The ozone concentration in the upper atmosphere averages $\pu{3.0 \times 10^13 molecules per cm3}$ in a region between $15$ and $\pu{35 km}$ high, at an average pressure of $\pu{0.001 atm}$ and temperature of $\pu{–25 °C}$. How thick would the ozone layer be, if these same molecules were at a pressure of $\pu{1.0 atm}$ and a temperature of $\pu{20 °C}$?

I understand that I need to use the ideal gas law, but I really have no idea how to approach the question, or even how to work with $\pu{molecules per cm3}$. I see that we have been given the values for temperature and pressure, but am unsure how to go about finding volume, and then the thickness.

I am also confused with how to deal with the two separate situations. Would you use $p = \pu{0.001 atm}$ or $p = \pu{1 atm}$. And would you use $T = \pu{248 K}$ or $T = \pu{293 K}$? And where do you find the value for the amount of substance?


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    $\begingroup$ Please share with us your thoughts and what exactly you find difficult in this problem. This network isn't about just doing the homework for you, but to help you to understand and proceed on your own when possible. $\endgroup$ – andselisk Nov 7 '17 at 5:59
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    $\begingroup$ Start with writing down the equation for ideal gas, then express the unknown entity (volume), unify the units, plug in the known values. Also, think about what formula from the geometry course can you utilize to find the thickness from the obtained volume. You can edit your question at any time and include your progress. $\endgroup$ – andselisk Nov 7 '17 at 6:15
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    $\begingroup$ Please be so kind and post the answer that whoever gave you the assignment provides. @andselisk and I are off by a factor of a thousand, and I believe the reason for that is that we interpreted the question differently. (If you are uncomfortable with the markup you can post a picture and notify me in the comments section of my answer, I'll rewrite it for you.) $\endgroup$ – Martin - マーチン Nov 7 '17 at 11:21
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    $\begingroup$ @Martin-マーチン The answer provided is 25 mm! Thank you so much for the help. $\endgroup$ – uyujc Nov 7 '17 at 11:59
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    $\begingroup$ You are welcome. Would it be possible for you to post the entire solution, it really hinges somewhat on the interpretation and I would like to see the whole explanation of it. If you found that exercise in a textbook, it would be nice if you could provide the source; thank you. $\endgroup$ – Martin - マーチン Nov 7 '17 at 12:02

Final conditions ($\pu{1.0 atm}$) give a hint at what altitude the thickness should be estimated – at the Earth's surface. Let's denote the unknown thickness as $h_x$, radius of the Earth as $r_\mathrm{E} = \pu{6371 km}$; also take $h_1 = \pu{15 km}$ and $h_2 = \pu{35 km}$ and put these dimensions on a rough 2D sketch:

enter image description here

Fig. 1. Schematic representation of two ozone layers at given altitudes.

To find $h_x$, we can use the volume of the new hypothetical ozone layer $V_x$ as the difference in volumes of two spheres with corresponding radii:

$$V_x = \frac{4 \pi (r_\mathrm{E} + h_x)^3}{3} - \frac{4 \pi r_\mathrm{E}^3}{3} \tag{1}$$

and express $h_x$ from (1) as follows:

$$h_x = \left(\frac{3V_x}{4 \pi} +r_\mathrm{E}^3\right)^{\frac{1}{3}} - r_\mathrm{E} \tag{2}$$

