Mixture of two ideal gases

Why is $\Delta H\_{\mathrm{mix}}=0$ in a mixture of ideal gases? If $\Delta H=\Delta U + \Delta PV $ and I know that in ideal gases there are no interactions between molecules so $\Delta Q=0$ and I consider $\Delta V=0$


How do I prove this? Do the gases need to have constant pressure so $\Delta H\_{\mathrm{mix}}=Q$? If not, I really don't understand how could it be $0$ otherwise.


Can someone please tell me if this correct?
So if $\Delta Q=0$ and $\Delta V=0$ then from the first law of thermodynamics I can say that $\Delta U=0$ so the temperature $T$ is constant. That would also tell me that pressure $P$ is constant.


From that I can say that $\Delta H\_{\mathrm{mix}}=Q$, so $\Delta H\_{\mathrm{mix}}=0$.


Is this correct?




I will give the general proof as to why $\Delta\_\mathrm{mix} H=0$ for a mixture of ideal gases. You will not be able to reach a conclusion with the equations you use, we need a bit of theory about the thermodynamics of solution. I will give some definitions, and then see what this definitions tell when we apply the ideal-gas model.


**1. Properties of a solution** There are three guys, all different, that represent different thermodynamic properties in a solution. For a general property $M$ we have:


* $M(p, T, y\_1, ..., y\_N)$ is the thermodynamic property of a **solution** as a whole. It is a function of pressure $p$, temperature $T$, and all molar fractions ($N-1$ becase one is linearly dependent).
* $M\_i(p, T)$ is the thermodynamic property of species $i$ as pure. It is a function of pressure $p$, temperature $T$.
* $\bar{M}\_i(p, T, y\_1, ..., y\_N)$ is the **partial molar property** of species $i$ in a solution. It is a function of presure $p$, temperature $T$, and all molar fractions ($N-1$ also). This is the new guy in solution thermodynamics and is **defined** as


$$ \bar{M}\_\mathrm{i} := \left[\frac{\partial (nM)}{\partial n\_\mathrm{i}}\right]\_
 {p, T, n\_\mathrm{j\neq i}} \tag{1} $$


where the symbol $j \neq i $ says that the differentiation must be done keeping all the amounts constant, with the exception of $n\_i$. Eq. (1) seems obscure but is simple to understand. Say the property $M$ is the molar volume, so $nV$ will be the total volume. $\bar{V}\_i$ will say how much the total volume $nV$ changes when we add a differential amount of species $n\_i$ in the solution. Partial molar properties were introduced by Gilbert N. Lewis, who also introduced fugacity and the model of ideal solutions, so I am just repeating what he said.


**2. Summability relation** Now, in solution thermodynamics, every thermodynamic property in a solution will **not** be just a function of two intensive variables. It will be a function of the different amounts of the species present in a solution. You may imagine that a mixture of hexane and chlorobenzene will have different properties whether the molar fraction of hexane is $0.1$ or $0.9$. Hence, the total differential of the property $M$ is now
\begin{align}
 \mathrm{d}(nM) &=
 \left[\frac{\partial (nM)}{\partial p}\right]\_{T, n} \mathrm{d}p +
 \left[\frac{\partial (nM)}{\partial T}\right]\_{p, n} \mathrm{d}T +
 \sum\_\mathrm{i} \left[\frac{\partial (nM)}{\partial n\_\mathrm{i}}\right]\_
 {p, T, n\_\mathrm{j\neq i}} \mathrm{d}n\_\mathrm{i} \\
 &= 
 \left[\frac{\partial (nM)}{\partial p}\right]\_{T, n} \mathrm{d}p +
 \left[\frac{\partial (nM)}{\partial T}\right]\_{p, n} \mathrm{d}T +
 \sum\_\mathrm{i} \bar{M}\_\mathrm{i} \mathrm{d}n\_\mathrm{i} \tag{2} 
\end{align}
where we used Eq. (1) to replace the partial derivative. The new term here is the final sum. If we: (1) open up $\mathrm{d}(nM)$, and (2) use that $n\_\mathrm{i} = x\_\mathrm{i} n$ or in a total differential form $\mathrm{d}n\_\mathrm{i} = n\mathrm{d}x\_\mathrm{i} + x\_\mathrm{i}\mathrm{d}n$ we get two equations when we group the terms with $n$ and with $\mathrm{d}n$
\begin{align}
 & n\mathrm{d}M + M\mathrm{d}n =
 n\left(\frac{\partial M}{\partial p}\right)\_{T} \mathrm{d}p +
 n\left(\frac{\partial M}{\partial T}\right)\_{p} \mathrm{d}T +
 \sum\_\mathrm{i} \bar{M}\_\mathrm{i} (n\mathrm{d}x\_\mathrm{i} + x\_\mathrm{i}\mathrm{d}n) \\
 & n\mathrm{d}M + M\mathrm{d}n - 
 n\left(\frac{\partial M}{\partial p}\right)\_{T} \mathrm{d}p - 
 n\left(\frac{\partial M}{\partial T}\right)\_{p} \mathrm{d}T -
 n\sum\_\mathrm{i} \bar{M}\_\mathrm{i} \mathrm{d}x\_\mathrm{i} - \sum\_\mathrm{i} \bar{M}\_\mathrm{i} x\_\mathrm{i}\mathrm{d}n = 0 \\
 & \left[M - \sum\_\mathrm{i} x\_\mathrm{i}\bar{M}\_\mathrm{i} \right] \color{blue}{\mathrm{d}n} +
 \left[\mathrm{d}M -
 \left(\frac{\partial M}{\partial p}\right)\_{T} \mathrm{d}p - 
 \left(\frac{\partial M}{\partial T}\right)\_{p} \mathrm{d}T -
 \sum\_\mathrm{i} \bar{M}\_\mathrm{i} \mathrm{d}x\_\mathrm{i} \right]\color{blue}{n} = 0 \tag{3}
\end{align}
For a system of a total amount $n$ and a total variation of $\mathrm{d}n$, two equalities rise in Eq. (3) in order to be zero. The second one, with a bit more of manipulation, gives the famous Gibbs-Duhem equation. The first one is the **summability relation**
$$ M = \sum\_\mathrm{i} x\_\mathrm{i}\bar{M}\_\mathrm{i} \tag{4} $$
This equation is also simple to understand. It is a connection between the property of the solution $M$, and all the partial molar properties. For example, if we have access in a mixture of hexane$(1)$+chlorobenzene$(2)$ to $\bar{V}\_1$ and $\bar{V}\_2$, then applying Eq. (4) will give us the molar volume of the solution in the whole range of composition, by simple algebra.


