Problem28
Section 13.3: The Ideal Gas Law

Question

Calculate the number of moles in the 2.00-L volume of air in the lungs of the average person. Note that the air is at (body temperature).

Question by OpenStax is licensed under CC BY 4.0

Final Answer

0.0786 mol0.0786\textrm{ mol}

Solution video

1:36

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Video Transcript

This is College Physics Answers with Shaun Dychko. We're going to calculate the number of moles of air molecules in the lungs of a person. So the temperature in the lungs is the core body temperature 37.0 degrees Celsius which we have to convert into kelvin in order to use it in our ideal gas law formula. So we add 273.15 to get a temperature of 310.15 kelvin. The volume we're told is two liters which we have to convert into cubic meters because we always want MKS units for formulas - meters, kilograms, and seconds. So we multiply by one cubic meter for every 1000 liters which is two times ten to the minus three cubic meters. The pressure we assume is about one atmosphere. If it was more than an atmosphere then they would be exhaling and if it was less than an atmosphere they'd be inhaling. So this is just the point between exhaling and inhaling which is the lungs are just filled up. And so the pressure is one atmosphere which we need to write in MKS units which is 1.013 times ten to the five newtons per square meter. So pressure times volume equals the number of moles times the universal gas constant times temperature and we can solve this for N by dividing both sides by R T So N is P V divided by R T. So that's 1.013 times ten to the five newtons per square meter pressure times two times ten to the minus three cubic meters ivided by 8.31 joules per mole kelvin and multiply that by 310.15 kelvin, and that's 0.0786 moles.

Comments

hayrogers5·
Hi, When presented with this question. How do you know that 1 atm is the pressure of lungs. Is this something that you are supposed to just know? When I set up my formula, since the question doesn't say anything about pressure, how do you know to assume the pressure is 1 atm?
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ShaunDychko·
Hello, good question. At 0:41 in the video I explain that if it was more than 1 atm then they would be exhaling, whereas if it was less than 1 atm then they would be inhaling and the question implies that the air is stationary in the lungs, in which case it must be at 1 atm. All the best, Shaun
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hayrogers5·
Hi, When presented with this question. How do you know that 1 atm is the pressure of lungs. Is this something that you are supposed to just know? When I set up my formula, since the question doesn't say anything about pressure, how do you know to assume the pressure is 1 atm?
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kkm_2009·
Hi, am i missing something? In the equation you are multiplying R and T but when you entered it in the calculator you divided by T?
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ShaunDychko·
Hi kkm, thank you for the question. The issue is a personal preference about how to type this into the calculator. To save some button pushing I don't usually enclose the denominator in brackets. If there were brackets for the denominator, then you would see R and T _multiplied_. However, without brackets, and knowing the calculator proceeds from left to right, I divide by each factor in the denominator, one after the other. As an example to explain: 12 / (6 \* 2) = 12 / 6 / 2 = 1. For 12 / 6 / 2, the calculator evaluates 12 / 6 first, which is "2", and then proceeds to divide that result by 2, giving the answer 1 We're on the same page, but just enter it differently in the calculator. All the best, Shaun
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CRainer·
Hi Shaun! Thank you for answering my previous question, but I am still confused with when to use 1 ATM vs 1.013x105 N/m2 as in the previous question you said use 1 ATM but now are saying to use MKS units and shouldn't use 1ATM. How am I supposed to choose which unit to use?
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ShaunDychko·
Hi Christina, Just for reference for other students, your original question was at https://collegephysicsanswers.com/openstax-solutions/suppose-gas-filled-incandescent-light-bulb-manufactured-so-gas-inside-bulb. The question is how to know which units to use. "atm" vs "N/m^2". It's always correct to use "N/m^2", so if you want a single never-fail rule, then go with "N/m^2". However, it's easier to conceptualize a pressure expressed in units of "atm" since we can say "ah, so 1.5 atm is one and a half times the air pressure I experience at sea level". We have experience with "atm", so it's nice, although not any more correct, to use "atm" when possible. You'll see many of the solutions here using "atm". So the question becomes - when is it OK to use "atm" units? The answer is that one needs to look at the formula and see whether a factor in units of "atm" is being combined with any other number with different units. Often, as in problem 24, the pressure, in "atm", is being multiplied by a dimensionless factor (it has no units) in which case "atm" is fine. It was being multiplied by Kelvin divided by Kelvin, which therefore has no units since they cancel. Hope this helps, Shaun
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mandyjohnsonhulsey·
The answer you show is wrong from the video... .0786 in video but solution give is .786 mol big difference.
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ShaunDychko·
Thank you so much @mandyjohnsonhulsey! I have made the correction.
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