Combined Gas Law - example problems"},"lengthSeconds":"1457","ownerProfileUrl":"http://www.youtube.c

Okay, so we look at Boyle's law, pressure and volume, Charles' law, volume and temperature, Guy Lusak's law with our pressure and temperature, and now we're going to put the three together and it's called the

combined

p gas law, v and t, all three are equal to constant but in this case I can change two of them and not just one. I can change two of them and then look and see how the third variable is affected. The graph will always have this type of shape, but at different temperatures, the whole thing. The graph shifts, we call these isotherms, so along this mark basically pressure and volume we have the same temperature. what we have is constant and just like the previous one, since we have temperature, we will still have to be centigrade plus 273 to get our kelvin temperature, the thing is that we can change more than one thing, so it is not that easy. all the time to see how the final answer will play out, so we have a balloon filled with gas that is 30 liters, what would that volume be?

So let's call our v1 312 k our temperature t1 and we have a pressure p1 of 162.1 kpa, okay and then it says what the volume would be, so it's going to be a v2 at standard temperature and pressure stp. Now the only way this works is that I have five numbers and I solve the sixth. I have six things in total. together, so I have to have five of them, you could say well, I only have three stp numbers, however, they are two stp numbers, what is the standard temperature, the standard temperature is zero degrees Celsius, the standard pressure is a atmosphere, okay, that's what stp is, however, this problem? well we always have to change the temperature to kelvin so I'm not going to use degrees centigrade not to mention dividing by zero wouldn't work that needs to be changed to add my 273 so my temperature here will be 273 k and if you remember. from our conversion sheet, one atmosphere is equal to 101.325 kpa, I don't have to do much as far as conversions because it is one atmosphere, if it was 2.3 atmospheres then I would have to calculate how many kpa it was, but since it is only one I can go directly to it so that then it becomes my t2 and this becomes my p2, so I actually have five numbers even though they weren't actually written in the problem, so I write my equation p1 v1 over t1 p2 v2 over t2, go ahead. and connect everything so that p1 162.1 kpa my v1 30 liters t1 312 k equals two my new pressure 101.325 kpa I don't know my new volume and I have 273 k for my new temperature, so five numbers, a variable yet to be found , how can I do it? solve something like this well since this is a fraction to get rid of the fractions I need to multiply by the reciprocal and I have to do it on both sides so here on this side the 273 cancel out the 101 cancel out the only thing that goes is what's left of that side is the v2 over here the kelvins will cancel the kpas will cancel and the only unit that will be left will be liters the l times liters now this is all multiply and divide so it doesn't matter which one you do first so 273 multiplied by 162.1 multiplied by 30 we divided by 101,325 we divided by 312 and I ended up with my v2 equal to basically 42 liters, the result was 41,995.

I can round that up to 42 liters, so my first answer was 42 liters, so that's what it takes to solve a

combined

gas law problem, put in all five, solve the missing one. Any questions about that one, well, problem two will basically do the same thing, but this time they gave me the five numbers, so make it a little easier. here is my first pressure 150 kpa 27 degrees celsius so I will have to convert it to kelvin first volume 4 liters new pressure 600 kpa new temperature 127 degrees celsius find the new volume, very similar to problem one, but I can see everything five numbers and another time like before 27 degrees plus 273 that's 300k so that will be my t1 my t2 will be 127 degrees plus 273 and that ends up being 400 400k write my equation connect them all so let's look at 150 kpa and 4 liters more than 300k and that will equal 600 kpa new v2 over 400 k again I want the v2 alone so on both sides multiply by the reciprocal of what is times the v2 400k over 600 kpa there 400k over 600 kpa on this side k's get The canceled kpas are cancelled. , leave me just in liters here, these things cancel out, so I'm just going to have my v2 and when I'm done with everything, basically 1,333 liters will be my v2, which is four thirds of a liter by the way. so v2 is equal to 1.333 leaders normal problem five numbers enter the five numbers solve the missing one stp problem you only see three numbers but actually there are still five numbers there the temperatures have to be in kelvin use the reciprocal to solve it okay, problem three It doesn't matter, I can solve for volume, I can solve for pressure, basically I'm going to do the same thing, here's my volume, pressure doesn't matter what units I'm in as long as they're the same, here's my first temperature, I have to change that to kelvin new temperature new volume find the new pressure so p1v1 over t1 p2 v2 over t2 let's start putting things together we have the first pressure 760 millimeters mercury first volume 15 liters temperature okay, add 273 here, that ends up being 240k, so ended here we have 240k okay now the other one well I don't know the new pressure p2 new volume 25 liters and new temperature well 87 87 degrees plus 273 is 360 k so here goes 360 k that is our t2 again we multiply both sides by the reciprocal to get the p2 by itself, so we're going to have 360 ​​k over 25 liters on both sides and when we go ahead and multiply everything and divide 360 ​​by 760 by 15, divide by 25 and divide by 240, we end up up with p2 equals 684 millimeters of mercury, so in that case I solved p2, any question is fine, how about we solve for temperature?

