OpenStax College Physics for AP® Courses, Chapter 13, Problem 34 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. A high-pressure gas cylinder has a volume of 50 liters and it's full of a toxic gas and it suddenly has a small leak out of a valve. The initial pressure in it is 1.4 times 10 to the 7 newtons per square meter and we assume that this pressure is an absolute pressure. So we can plug it directly into our ideal gas law formula. And we make that assumption because in Part C it says: to what temperature must have tank be cooled to reduce the pressure to 1 atmosphere? And so if we were talking about gauge pressure then that would mean what temperature should the tank be cooled to reduce the pressure to zero gauge pressure which would be one atmosphere absolute pressure. So that's what I'm using to make the assumption that this number given is an absolute pressure. So the initial temperature is 25 degrees Celsius which we have to convert into kelvin by adding 273.15 and we do the same to the final temperature and this works out to 298.15 kelvin initial temperature and 194.65 kelvin final temperature. The initial volume we convert into cubic meters by multiplying by one cubic meter for every 1000 liters. So in Part A it says: what is the final pressure in the tank assuming that not much gas leaks out? So, initially, we have initial pressure times volume which does not get a subscript because the tank's volume is the same in both cases. The tank is not being squished or expanding. So there's no need for a V I in a V F here. And initially we have P I V equals the initial number of moles times the ideal or the universal gas constant times the absolute temperature initially. And we have P F V equals NFRTF in the final case. Now since we're told that the leak loses a negligible amount of gas that means that the initial number of moles is approximately equal to the final number of moles, and so we'll just call it n with no subscript. So we divide these corresponding sides of these two equations because we're solving for P F here and we'll substitute N in place of N F and N I. And so we have the V's canceling and we have the NR's canceling, and then we multiply both sides by the initial pressure. And we find that the final pressure then is the initial pressure times the final temperature divided by the initial temperature. So that's 1.4 times 10 to the 7 newtons per square meter times 194.65 kelvin divided by 298.15 kelvin and this is 9.14 times 10 to the 6 newtons per square meter. Now in Part B we're told to suppose that some significant amount of gas does escape - suppose one tenth of it escapes - in that case the number of moles in the final case is the initial number of moles minus one tenth of it that has disappeared out of the tank leaving us with nine tenths left over which is 0.9. So N F is 0.9 N I. So making the same sort of ratio between these two sides of these formulas here we have P F V equals N F which is 0.9 times N I times R T F and dividing by P I V equals N I R T I. And so we have these common factors canceling here and then multiplying both sides by P I we get that final pressure is 0.9 times initial pressure times final temperature over initial temperature. So we expect an answer of 0.9 times the answer we got in Part A. But I just plug in all the numbers given to us in the question rather than using an answer from a previous part since we may have made a mistake in this calculation and we don't want that mistake to ruin our answer for the next question too. So this all works out to 8.23 times 10 to the 6 Pascal's, Pascal's being an abbreviation for newtons per square meter. Now in Part C were asked: to what temperature must the tank be cooled to reduce the pressure to one atmosphere? And we're assuming that not much gas has leaked out so that in that case N I and N F are both the same so we just call it N. And so we're dividing the two sides again but we're putting this side over on the left here to begin with because we know we want to solve for T F. So we have N R T F divided by N R T I equals P F V divided by P I V. The NR's cancel and then we multiply both sides by initial temperature. The V's also cancel and we are left with final temperature is initial temperature times the ratio of final to initial pressure. So that's 298.15 kelvin times one atmosphere is the final pressure we're told and convert it into Pascal's by multiplying by 1.013 times 10 to the 5 Pascal's per atmosphere divided by 1.4 times 10 to the 7 newtons per square meter initial pressure. And this is 2.16 kelvin. And Part D asks us: is it a practical solution to cool the tank to this temperature to prevent leaking? Because once the tank has an absolute pressure of 1 atmosphere there would be no gas leaking out because the pressure will be the same as the atmospheric pressure pushing in. And it's not practical though to cool such a large mass to 2.16 kelvin. So it's not a practical solution.