From ideal gas law for a relatively thin gas layer without temperature and pressure gradients:

$$V_x = \frac{nRT_x}{p_x} \tag{3}$$

where $n$ is the amount of gas in ozone layer, $T_x = \pu{20 ^\circ C}$ and $p_x = \pu{1 atm}$ (new conditions). Let's also denote ozone concentration as $N = \pu{3.0e13 cm-3}$, then total amount of ozone is to be found via initial volume $V_0$ and Avogadro constant $N_\mathrm{A}$ as

$$n = V_0 \times \frac{N}{N_\mathrm{A}} \tag{4}$$

Initial volume can also be found from Fig. 1 using the difference in volumes of the corresponding spheres with radii $r_\mathrm{E} + h_1$ and $r_\mathrm{E} + h_2$:

$$V_0 = \frac{4 \pi (r_\mathrm{E} + h_2)^3}{3} - \frac{4 \pi (r_\mathrm{E} + h_1)^3}{3} = \frac{4 \pi}{3}((r_\mathrm{E} + h_2)^3 - (r_\mathrm{E} + h_1)^3) \tag{5}$$

Now all we have to do is carefully plug in the values with the corresponding units carefully and find all the unknown variables. From (5):

$$V_0 = \frac{4 \pi}{3}((\pu{6371 km} + \pu{35 km})^3 - (\pu{6371 km} + \pu{15 km})^3) = \pu{1.03e10 km3} = \pu{1.03e25 cm3} \tag{6}$$

Now let's find the amount of ozone using $V_0$ and equation (4):

$$n = \pu{3.0e13 cm-3} \times \frac{\pu{1.03e25 cm3}}{\pu{6.023e23 mol-1}} = \pu{5.12e14 mol} \tag{7}$$

and, subsequently, the unknown volume $V_x$ using (3):

$$V_x = \frac{\pu{5.12e14 mol} \cdot \pu{8.314 J mol-1 K-1} \cdot \pu{293 K}}{\pu{101325 Pa}} = \pu{1.23e13 m3} = \pu{1.23e4 km3} \tag{8}$$

Finally, plugging all the values in (2), we can find the thickness:

$$h_x = \left(\frac{3 \cdot \pu{1.23e4 km3}}{4 \pi} + (\pu{6371 km})^3\right)^{\frac{1}{3}} - \pu{6371 km} = \pu{2.4e-5 km} = \pu{24 mm} \tag{9}$$

The result is somewhat close to the fact posted by NASA:

Around the world, the ozone layer averages about 3 millimeters (1/8 inch) thick, approximately the same as two pennies stacked one on top of the other.

  • $\begingroup$ When you forget to hit post and then someone else is faster... now I wonder why I have different results... well, I'm going to have to read your post then. $\endgroup$ – Martin - マーチン Nov 7 '17 at 10:51
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    $\begingroup$ @Martin-マーチン This site really needs a function that many messengers have: "X is typing..." :) You relied on ideal gas equation to determine the amount, whereas I ditched it and used ozone number concentration instead. For some reason I didn't find an idea of applying the ideal gas law when there is probably a reasonably high gradient of temperature and pressure appealing. $\endgroup$ – andselisk Nov 7 '17 at 10:59
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    $\begingroup$ Yeah, I agree. Initially I didn't want to bring the spherical shell into the answer, but then I thought I still should do that (then I was distracted, haha). I found that I am off by a factor of 1000. That is also the discrepancy that I found when I calculated the particle density from the ideal gas. Now that implies that this question is open for interpretation... gosh I hate when that happens. Anyway, you got my vote :D $\endgroup$ – Martin - マーチン Nov 7 '17 at 11:12
  • $\begingroup$ @Martin-マーチン +1 too, Chemistry Votes Exchange:) Maybe the OP eventually can comment on the situation or submit the work and get an answer (probably, a different one – that'd be fun). $\endgroup$ – andselisk Nov 7 '17 at 11:16
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    $\begingroup$ @Martin-マーチン When using ideal gas law for determining $n$, I also arrived at $\pu{23.8 m}$. Either way, I think both approaches are valid depending on the question's interpretation. $\endgroup$ – andselisk Nov 7 '17 at 11:33

The number concentration ($C$, or number density $\rho_N$) is not really relevant for this exercise. (For more on this, see at the bottom.)

