# What is the entropy of mixing a gas into the whole atmosphere?

Suppose I release 1 mole of pure argon into the atmosphere. What is the entropy of mixing? The equations I have found seem to only apply to the case of a fixed volume. Should I use an estimate of the volume of the whole atmosphere? Some sort of infinite approximation?

Performing an actual calculation could be quite complicated, even if we use the ideal gas approximation, because the atmosphere has enormous variations in both temperature and density.

To simplify the problem, let's think of the atmosphere as an ideal gas of uniform temperature and density. Then we need to determine how the entropy of the argon released into the atmosphere increases as a result of this release.

Since we're doing an ideal gas approximation, the argon doesn't "know" about the other atmospheric gases – the only thing that affects its entropy is the increase in volume. But, when doing an entropy calculation, we do need to account for the fact that the atmosphere already contains 0.93% Ar. Effectively, the presence of the existing argon removes a fraction of the configurational possibilities, since if we replace argon with argon, we don't change the entropy. Hence the entropy increase is calculated from the effective configurational volume that remains for the argon relased from the box.

Specifically, for the purpose of calculating entropy, the effective final volume for the argon released from the box is:

$$V_\mathrm f = V_\text {atmosphere} - V_\text{Ar in atmosphere}+ V_\text {Ar in box}\approx V_\text {atmosphere} - V_\text {Ar in atmosphere} = (1- x_\text {Ar})V_\text {atmosphere},$$ where $$x_\text {Ar}$$ is the mole fraction of argon in the atmosphere, which is 0.93% = 0.0093.

Proceeding, we start with the following general expression for a pure substance (again, the ideal gas approximation allows us to ignore the non-argon molecules):

$$\left(\frac{\partial S}{\partial V}\right)_{\!T} =\frac{\alpha}{\kappa},$$ where $$\alpha$$ is the coefficient of thermal expansion, and $$\kappa$$ is the isothermal compressibility.

For an ideal gas, $$\alpha = \frac{1}{T}$$ and $$\kappa = \frac{1}{p}$$, hence:

$$\left(\frac{\partial S}{\partial V}\right)_{\!T} =\frac{p}{T}$$

If we additionaly assume the expansion is isothermal (i.e., T of the box is the same as that of the atmosphere--which we've already assumed is of uniform T) we obtain, for an ideal gas:

$$\mathrm dS = \frac{p}{T}\,\mathrm dV = \frac{n R}{V}\,\mathrm dV \Rightarrow \Delta S = n R \int\limits_{V_\mathrm i}^{V_\mathrm f}\frac{\mathrm dV}{V} = nR \ln\frac{V_\mathrm f}{V_\mathrm i}=nR \ln\frac{0.9907 \,V_\text {atmosphere}}{V_\mathrm i}$$

where $$V_\mathrm i$$ is the initial volume available to the argon, and $$V_\mathrm f$$ is the final volume available to the argon.

As you can see, releasing a mole of argon into the atmosphere will result in a significant entropy increase, once the argon is fully dispersed.

Note that, in the case where the atmosphere is 100% Ar: $$V_f = V_\text {atmosphere} - V_\text{Ar in atmosphere}+ V_\text {Ar in box}=V_{\text{Ar in box}} \Rightarrow \frac{V_f}{V_i}=1\Rightarrow \Delta S = 0, \text{ as expected.}$$

We can also calculate the corresponding free energy change using:

$$\Delta G = \Delta H - T\,\Delta S$$

With the ideal gas approximation, there is no enthalpy of mixing, hence:

$$\Delta G = -T\,\Delta S = -nRT \ln\frac{V_\mathrm f}{V_\mathrm i}.$$

If we assume STP ($$273.15\ \mathrm K$$, $$1\ \mathrm{bar}$$), and take the volume of the atmosphere to be the area of the earth's surface times, say, 50 kilometers*, then we have:

$$\Delta S = (1 \ \mathrm{mol})\ R\ \ln\frac{2.53 \times 10^{22}\ \mathrm L}{22.7\ \mathrm L} = 0.403\ \frac{\mathrm{kJ}}{\mathrm K}$$

$$\Delta G = - (1 \ \mathrm{mol})\ R\ (273.15\ \mathrm K) \ln\frac{2.53 \times 10^{22}\ \mathrm L}{22.7\ \mathrm L} = -110\ \mathrm{kJ}$$

*This is all arbitrary, since we're unrealistically assuming constant pressure and temperature; but I needed to pick a number for the calculation, so I chose the top of the stratosphere ($$50\ \mathrm{km}$$). $$94\ \%$$ of the atmosphere's gases are found below this altitude, so about $$94\ \%$$ of the argon would end up there as well.

