Gay Lussac's Law Practice Problems

Okay, so let's do a bunch of

practice

problems

with Gay Lussac's law just to review laws written like that. It is assumed that some kind of change is occurring, so we have p1 and t1, which are the initial pressures and temperatures 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 gas can is 235 kilo Pascal when set to room temperature 20 degrees Celsius If the can is hot at 48 degrees Celsius, what will be the new pressure inside the can?

So let's use the equation and how I'm attacking these gas law

problems

. The first thing I like to do is make sure I know what everything is. My variables are fine, so let's figure this out. We have an initial pressure before the change, so p1 will be 235 kPa, so we know this guy well here when he is at room temperature, so this is the initial temperature of 20 degrees Celsius. and then the can is heated to 48 degrees Celsius, so there's our change, we go from 20 to 48, so our second temperature, our final temperature will be 48 degrees Celsius, so we know these guys and it will be p2 the new pressure that Let's solve OK, so there are variables here.

You might be thinking you can remember this every time 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. Okay, so now before we plug anything in, let's put these Celsius temperatures in Kelvin. Okay, so for t1 we're going to take 273 and we're going to add 20 to it, so now we're going to have 293 Kelvin for t1 and t2 is going to be 273 plus 48, so it's going to be 321 Kelvin. Okay, now that we have all of these variables set up, let's rearrange this equation to get p2 on its own.

Okay, so the first thing I want What I do to get p2 by itself is I take t2 out of the denominator here and I'll do that by multiplying both sides by t2. Now I have t2 here below and above this fraction so they both cancel and we can rewrite this as t2 multiplied by p1 divided by t1 equals p2. Now some of you might not like that p2 is on the right side so I'm just going to reverse it, but it's actually the same equation either way so this way just say that p2 is equal to t2 times p1 divided by t1, okay, here we go, now p2 is alone and we are ready to plug in the variables, okay, so p2 is equal to t2 321 Kelvin times p1 235 kilopascals divided by t1, which is 293 Kelvin, okay so let's move on Go ahead and plug that into our calculators and the answer we're going to get.

I'm going to round this to three significant figures which is 257. Now what are the units I'm going to use? Well, it depends on what. It cancels out here, obviously, we know this is going to be pressure, so we could say, Okay, you know it's going to be kPa, but let's show how we're going to do the unit. Cancellations here, we have Kelvin up here and Calvin down here. I mean, down here, okay, so we'll get rid of the kelvin and we'll be left with the kilo Pascal, which is obviously a unit of pressure, so we know our final answer is going to be 257 kilo. pascals and again I rounded this answer with my calculators, but I spit out a very long number for this, but I rounded it to three significant figures because there are three significant figures in each of the numbers I started with, okay, this is how we do it .

So let's look at it a little more. I'm going to look at three problems in total and each one of them will be a little more complicated than the last, okay, so here we are talking about a tire, a car tire that has a pressure of two point three eight ATM to twenty-five point three degrees Celsius if the internal pressure reaches four point three eight atm the tire will explode well, how hot will the tire have to be for this explosion to occur and then report the temperature in degrees Celsius? Well, again let's start with this equation.

The first thing we're going to do is clarify our four variables, so we start with this initial pressure and this initial temperature and then we say let's go. We're going to increase the temperature, which will 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 that we're going to have to finish converting it to Kelvin in one minute before we can use it in the equation. Okay, now if the internal pressure reaches four point eight, that's where the pressure will be after the change, there's our p2 four point eight atm. what will the temperature be how hot would it have to be for the temperature to rise that much, then t2 will be the variable where itself or so we know p1 we know P two we know that t1 and t2 are going to be be the variable that we are solving now again , let's convert our Celsius temperature here, which is t1, let's convert to Kelvin temperatures that we can use for gases, okay, again, we take 273, this is t1, we take 273 and let's add 25.3 to this, okay, now remember the rules about adding and subtracting with significant figures.

We look at which of the numbers has some accessible places or the few significant figures on the right side. Okay, this has a decimal. 273 doesn't have Okay, that means I'm going to draw a line down here and I'm going to round up to this digit. I look at the three here so that the eight stays the same, but remember that we have to round using meaningful addition and subtraction. numbers rule, so two hundred ninety-eight point three, cut here, will become just 298 Kelvin and that's what I'm going to use. 41 Okay now, solving this equation for T 2 is going to be a little more difficult because it's in the denominator.

I'm going to have to follow several steps, okay, so be patient. I will explain each one and after solving this problem completely in the next problem I will show you a shortcut. It's probably a shortcut that your teacher your textbook hasn't shown you, but it's a way to take all of these steps that I'm about to follow and condense them into two, but first I want to show you how to do it in a longer way. the correct way, so t2 the first thing I want to do is get it out of the background. I can't solve it if it's at the bottom, so I'll get it off the bottom by multiplying both sides by t2.

Okay, now it's down here, it's up here, so it cancels out and I can rewrite this as t2 times p1 divided by t1 equals p2. Okay, now it's up, but it's far from a loan and we want to get it on our own. To solve it, okay, so I want to get p1 and t1 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 hell of it, p1 here. and divide this by p1 here put p1 at the bottom so now p1 over p1 both cancel 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 I just have to get this t1 at the bottom here, so I'll multiply both sides T 1 over T 1 by this t1 at the top T 1 at the bottom, they cancel and I'm left with t2 equals p2 times t1 divided by p1 .

Okay now, we're ready to plug in our numbers here, so we'll set t2 equal to p2 4.08 ATM times t1 298 Kelvin and divide it by p1 2.38 ATM. Now notice that each of the numbers that I'm entering here has three significant digits, so when I put this into my calculator and do the calculations, the number that I get, the answer that I get, I'm going to round it to three significant figures, it's okay, and it turns out that after I put that in the calculator, I round it up, I get 511, what are my units? Well it depends on what cancels out here ATM at the top ATM at the bottom Schoop both cancel out I'm going with Calvin so it's going to be 511 Kelvin okay and that makes sense because we said 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 out these units just to make sure we set everything up carefully, but we're not done yet.

We are not done because the problem asks us to report the temperature in Celsius, not Kelvin. Okay, let's come back here and see how we do it. We add 273 to Celsius to get Kelvin. To get Celsius, we'll take the Kelvin Temperature and subtract 273, so we'll take 511 minus 200 in the third set and get 238 degrees Celsius. That's very hot, but it's possible. It is possible for a tire to get that hot. It's unlikely, but it's possible. So there's our final answer, we don't have to worry about rounding to significant figures because both 511 and 273 have the same number of significant digits here, so we're set to 38 degrees Celsius.

This is the answer to this, okay, 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. Well, here's our last problem.