You have correctly identified that you will need the ideal gas law (obviously as an approximation): $$pV = nRT$$

For the ozone layer you have been given the pressure $p_\text{layer}$, temperature $T_\text{layer}$, and the thickness $h_\text{layer}$: \begin{align} p_\text{layer} &= \pu{0.001 atm}\\ T_\text{layer} &= \pu{-25 ^\circ C} = \pu{248.15 K}\\ h_\text{layer} &= |\pu{15 km} - \pu{35 km}| = \pu{20000 m} \end{align}

You are asked to calculate the thickness $h_\text{std}$ at $p_\text{std} = \pu{1 atm}$ and $T_\text{std} = \pu{20 ^\circ C} = \pu{293.15 K}$ ($\text{std}$ is short for standard).
The exercise clearly states "same molecules", therefore $n_\text{layer} = n_\text{std} = n$. You can now assume that you are simply compressing the gas, assuming in both cases the same area $A$, therefore you can express the volume as $$V = A\cdot h.$$

With this you can form the following relations: \begin{align} p_\text{layer}Ah_\text{layer} &= nRT_\text{layer} &\Leftrightarrow&& \frac{p_\text{layer}h_\text{layer}}{T_\text{layer}} &= \frac{nR}{A}\\ p_\text{std}Ah_\text{std} &= nRT_\text{std} &\Leftrightarrow&& \frac{p_\text{std}h_\text{std}}{T_\text{std}} &= \frac{nR}{A}\\ \end{align}

Since the right sides equal each other you arrive at \begin{align} \frac{p_\text{layer}h_\text{layer}}{T_\text{layer}} &=\frac{p_\text{std}h_\text{std}}{T_\text{std}},\\ h_\text{std} &= \frac{p_\text{layer}}{p_\text{std}}\cdot \frac{T_\text{std}}{T_\text{layer}}\cdot h_\text{layer}. \end{align}

Fill in the known values and the thickness is $h_\text{std}\approx\pu{24 m}$.

What should be clear from that exercise is, that it is a very crude approximation. It does not take into account, that the layer is a spherical shell. In that case you would have to calculate the volume differently, and you would have to know $r_\text{earth}=\pu{6371 km}$ (according to a quick Google search): \begin{align} V_\text{layer} &= \frac{4}{3}\pi\left[ (r_\text{earth} + \pu{35 km})^3 - (r_\text{earth} + \pu{15 km})^3\right]\\ &\approx 4\pi (r_\text{earth}+\pu{35km})^2\cdot\pu{20 km}\\ &\approx \pu{1E10 km3}\\[2ex] V_\text{std} &= \frac{4}{3}\pi\left[ (r_\text{earth} + h_\text{std})^3 - r_\text{earth}^3\right]\\ &\approx 4\pi (r_\text{earth} + h_\text{std})^2h_\text{std}& (r_\text{earth}>>>h_\text{std})\\ &\approx 4\pi r_\text{earth}^2h_\text{std}\\ \end{align} And with that calculation we would get to the thickness being $h_\text{std}\approx\pu{23 m}$.

Since the ozone layer is in no way a pure substance, the ideal gas also is only a crude approximation as you are treating everything the same. That that is not the case is even given from the values of the exercise.

You can calculate the number concentration $C$ also from the ideal gas: \begin{align} && C &= \frac{N}{V} \\ && pV &= nRT \\ && N&= n\cdot N_\mathrm{A}\\[2ex] \implies&& C &= \frac{p\cdot N_\mathrm{A}}{RT}\\ &&C_\text{layer} &\approx \frac{\pu{0.001 atm}\cdot\pu{6.023E23 mol-1}}{ \pu{82.057 cm3 atm K−1 mol−1}\cdot\pu{248.15 K}}\\ &&C_\text{layer} &\approx \pu{3E16 cm-3} \end{align}

From the given number concentration of ozone $C_\text{ozone} = \pu{3.0E13 cm-3}$ you can see that only every thousandth molecule in that layer is ozone.


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