Video: Partial Pressure of a Gas

Air is 21% oxygen. Find the minimum atmospheric pressure that gives a relatively safe partial pressure of oxygen of 0.16 atm. What is the minimum pressure that gives a partial pressure of oxygen above the quickly fatal level of 0.060 atm?

08:15

Video Transcript

Air is 21 percent oxygen. Find the minimum atmospheric pressure that gives a relatively safe partial pressure of oxygen of 0.16 atmospheres. What is the minimum pressure that gives a partial pressure of oxygen above the quickly fatal level of 0.060 atmospheres?

So in this question, weโ€™ve been told firstly that air is 21 percent oxygen. What we need to do is to find the minimum atmospheric pressure that gives a relatively safe partial pressure of oxygen of 0.16 atmospheres. Then we need to find the minimum pressure that gives a partial pressure of oxygen above whatโ€™s known as the quickly fatal level of 0.060 atmospheres. To do this, we first need to have a look at Daltonโ€™s law of partial pressures. Daltonโ€™s law tells us that the total pressure of a mixture of gases ๐‘ sub total is equal to the sum of all the partial pressures of the gases that make up that mixture. So in this equation, weโ€™ve got ๐‘› gases making up the mixture. And the total pressure of this mixture is the sum of the partial pressure of the first gas plus the partial pressure of the second gas plus so on and so forth until we get to the partial pressure of the final gas, the ๐‘›th gas.

Now we take each one of these gases in the mixture to be an ideal gas, which means that they follow the ideal gas equation. The ideal gas equation tells us that the pressure exerted by the gas multiplied by the volume occupied by the gas is equal to the number of moles of gas that we have multiplied by the molar gas constant multiplied by the temperature of the gas. Also as a fun thing, you could keep a tally of the number of times I say gas in this video because I have a feeling thatโ€™s gonna be a fair few times. Anyways, so we can apply the ideal gas equation to each one of these partial pressures here. To do this, we need to rearrange the ideal gas equation first to isolate the pressure. And we can do this by dividing both sides of the equation by the volume ๐‘‰. This gives us that ๐‘ is equal to ๐‘›๐‘…๐‘‡ over ๐‘‰. And then we can take this expression and substitute it in into Daltonโ€™s law.

So instead of ๐‘ sub one, ๐‘ sub two, and ๐‘ sub ๐‘›, weโ€™ve substituted it with the right-hand side of this equation: ๐‘›๐‘…๐‘‡ over ๐‘‰ sub one and so on and so forth. So in other words, this is the partial pressure of the first gas, this is the partial pressure of the second, and so on and so forth, and this is the partial pressure of the ๐‘›th gas. Now in this question, weโ€™re studying the atmosphere; weโ€™re studying the air. So we can make a few assumptions that are fairly accurate. Firstly, we can assume that the gas is a well enough mix together that the temperatures of all the gases is the same, because letโ€™s say weโ€™re looking at a certain region of air; letโ€™s say weโ€™re looking at some air in this arbitrary block that Iโ€™m drawing on the screen right now.

Letโ€™s say that that particular region of air consists of three different kinds of gases, and letโ€™s say that this region represents the atmosphere fairly well. So weโ€™ve got the gas thatโ€™s made up of orange particles, the one thatโ€™s made up of pink particles, and the one thatโ€™s made up of green particles. And we can see that theyโ€™re fairly well mixed together. So the temperatures of all the gases should be the same, because thereโ€™s no particular way to increase the temperature of one of these gases in the mixture without increasing the temperature of the others. After all, a temperature change comes about due to the change in kinetic energy of the particles. So say we could increase the temperature of one of the gases somehow. Letโ€™s say we could do it for the orange particles. Well then those orange particles would have larger kinetic energies, so they would be flying about a lot more. But because the orange gases mix together so well with the other gases, these orange particles that are now flying about at larger speeds would collide with the other particles thus transferring some of the kinetic energy to the other particles.

