Buoyancy

So guys in this video, we're gonna take a look at buoyancy with an air. Now in analytical chemistry, we're trying to minimize as many factors as possible in order to ensure that we get the highest level of precision and accuracy within our measurements. Now, one of the factors that we have to take into effect is buoyancy, buoyancy is the upward force exerted on an object in a liquid or gas here. Since we're looking at buoyancy and air, or looking how buoyancy works within a gas. Now, even as simple as weighing an object on an analytical balance has some level of uncertainty associated with it. Remember there are two forces at work. There is the downward force upon the object as a result of gravity. We can use the variables A or G. To represent it With a value of 9.8 m over second squared. But here, when we put object on an analytical balance, we have to take into account that there is airflow. This airflow will distort our true mass of our object. This airflow is buoyancy, it's the upward force upon the object. So even when you put it in a closed analytical balance, there is some level of uncertainty associated with the mass that you're recording here. The equation below will allow us to calculate the true mass of the object as though exists within a vacuum. If it exists within a vacuum, this eliminates air flow entirely. So we'll get the actual mass of the object. If we take a look here at this buoyancy equation, we're gonna say m here represents our true mass. This m here is our apparent mass. The mass that you read off of the analytical balance. Once you wade, you're an elite D. A. Represents the density of air Which has a density of .0012g over male leaders. When the pressure is one bar and the temperature is 25°C, D. W. represents the density of our calibration weights. And here, if it's a standard calibration weight, its density will be eight g per mil leader. But you have to pay very close attention to the question because different types of metals and alloys can be used in place of a standard calibration weight, they each would have their own different density. So if it was using something that was not a standard calibration weight, they would give you a new density to input into the formula finally, D just represents the density of the weight object. Yeah. So remember anytime we wait anything, there is a level of uncertainty associated with airflow. There's that upward force that buoyancy force that will distort our true mass. This equation allows us to weigh the mass as though it exists within a vacuum, which eliminates the whole idea of airflow. Now that we've seen this equation will apply to the example that we see below. So click on the next video and see how we approach that problem

Yeah. So the pipettes that you'll be using in your analytical labs come from the manufacturer as usually class a pipettes. But once you obtain them there is some level of uncertainty associated with them, you have to first calibrate them before we can use them in experiments. So here we're gonna say convenient method and calibrating pipettes is to weigh the water delivered from them by using the density of water at a given temperature. We can determine the volume of density with greater accuracy. So here it says, assume a 50 millimeter pipette is in need millim calibration. An empty flask weighs 49.563g. When water delivered from the pipette is added to the empty flask, the new mass is recorded as 69.618g were asked what is the mass of water delivered? Now here they're saying the density of the standard weights is determined to be 8.40 g per mil leader. Remember in the equation above, we said that the standard weight is 8.0 g per mil leader in this question, we must be using a different type of metal or alloy for those calibration weights, which is why it's a new number. So again, always be careful if they're giving you a new density for the calibration weights, if they do use that one. So we're gonna say here em equals and prime times one minus D. A over D. W Divide by 1 - D. A over D. We need to first find the apparent mass. Well, when the flask is filled with water, The weight of the water and the flask together is 69.618g. When the flask is empty, it weighs that much. When we subtract them, That'll give me the mass of just the water. So that comes out to 20.055 g. So that's our apparent mass. So times one over. So d. is the density of air which is .0012 g per meal leader. The density of the calibration weight we're g is 8.40 g per male leader, Divided by 1 -4 again divided by the density of water. Here, they don't tell us the exact temperature. The density of water can change with changing temperatures but most of them are around one g per mil leader. So we're gonna use one grand per millimeter for the density of water. So we just have to solve for true mass now. So it's 20.055g. Times when I work all this out in here, It gives me a .999857, divide by the bottom. When I subtract this value here from one, what does it give me? Give me 10.9988. Then what do we get at the end? So at the end we're gonna get 20.079 g. So notice are apparent mass. When we read it from the analytical balance was that number. But the true mass. when we take into account air flow gives us this new mass for our water. So again, this is a great way in order to find the true mass of an object, and also a good way to calibrate the pipettes that you'll be using extensively within your analytical labs. So now that we've seen this will take a look at buoyancy when it's found within fluids.

