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Gay Lussac's Law Practice Problems

Gay Lussac's Law Practice Problems
okay so let's do a bunch of

practice

problems

with gay

lussac

's law just to review the laws written like this it assumes that some sort of change is taking place so we have p1 and t1 which are the initial pressures and temperature of the gases then a change occurs and then we have P 2 and T 2 which are the pressure and the temperature after this change okay so here's a

practice

problem the pressure and a sealed can of gas is 235 kilo Pascal's when it sets at room temperature 20
gay lussac s law practice problems
degrees Celsius if the can is warm to 48 degrees Celsius what will the new pressure inside the can be so we're going to use the equation and what I'm attacking these gas law

problems

the first thing that I like to do is make sure that I know what all my variables are okay so let's figure this out we have an initial pressure before the change so p1 is going to be 235 kPa so we know this guy up here all right when it sits at room temperature so this is the initial temperature 20
degrees Celsius okay and then the can is warm to 48 degrees Celsius so there's our change we go from 20 to 48 so our second temperature our final temperature is going to be 48 degrees Celsius so we know these guys and it's going to be p2 the new pressure that we're going to be solving for okay so here are variables now you may be thinking you may remember this whenever we use gases we need to convert these temperatures into Kelvin temperatures and we do that here by taking the
Celsius temperature 20 or 48 and adding 273 to it okay so now before we plug anything in let's get these Celsius temperatures into Kelvin okay so for t1 we're going to take 273 and we're going to add 20 to it and so now we're going to have 293 Kelvin for t1 and t2 is going to be 273 plus 48 so this is going to be 321 Kelvin okay so now that we've got these variables all set let's go about rearranging this equation to get p2 by itself okay so the first thing that I want to
do to get p2 by itself is get t2 here out of the denominator and I'm going to do that by multiplying both sides by t2 now I have t2 on the bottom here and on the top of this fraction so they both cancel out and I can rewrite this as t2 times p1 divided by t1 equals p2 now some of you guys might not like that p2 is on the right-hand side so I'll just flip it but really it's the same equation either way so this way we'll just say p2 equals t2 times p1 divided by t1 okay so there we
go now p2 is by itself and we're ready to plug in the variables okay so p2 equals t2 321 Kelvin times p1 235 kilo pascals divided by t1 which is 293 Kelvin okay so we go ahead we plug that into our calculators and the answer that we're going to get out I'm going to round this to three significant figures is 257 now what are the units I'm going to use well it depends on what cancels out here obviously we know this is going to be pressure so we could just say all right well you
gay lussac s law practice problems
know it's going to be kPa but let's just show how we'll do the unit cancels here we have Kelvin up here and calvin up here down i mean down here on the bottom okay so we're going to get rid of the kelvin and we're going to be left with kilo Pascal's which obviously is a unit of pressure so we know that our final answer is going to be 257 kilo pascals and again I rounded this answer my calculators but a very spit out a very long number for this but I rounded it to three
significant figures because there are three significant figures in each one of the numbers that I started with okay so that's how we do that let's look at some more I'm going to look at three

problems

total and they're each each one's going to be a little bit trickier than the one before it okay so here we are talking about a tire a car tire that has a pressure of two point three eight ATM at twenty five point three degrees Celsius if the pressure inside reaches four point oh
eight atm the tire will explode alright how hot will the tire have to be for this explosion to happen and then report the temperature in degrees Celsius okay so again we're going to be starting with this equation the first thing let's do is let's get our four variables straight so we're starting with this initial pressure and this initial initial temperature and then we're saying that we're going to going to raise the temperature which is going to cause this increase in
pressure okay so our initial pressure is going to be two point three eight atm and our initial temperature is going to be twenty five point three degrees Celsius but remember we're going to have to end up converting that into Kelvin in a minute before we can use it in the equation okay now if the pressure inside reaches four point oh eight so that's where the pressure is going to be after the change there's our p2 four point oh eight atm what's the temperature going to be how hot
would it have to be for the temp for the pressure to get that high so t2 is going to be the variable where himself or so we know p1 we know P two we know t1 and t2 is going to be the variable that we're solving for all right now again let's convert our Celsius temperature here that's t1 let's convert into Kelvin temperatures that we can use it for gases okay again we take 273 this is t1 we take 273 and we're going to add 25.3 to this okay now remember the rules about adding
gay lussac s law practice problems
and subtracting with significant figures we look which of the numbers has a few accessible places or the few significant figures on the right-hand side all right this has one decimal place 273 doesn't have any okay so that means I'm going to draw a line down here and I'm going to round to this digit I look over here to the three so the eight stays the same but remember we have to round using the addition and subtraction significant figures rule so two hundred ninety eight point three
cut it off here is going to turn into just 298 Kelvin and that is what I'm going to use 41 all right now solving this equation for T 2 is going to be a little bit harder because it's in the denominator I'm going to have to go through a number of steps okay so bear with me I'll explain each one then after I do this problem out completely in the next problem I'll show you a shortcut it's it's probably shortcut that your teacher your textbook hasn't showed you but
it's a way to take all these steps I'm about to do and condense them into two all right but first I want to show you how to do it on sort of the long way the correct way okay so t2 first thing I want to do is I want to get it out of the bottom I can't solve for it if it's in the bottom so I'll get it out of the bottom by multiplying both sides by t2 okay so now it's on the bottom here it's on the top here so it cancels out and I can rewrite this as t2 times p1 divided
by t1 equals p2 okay now it's on the top but it's far from a loan and we want to get it alone in order to solve for it okay so I want to get p1 and t1 both on the other side I don't know it doesn't really matter what we do but let's divide both sides by P one just for the heck of it p1 here and divide this by p1 here put p1 on the bottom so now the p1 over the p1 they both cancel out and I can rewrite this as t2 divided by t1 equals p2 divided by p1 we're almost there one
more step to get t2 by itself gotta get this t1 on the bottom here so I'll multiply both sides T 1 over T 1 times this t1 on the top T 1 on the bottom they cancel out and I'm left with t2 equals p2 times t1 divided by p1 whoo there we go all right now we're ready to plug our numbers in here so we'll do t2 equals p2 4.08 ATM times t1 298 Kelvin and divide that by p1 2.38 ATM now notice that each of the numbers I'm putting in here have three significant digits so when I put
this into my calculator and do the math the number that I get the answer I get I'm going to round that to three significant figures okay and it turns out that after I put that in the calculator I round it out I get 511 what are my units well it depends what cancels out here ATM on the top ATM on the bottom Schoop they both cancel out I'm left with calvin so it's going to be 511 Kelvin all right and that makes sense because we said that we were looking for a temperature obviously T 2
is going to be Kelvin it's going to be a temperature but it's still important to cancel these units just so we can make sure that we set everything up carefully but we're not done we're not done because the problem asks us to report the temperature in degree Celsius not in Kelvin okay so let's go back here and look at how we do that well we added 273 to the Celsius to get Kelvin to get the Celsius we'll take the Kelvin temperature and we'll subtract 273 so we're
going to take 511 minus 200 set three and we're going to get 238 degrees Celsius darn that's hot but it's possible it's possible for a tire to get that hot it's unlikely but it's possible so there is our final answer we don't have to worry about rounding with significant figures because both 511 and 273 have the same number of significant digits here so we're set to 38 degrees Celsius is the answer to this all right one more problem and here I'm going to show
you how to solve for something that's in the denominator but how to do it in just two steps so you don't have to keep multiplying and dividing both sides over and over again okay here's our last problem