Therefore, assuming that this gas is in equilibrium, which it should be, the temperatures of all these gases should be the same. So we can safely say that the temperatures of all these gases is the same: ๐‘‡ one is equal to ๐‘‡ two is equal to so on and so forth is equal to ๐‘‡๐‘›. And we can just call this ๐‘‡ to save us writing all the subscripts. Now we can apply the same kind of logic to the volumes occupied by these gases. Weโ€™re imagining a chunk of atmosphere here when weโ€™re looking at the blue box, but equally we could consider the entire atmosphere if we wanted to. Again, weโ€™re assuming that all these gases are well mixed and spread out over the entire atmosphere. So this little box that weโ€™re looking at could be drawn in any part of the atmosphere, and it should still be a good representative of the entire atmosphere. In other words, because each one of these gases is well distributed over the entire volume, each gas must therefore occupy the same volume, regardless of whether weโ€™re looking at once again a little box or if weโ€™re studying the entire atmosphere.

Basically what we can say then is that this time the volume occupied by each one of these gases is the same, and weโ€™ll call this ๐‘‰ to save us writing the subscript once again. Now of course the value of ๐‘… is just a constant. So ๐‘… sub one is equal to ๐‘… sub two is equal to ๐‘… sub ๐‘›. But thatโ€™s a given anyway because again itโ€™s the molar gas constant. So itโ€™s the same for all gases, which means that we can actually factorise this expression here, which means that we can simplify the ๐‘ subtotal expression that weโ€™ve got on the left by saying that ๐‘ subtotal is equal to ๐‘› one multiplied by ๐‘…๐‘‡ over ๐‘‰ because as weโ€™ve said earlier the values of ๐‘‡, ๐‘‰, and ๐‘… are the same for all gases. And so weโ€™ve got ๐‘› one multiplied by ๐‘…๐‘‡ over ๐‘‰ plus ๐‘› two multiplied by ๐‘…๐‘‡ over ๐‘‰ plus so on and so forth until we get to ๐‘› sub ๐‘› multiplied by ๐‘…๐‘‡ over ๐‘‰.

Oh and by the way, the larger ๐‘› represents the number of moles of gas, whereas the subscript ๐‘› just represents that itโ€™s the ๐‘›th gas in the mixture. By context, we should realise the these are not the same thing. Anyway, so at this point, we can factorise this expression. We can say that this expression tells us that ๐‘ subtotal is equal to ๐‘…๐‘‡ over ๐‘‰ multiplied by ๐‘› one plus ๐‘› two plus so on and so forth until we get to the final gas. In essence, the partial pressure of a gas is only dependent on the number of moles of that gas within the mixture, or it only depends on how much of that gas there is in the mixture. Therefore, if weโ€™ve been told that air is 21 percent oxygen, then the partial pressure of oxygen will be 21 percent of the pressure of the air. So letโ€™s say for our expression that ๐‘› sub one is the number of moles of oxygen that we have. In other words, the subscript one refers to oxygen. Maybe the subscript two could refer to like another gas in the atmosphere such as nitrogen. Three could be something like carbon dioxide, and so on and so forth. But letโ€™s say that ๐‘› sub one represents the number of moles of oxygen.

So we can then say that ๐‘ sub one, the partial pressure of oxygen, is equal to 0.21 times the total pressure. Because as weโ€™ve just said, itโ€™s 21 percent of the total pressure. And once again, the reason for that is that the partial pressure of any gas only depends on how much of that gas there is in the total mixture. So using this expression here, we can calculate the total pressures that weโ€™ve been asked to find in the question. To do this, we need to rearrange first to say that the total pressure is equal to ๐‘ sub one, the partial pressure of oxygen, divided by 0.21. Now in the first instance, weโ€™ve been told that the partial pressure of oxygen is 0.16 atmospheres. So the total pressure, which is also the minimum atmospheric pressure, as weโ€™re trying to find is equal to 0.16 atmospheres divided by 0.21. And so we find that to two significant figures, the total pressure is 0.76 atmospheres. And we give our answer to two significant figures because all of the values that weโ€™ve used so far in the calculation have been given to two significant figures. So we can write our answer down here.

Now we can move on to the second part of the question. This time weโ€™re looking to find the minimum pressure that gives a partial pressure of oxygen of 0.060 atmospheres. And this partial pressure is known as the quickly fatal level, because any partial pressure of oxygen below this will be quickly fatal. And so once again we find that the total pressure is 0.060 atmospheres divided by 0.21, which once again to two significant figures happens to be 0.29 atmospheres. And at this point, weโ€™ve answered our question completely.

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