in this video, we're taking a look at buoyancy within fluids. Now, when we're taking into account buoyancy within fluids, we're looking at Archimedes principle, which states that the buoyancy force, so that's the upward force acting upon an object um is equal to the weight of the fluid displaced by the object. So buoyancy force equals weight of liquid displaced. Remember when we're trying to place an object within water, there are three scenarios that are possible. We could have one scenario where the object itself floats on top of the water partially or or like nearly um completely floats on water. We have one where it is exactly in the middle and then one where it sinks all the way to the bottom. So these are three scenarios Now, remember, ultimately, and objects buoyancy within a fluid is determined by its density. So, in the first object, we can say here that P. W. Represents the density of water, right? And P. A. Represents the objects density here, the object is floating on the water. That's because the density of the water must be greater than the density of the object. In this one, it's floating directly in the middle of the fluid. In this case they must be equal to each other. So density of water equals the density of the object. We would coin this neutral buoyancy and then finally the object sinks all the way down to the bottom, that's because the density of water must be less than the density of the object. When we're dealing with Archimedes principle. The equation that we're dealing with is f, which is the force of the of the uh of the object or the actually the buoyancy force? The upward force, sorry, equals mass Equals mass of the liquid times gravitational field strength. So that's just acceleration due to gravity. So that's 9.8 m over second squared And that equals the density of the liquid times the volume of the liquid displaced. And again, we have the gravity due to acceleration 9.8 m over second squared. Now, in each of these objects, we always have two forces at work. We have the force that's pushing up. So this is our force due to the to the buoyancy force. And then you have another force due to gravity. Those are the two forces acting upon these objects here. When it's floating on the surface, we'd say that the force, the buoyancy force is equal to the gravitational force in this one where it's exactly equal in the middle. We would say in this case that the buoyancy force must be greater than the gravitational force. And lastly, when it's sunken all the way down to the bottom, the force that's pushing it down acting upon it, the gravitational force must be greater. So again, we have two scenarios happening here where we can look at the density of the object versus the density of the water, which ultimately determines the buoyancy of the object overall. And then we can take into account the forces at work. Remember the buoyancy force is always the force that's pushing upward, trying to cause the object to float, and it fights against gravity, which pushes downward. So important to remember the Archimedes principle, which will be utilizing when answering questions, dealing with objects suspended in fluid, and then understanding the relationship between the density of the objects. As we work more and more towards Archimedes principles, we'll take a look at examples. So on the one below, we're dealing with the typical Archimedes principle type of question. Click on the following video to see how I approach that question. Yeah.