**3. Gibbs theorem** It would be nice if we could get values for the partial molar properties. There are two approaches to this. The first one is going to the laboratory, and the second one is to impose a model behavior for the fluid. For the ideal gas model, we can get all the molar properties with the addition of Gibbs theorem:




---


A partial molar property (with the exception of molar volume) of a species in a mixture of ideal gases at a pressure $p$ and $T$, is equal to the molar property as a pure ideal gas at temperature $T$ **but** at the partial pressure in the mixture.




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Mathematically, this is
$$ \bar{M}\_\mathrm{i}^\mathrm{ig}(p,T) = M\_\mathrm{i}^\mathrm{ig}(p\_\mathrm{i},T) \hspace{1 cm}
 M \neq V \tag{5} $$


**4. Enthalpy of mixing** Now we apply Eq. (5) for the enthalpy. But this case is really easy, because for ideal gases, the enthalpy of a pure species is **independent** of pressure. It is only a function of temperature, so we can evaluate the pressure at any point we want and the result will not change
$$ \bar{H}\_\mathrm{i}^\mathrm{ig}(p,T) = H\_\mathrm{i}^\mathrm{ig}(\color{blue}{p\_i},T) = 
 H\_\mathrm{i}^\mathrm{ig}(\color{blue}{p},T) \tag{6} $$
Combining Eqs. (4) and (6) when $M=H$
\begin{align}
 H^\mathrm{ig} &= \sum\_\mathrm{i} x\_\mathrm{i} \bar{H}\_i^\mathrm{ig} \\
 H^\mathrm{ig} &= \sum\_\mathrm{i} x\_\mathrm{i} H\_\mathrm{i}^\mathrm{ig}(p,T) \\
 H^\mathrm{ig} - \sum\_\mathrm{i} x\_\mathrm{i} H\_\mathrm{i}^\mathrm{ig}(p,T) &= 0
 \rightarrow
 \boxed{\Delta\_\mathrm{mix}H := H^\mathrm{ig} - \sum\_\mathrm{i} x\_\mathrm{i} H\_\mathrm{i}^\mathrm{ig}(p,T) = 0}
 \tag{7}
\end{align}
The left side in Eq. (7) is the **definition** of the enthalpy of mixing, and for ideal gases it is equal to zero. In fact, for any property $M$, its value of mixing is **defined** as
$$ \boxed{\Delta\_\mathrm{mix} M = M -
 \sum\_\mathrm{i} x\_\mathrm{i} M\_\mathrm{i}(p,T)} \tag{8} $$


As you see, for a mixing process, the change in the property is not just some typical subtraction like "$H\_2 - H\_1$".


You had some doubrs about the pressure in a process of mixing. I think Eq. (8) will clear them. For example, if you *could* have an enthalpy-meter, you would have to:


1. Measure each enthalpy of the pure species to get the values of $H\_\mathrm{i}(p,T)$. This measurement must be done with every species at $p$ and $T$.
2. Measure the quantities being mixed, so you know also the values of the $x\_\mathrm{i}$'s.
3. Mix the species, so you get the solution, and measure the value $H(p,T, x\_1, x\_2, ..., x\_N)$. This measurement, **also**, needs to be done with the solution at $p$ and $T$.


Thus, a mixing process is just made by changing the different amounts of the species. The pressure, **regarding** the enthalpy of mixing, refers to the same value before and after. You can imagine a process of mixing where the pressure may not be constant. However, if you can guarantee experimentally that the final state of the solution is pressure $p$, your results will still be accurate because the enthalpy is a state function.


Typically, heat experiments are done to get enthalpy data of solution, and volumetric experiments are done to get molar volume data of solution. These are the easiest ways to calculate partial molar properties.


You may ask why we arbitrarily relate partial molar properties with pressure $p$ and $T$. We could have chosen molar volume $V$ and entropy $S$ to define the partial derivative in Eq. (1). This is because of practical convenience, since $p$ and $T$ can easily be manipulated experimentally.


As an exercise, you can do the same with the molar volume $V$, and reach that it is also zero. For the Gibbs energy $G$ and entropy $S$ you will not get a zero value.


**References**


* **J. M. Smith, H. C. Van Ness, M. M. Abbott**, "Introduction to Chemical Engineering Thermodynamics", 7th ed., Mc-Graw Hill, 2005.
* **J. M. Prausnitz, R. N. Lichtenhaler, E. G. de Azevedo**, "Molecular Thermodynamics of Fluid-Phase Equilibria, 3rd ed., Prentice Hall PTR, 1999.


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