Thermal Dependency - Video Tutorials & Practice Problems

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Thermal Expansion

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concept

Thermal Expansion

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with every calculation and measurement, we have to be sure to minimize all outside factors so that our data isn't corrupted. Now one of the outside factors we have to be in mind of is thermal dependency. Now we're gonna say the process of calibration is the process of measuring the actual quantity of mass, volume or other type of of chemical measurements that relate to what we observe on an analytical scale. So again we're trying to be as accurate and precise as possible with all our calculations so we have to make sure that we eliminate or at least try to minimize as much as possible any outside factors. Now, the idea of temperature comes into play here, we're gonna say in general when taking the measurements of a solution, we have to take into account any type of thermal expansion that could somehow influence either our solution or our instrumentation. So the temperature of the room, the temperature of the instruments themselves can have an effect on the volume that's being recorded with every type of um lab process that we're undergoing. Now we're gonna say here we can correct for this thermal expansion by using our formula. So here we're gonna say that C. Prime represents the concentration of our solution divided by D. Prime which is the density of our solution. It equals the new concentration divided by the new density of our solution. Now if we take a look here We look at this graph in this graph, we have the temperature of our solution or temperature of water at different numbers all the way from 10°C to 30°C. Now, if you take a look as we're going from 10-30, we can see that our temperature is clearly increasing. Now look at the density of water as the temperature changes as it increases, what's happening to our density while we're starting up here with 0.99970 to six. And as you go down, you should notice that the density is decreasing as our temperature is increasing later on, we'll talk about why this is so, but just remember for right now that the trend is, as our temperature is increasing, our density will decrease The next two columns. So here, what we have is at temperature shown. So we're recording the volume of a container or recording the volume from some type of instrument at a given temperature. This would be the volume of the water for every one g. So let's say that we're taking uh volumetric analysis of water delivered at 10°C, we'd say at 10°C, we would basically deliver 1.14 mls of water for every one g of water. But here's the thing, when it comes to our instruments, all the glassware is usually calibrated to around 20 degrees Celsius. So here we have corrected the volume delivered to 20 degrees Celsius because that's the temperature in which we calibrate our glassware. So in actuality it's really delivering 1.15 mls for every one g of water. So remember the column on the left? This column here is a temperature um the volume record when we know a specific temperature. And this is us correcting for it based on the calibration temperature of most glass wares within your analytical lab. Now keep in mind this chart as we go and venture off into thermal expansion. No, you don't have to memorize all the numbers. Just remember the general trends that are associated with it. If you have questions dealing with this, you would be given values in some way or another. Either through the question itself or given this chart on a separate sheet of paper. So guys, now that we've covered this, move on to the example that's given on the bottom, see if you can tackle that question which relates polarity with parts Pavilion and also talks about this idea of thermal expansion. What effect does it have on the polarity Once you've done that, come back and see how I approach the same exact question

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example

Thermal Expansion

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Here. It states that Massachusetts limits the amount of lead in drinking water to 219 parts per billion. Part a states express the total in polarity and part B states, what will happen to the polarity of the solution as the temperature increases? Will it increase decrease or remain constant? All right. So let's first tackle part a And remember here they're giving us parts per billion. Remember parts per billion. One Parts per billion means one microgram over one liter. So, by telling us that we have 219 parts per billion of lead. It's really saying that we have 219 micrograms of lead for every one leader, we need more clarity and more clarity is just simply moles over leaders. So all we have to do here is change Michael grams, two g and then from grams to moles. So we want to get rid of micrograms. So micrograms go on the bottom one. Microgram is 10 to the negative six g. So micrograms are gone. Then we're going to say here for every one mole of lead. We have 207.2 g as the massive lead according to the periodic table. So at the end here we'll have moles over leaders which will be our polarity. So that comes out to being approximately 1.6 times 10 to the negative six Mohler of lead for part B. Now here we're asked what's gonna happen to the polarity as the temperature increases? Remember polarity equals moles of our saw you divided by leaders of solution now as you increase the temperature that's going to cause an increase in your volume. So what's happening here is the leaders on the bottom it's going to expand. It's gonna get become a larger value while the moles stay constant. So your top portion is staying the same as your bottom is increasing. So that means overall your polarity will decrease. So we'd expect similarity to to decrease as the temperature increases, the same thing is happening here with density. Remember we said as the temperature was increasing, we can see that our density is dropping. That's because density equals mass over volume. Again, by increasing the temperature, you're increasing your volume as your top portion stays constant. So as a result of this, your density drops. So just remember some of the fundamentals when it comes to thermal expansion. Remember we're trying to minimize all outside factors to make sure our data is not corrupted. So we have to take into account that concentration and density of a solution is determined and affected by changes in temperature. Now that we've seen this process will move on to other examples dealing with thermal dependency

Thermal Expansion Calculations

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example

Thermal Dependency Calculations

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So here it states if a 0.2135 moller a quick solution is prepared at 21 degrees Celsius. Where the density is given as 210.9979955 g per meal leader. What is the new concentration If the same experiment is performed a month later when the temperature is now 26°C with this new density. All right. So we're talking about different temperatures and as a result our density changes. So our formula here we're gonna have is concentration of my solution prime divided by density. Prime equals concentration over density. So we have our initial concentration as .02135 Molar. That's associated with this density here. So 0.9979955 g per millimeter equals while we're asked to find what the new concentration is. So c divided by 0.9967867 g per milliliter here we'll cross multiply these two numbers together and then these two numbers all these two things together. So at this point it gives me .9979955g/l times. My new concentration equals when I multiply this and this together. We're gonna get 0.0 to 1 to 81396045 grams per million times more clarity, Divide both sides now by .9979955 g per mil leader. Okay so this cancels out with this. My grams per millimeter units cancel out. So my answer here will be left in polarity. So my new concentration for my solution will be .0-13 to moller. Now this answer itself makes sense because we know here when the temperature was 21°C, our concentration was .02135. And from our previous video, we said that if our temperature is going to increase, that's going to cause an increase in my volume. And because polarity equals moles over volume that means that my polarity should decrease. And that's what we're seeing here. We're seeing that our polarity has indeed decreased. And now it's a new number of 0.2132 molar. Now that we've seen this example, come back and see if you can take a look at the next example here, see if you can figure out what the answer is to this particular practice question. So again, guys, attempted on your own, come back and see if you get the same answer as I do

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Problem

Problem

The mass of an empty container at 28 ^{o}C is 83.93 g. The mass of the container when filled with water from a 25-mL pipet is 108.70 g. Calculate the true volume of delivered water by the calibrated pipet.