Okay here in this question it says a wooden block with measurements of 0.15 by 0.44 by 0.56 m is afloat on a lake. If it is submerged by 0.31 m. What is its mass? Alright, so here it's telling us that it is a float so it's floating on the lake. That tells me that force buoyancy force is equal to mass times the gravitational field strength. If they had said that it was that neutral buoyancy then we'd say that the buoyancy force would be greater than mass times gravity. And if it had sunken to the bottom then that would mean that the force, the buoyancy force would be less than mass times gravity. Yeah, again because it's floating we're gonna say they're equal to each other. Remember here that force would then equal volume of water displaced. Which is why we continue and say that this equals the density of the liquid times the volume of the liquid displaced times. Again, gravitational field strength, this portion I'm gonna bring down so mass equals gravity. Mass times gravity equals density times volume times gravity. Both of them have the same gravity involved. So you can just eliminate that. So the equation I'm gonna use is gonna simply become mass equals density, times volume here. Alright, so we have length times length, times length length Q will give me volume. So volume equals .15 m Times .44m times .56 m. So when I multiply all those together I get .03696 m cubed here. The density of the liquid which is water is just one g per millimeter. But here I have meters cubed. I'm gonna have to convert meters cubed into centimeters cubed because remember a centimeters cubed is equal to a middle leader. So we're gonna put meters here on the bottom centimeters here on top one, senti is 10 to the negative two. We have to cube this entire thing that the meters cubes can cancel out. And I'll have centimeters cubed at the end. So when I work that out, it really means that I have 0.3696 m cubed. Times one centimeters cubed over 10 to the negative six m cubed. So that's gonna give me 36960 centimeters cubed. Take that and plug it in here. 36960 centimeters cubed. And what happens here is male leaders and centimeters cubes cancel out. So at the end I'll have the mass as 36,960 g. So that would be our answer. In terms of this question again, you have to determine is the object sinking floating or is it neutral buoyancy To determine which version of force equals two greater than or less than mass times gravity. From there we're able to isolate the one variable that we needed at the end which is mass. So continue onward. And looking at examples where we're dealing with buoyancy, whether it be in air or whether it be within fluids. When it's dealing with air, we have to take into account air flow. That's when we use the buoyancy equation to find the true mass. In this set of questions, we're dealing with buoyancy in fluids. In that way we'd have to use the Archimedes principle. So remember the distinction between the two forms of buoyancy, whether it's in air or whether it's in a fluid.

here, we're told that we have a small crystal of sucrose and it has a mass of 5.345 mg. We're told that the dimensions of the box like crystal were 2.20 millimeters by 1.36 millimeters by 1.12 millimeters. Where asked what is the density of the sucrose crystal expressed in grams per milliliter. So we want our density in grams per middle leader. So what we're gonna do first is we're gonna convert the milligrams given to us into g. So that's 5.345 mg. We're gonna stay here for every one mg. We have 10 to the negative three, That's 5.345 times 10 to the -3 g. So we're gonna put that here up top. Now we need to figure out our volume, we need milliliters. We'll realize here that they give us millimeters here on the bottom. And if we multiply all three together, that's gonna give me millimeters cubed. Also remember that if we can get centimeters cubed, that a centimeter cubed is equal to a millimeter millimeter. So what we're gonna do here is we're gonna multiply those dimensions altogether to get millimeters cubed, convert that to centimeters cubed. And that will represent the mls that we need at the end for our density. So we have 2.20 millimeters times 1.36 millimeters times 1.12 millimeters Together. They give me 3.35104 mm cubed. All we have to do now is convert millimeters cube two centimeters cubed, Remember the conversion is one centimeter is equal to 10 millimeters. Those millimeters can't cancel with the centimeters to millimeters cube. So I'm gonna keep the entire thing. So that's gonna give me 3.35104 mm cubed. one cubed is just one 10 Cubed is 1000. So here those cancel out. So we'll get cm cubed which is the same thing as ml. So that comes out to 3.35104 times 10 to the -3 middle leaders. So when we punch that in we get 1.595 grams per milliliter. Which if we want three sig figs we get 1.60 g per milliliter of this box like crystal and its density. So take this approach and what we've learned thus far And see if you can attempt a practice question here on the bottom. Again attempted on your own. If you get stuck, just come back to the next video and see how I approach answering that question. Good luck guys.

An empty container weighing 73.190 g is filled with an unknown liquid and the combined mass is recorded as 87.308 g. The container was then emptied and filled with water and recorded a new mass of 88.442 g at a temperature of 19 ^oC. (d = 1.0027 g/mL). Calculate the density of the unknown liquid. 