So we want to know a new temperature in degrees Celsius, it was in stp, so remember that the 240 milliliters is says it's stp, that means it's at zero degrees Celsius and one atmosphere, okay and then it goes to a new pressure of 1 .48 atmospheres and a new volume of 180 milliliters, so p1 v1 over t1 p2 v2 over t2 let's start plugging everything in. The first pressure was one. atmosphere first volume 240 milliliters no problem if it is in milliliters as long as both are in milliliters first temperature zero degrees Celsius 273 k remember that you always have to add 273 we cannot divide by zero anyway it is equal to k 1.48 atmospheres 180 milliliters and more t2 now, like in the previous

example

s, when we have something at the bottom that is not just multiplying division, okay, I need to do a cross multiplication, okay, so it will look like I'm basically going to have my atmosphere times 240 milliliters and then t2 equals at my 1.48 atmospheres 180 milliliters and then 273 k and now that I have the cross multiplied part all I have to do now is divide both sides by 1 atm of course dividing by 1 is not going to do anything and the 240 they do it. on this side, do it here again, one atmosphere and the 240 milliliters, so essentially I have 1.48 times 180 times 273 divided by 240 divided by 1.

So when you do all that, we'll end up with a t2 which is equal to 303 k but they don't want to know the temperature in kelvin, they want the temperature in degrees Celsius, so I still have to subtract 273 to put it back in degrees Celsius, so my final answer ends up being 30 degrees Celsius, so notice. how each of these, I'm tending to do them the same way, write your equation, enter all the numbers, sometimes I can just multiply or divide, sometimes I need to do the cross multiplication and then divide so the unknown is all by itself alone and now you have to make sure and read the problem did they want an answer in kelvin? they wanted an answer in degrees centigrade if I started in degrees centigrade I want to end in degrees centigrade if it says tell me in kelvin then I would leave it in kelvin so it depends on the last part.

Any questions about that had to ask a little more. Okay, the last one here doesn't have numbers, so like the previous ones, let's make some things up. I want to keep things as simple as possible to have volume. How would the volume change if the temperature of the air inside doubled while the pressure was reduced to half of what it was now? They didn't tell me anything. restrictions so you know what I'm going to do the temperature in kelvin makes it easier, I don't have to change it first and all it says is that the pressure needs to be reduced, so let's put some numbers in, okay t1, okay? we're already in kelvin, we want to double it, so let's say we start at 300k and then t2 we'll double it to 600k in kelvin, just double it, okay, we already know that if we're in degrees Celsius it doesn't quite work.

Okay, with the doubling, so let's leave it at Kelvin. What about the pressure? Alright. Hey, let's ease the pressure. Let's say it was two atmospheres and then we halved it, so now let's do the second atmosphere, which is always easier. because multiplied by one divided by one is not going to change things much, what we really want to know is what is going to happen to the volume, so let's start by saying that the volume was 10 liters. Well, 10 is a nice round number, now I have it. My five numbers and honestly guys it's not going to matter as long as I double whatever that is and whatever this is cut it in half and then we look at some round number for my volume just like we had p1 v1 over t1 p2 . v2 over t2 let's go ahead and plug all our stuff in so p1 two atmospheres v1 10 liters over 300k now we're going to go to p2 one atmosphere we don't know the new volume that's what we want to see how it changes and my new temperature 600k I want find the new volume so basically 600k in one atmosphere so this side cancels out completely but if I do that on one side I have to do that on the other side 600k on my 1 atmosphere and the reason I chose these numbers keep in mind what happened the 1 atmosphere well, that was canceled with the atmosphere it left me two 600k over 300k it leaves me two twice two times the ten is my v2 is 40 liters so it was a 10 it went to 40 Basically how changed by 4?

That's how the change was or we call it quadrupled, so if the temperature doubled and the pressure was halved, my volume quadrupled, so that's how it's a little more complicated when you combine it. gas law now the last thing I want to go over is combination, really when you think about it, the other three laws are all here because the other three laws had a particular condition, for

example

, start with the gas law of combination and go on with the temperature constant, okay, constant. temperature means that t1 is equal to t2 which cancels them out and now I have Boyle's law, p1v1 is equal to p2v2 if instead I said, hey, let's make it constant pressure, well, constant pressure means that p1 is equal to p2 and in that case, the pressures cancel out and I am left with hi-hats. law then the combined gas law actually has the other three laws every time I say this doesn't change or this is constant I throw out that letter so constant temperature I throw out the t's I keep Boyle's law constant pressure I take out the ps and I I stay with Charles's law Constant volume.

Take out the V's and I'm left with Guy Lucex's law, okay, any questions, okay.