pressure should drop. – Physther Oct 1 '16 at 12:57
• Both the temperature and the pressure increase. Volume is not the only thing that affects pressure. – Chet Miller Oct 1 '16 at 14:34
• Yeah... Of course... Unfortunately that's not what I get... I need to do more debugging. – Physther Oct 2 '16 at 8:09

It appears "saturation pressure" is defined by some to be the same as vapor pressure. Other, equally professional folks, define it strictly as the vapor pressure at the boiling point. Let's stick with vapor pressure.

Q1: According to LeChatlier's Principle, after the heat (a stress) is applied, the system will adjust to reach a new equilibrium. In your system, especially since the heat must be applied over a finite time, this adjustment will be quite rapid unless your container is very large.

Q2: The gas and the vapor occupy the same space, they each have the same volume independently of one another. They also have the same temperature.

Q3: I'm not sure what you're given but maybe this will help. Let's begin with, say, 1 L of water and 1 L of gas at 10°C. Now raise the temperature to 90°C. Let's start by assuming the volume of the gas remains constant. My calculations, using the IGL and the constant volume assumption, show that an additional 0.023 mol, or 0.42 g, of water entered the gas phase. That same amount left the 1000 g of liquid reducing its volume. I'll leave it to you to decide if that volume change is significant. Also note that at temperatures below the 90°C example here, the amount of liquid entering the vapor phase becomes significantly less than the value shown here.