• A. 0.92861 g/mL
• B. 0.92565 g/mL
• C. 0.92815 g/mL
• D. 0.93000 g/mL

here, we're told that a piece of concrete weighs 120 newtons. And when it's fully submerged it's apparent weight is 97 newtons determine the density of the water. If the volume of water displaced is 4200 centimeters cubed. Alright, so remember here that are buoyant buoyant force is f we're gonna say here that are buoyant force is equal to the weight of the water displaced. So that's equal to the density of the liquid itself. We have our volume of the water that is displaced. And then we have G which is our gravitational field strength. Now we're gonna say to figure out our buoyant force, we're gonna say buoyant force F is equal to the actual weight minus the apparent weight. So that's gonna be 100 and 20 newtons minus 97 newtons. Which gives us 23 newtons. So it's gonna be 23 newtons here, we're looking for the density of the liquid. Here, we have the volume of the liquid here, we're gonna have to convert it And that's because of the units of our gravitational field strength, which is 9.8 Newtons divided by kg. So what we're gonna do here is we're gonna convert our volume from centimeters cubed, two m cubed. And that's because remember, a newton is equal to meters times kilograms over second squared because meters are involved, we should change our volume two m. So we're gonna have 4200 centimeters cubed, we're gonna stay here for every one centimeter, it's 10 to the negative two m. Well you want centimeters cubes to cancel out. So we're gonna have to cube this whole thing. But what effect does that really have? That's really saying 4200 centimeters cubed times 10 to the negative six m cubed on top and one centimeters cubed on the bottom. Okay, so that's equal to 4.20 times 10 to the negative three m cubed here. Alright, so now what we're gonna do is we're going to divide both sides here By 4.20 times 10 to negative three m cubed and also the 9.8 newtons over kg. Okay, so do that on both sides here. So this cancels out with this are newton's will cancel out. So at the end we'll get our density Of this liquid as being 558.8 kg over m cubed here. If we wanted to do sick fix. This has if we let's just assume there's a decimal there. Let's say we have a decimal. There will say that has 36 fix. This has two sig figs. Let's say there's a decimal here. This has 466. So if we wanted to. 66 here we can say this is 5.6 times 10 to the two kg over meters cubed. So that'll be the density of our liquid? Remember we know that we're using this version of an equation because we're dealing with a solid object. Um and it's placing in a fluid. And when we're talking about fluids, we're gonna adopt Archimedes principle in order to figure out our answer. If we're dealing with just an analytic being weighed on a typical analytical balance, we all all we have to worry about is air flow, and in that case we'd use the buoyancy equation now that you've seen this example, but to see if you can tackle example to again, as always, if you're stuck, just come back and see how I approach on answering that question.

Okay. So here it says the density of propane and odorless hydrocarbon compound used in cooking is 0.9 to two g per milliliter were asked when a sample of it is placed on an analytical balance, a weight of 8.15 times 10 to the eight nanograms is obtained calculate the true mass of propane. All right. So, we're looking for the true mass. So that means we're looking for the mass of an analytic when we're weighing on an analytical balance. So now we're gonna have to use the buoyancy equation. So that is mass. True mass of my object is equal to the mass of it. When it's on the analytical balance, times one minus the density of air divided by the density of water. And then on the bottom we have 1 - the density of air divided by the density of the object or analyze itself. Now, we need to have the mass of the an elite in grams. It's given to us here in nanograms. So, just remember we have 8.15 times 10 to the eight nanograms. For every one nanogram, it's 10 to the negative nine g. So that means that the weight according to the analytical balance is 8.15 times 10 to the -1 g. We're gonna have one minus the density of air, which is 10.12 g per milliliter Divided by the density of water. And here would be 8.0 g per milliliter. Then on the bottom we're gonna divide by 1 - the density of air again. Yeah. And then it's gonna be our the density of the object itself. So here that would be the point 9 to 2 g per milliliter. Like we're told right here. So when I do 1 -1 the density of air divided by the density of water. I'm gonna get here .99985. The density units cancel out. Then we're gonna divide by 1 - the density of air, divided by the density of the object itself, gives me .998698. Now here, we're gonna want to answer to have three significant figures because 8.15 has three significant figures in it. When we do that, we get .81594g. Which at the end gives us .816g as the true mass of my object. Once I've taken into account airflow now finally take a look at the final practice example that's left on the bottom of the page. Attempt to do it on your own. But again, as always if you get stuck, don't worry, just come back and take a look at how I approach on answering that practice question. Mhm