Then again, because of the logarithm, the choice of atmospheric volume doesn't significantly affect the calculation, as you can see if you compare my answer with Buck Thorn's. Indeed, even if we allow the argon to expand to fill the observable universe ($$4 \times 10^{83} L$$), this only increases $$\Delta S$$ and $$\Delta G$$ by a factor of four! [$$\frac{83}{22}\approx 4$$.]

Calculation done in Mathematica (in Mathematica, ln is represented by "Log"): • Doesn't the atmosphere's initial concentration of ~1% play a role in approximating this? Intuitively it seems that it should reduce it considerably, for example if nitrogen was being released I would expect a lower number? – Doug McClean Mar 1 '20 at 13:19
• @DougMcClean That's a good point (& I upvoted your comment). It seems there should be some effect. For instance, suppose you had a, say, 10 L box of pure Ar at STP, and were standing inside an atmosphere-sized box of pure Ar at STP. If you opened the box, what would be $\Delta S$? The answer is zero. I.e., you could easily reverse the process by simply closing the box, which would put you back to your initial conditions. If I have a chance, I'll think about how to calculate the effect of 1% of Ar. In the meantime, I'll edit my question to include the assumption of no Ar in the atmosphere. – theorist Mar 1 '20 at 16:34
• I believe what you are referring to is a form of Gibbs' paradox: en.wikipedia.org/wiki/Gibbs_paradox – Buck Thorn Mar 1 '20 at 17:27
• @DougMcClean I've addressed your question about the effect of the ~1% Ar content of the atmosphere. As you can see, it makes little difference in the calculation. – theorist Mar 5 '20 at 6:05

To compute the entropy change, assuming the process is isothermal, you need to know the initial volume of the gas, in addition to the approximate final volume. The entropy of mixing in this case is nearly identical to the entropy change for the dispersed argon, since the volume increase for the atmospheric gases is negligible.

If one assumes the atmosphere has a thickness $$\mathrm{d}R$$ then its volume is $$\approx 4\pi R^2 \mathrm{d}R$$. The value of $$\mathrm{d}R$$ to use is unclear since the atmospheric gas concentration varies due to thermal and gravitational effects. One could use the boundary at which atmospheric effects become noticeable due to the presence of gas, $$\mathrm{d}R\approx \pu{120 km}$$ and a radius $$R\approx \pu{6365 km}$$, to estimate the atmospheric volume as $$\approx\pu{6.1e10 km^3}$$. Given the estimated mass of the atmosphere $$\pu{5.15e18 kg}$$ this results in an estimated average density of $$\approx\pu{0.084 kg/m^3}$$, well below the density of air at sea level ($$\approx\pu{1.2 kg/m^3}$$).

Using the above upper estimate of the atmospheric volume, we now need an estimate of the volume in the argon vessel. If the molar concentration is assumed identical in the atmosphere and in the argon vessel, and using an estimate of $$\approx\pu{29 g/mol}$$ for the average molecular weight of the atmospheric gas, then the initial volume of argon can be estimated as $$\approx\pu{ 0.024 m^3}$$ (approximately the ideal molar volume at STP). Using the standard formula for the molar entropy change of an ideal gas under an isothermal expansion gives

\begin{align} \Delta _{\text{mix}} S_\text{m} &= R \log\left(\frac{V_\text{f}}{V_\text{i}} \right) \\ &= \pu{ 8.3 J/K } \log \left( \frac{\pu{6.1e10 km^3}}{\pu{ 0.024 m^3}} \right) \\ &=\pu{ 410 J/K }\end{align}

Alternately considering the estimated mass of the atmosphere and assuming the final density to be uniformly $$\pu{1.2 kg/m^3}$$ then the relevant final volume is $$\approx\pu{ 4.3e9 km^3}$$. Then the standard formula for the entropy change gives

\begin{align} \Delta_{\text{mix}} S_\text{m} &= \pu{ 8.3 J/K } \log \left( \frac{\pu{ 4.3e9 km^3}}{\pu{ 0.024 m^3}} \right) \\ &=\pu{ 390 J/K }\end{align}

Clearly the very different estimates of the final volume of the atmosphere do not result in a great difference in the final estimated entropy change. You can thank the logarithm for that.

• Note a slightly smaller estimate of the final volume ($\pu{4.1e9 km3}$) is obtained if the final volume is limited to a layer 8 km thick. – Buck Thorn Jun 28 '20 at 8:39