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Boyle's Law Practice Problems

Jun 04, 2021
In this video we are going to talk about Boyle's law, so what exactly is Boyle's law? Well, let me give you an illustration first, so imagine if we have a column where the volume can be adjusted and let's say there are some gas particles inside it. I'm not going to put too many, maybe like seven gas particles. Now let's say that if we apply a force to compress this container to decrease the volume, now the volume decreases considerably and we will still have the same seven. molecules, so what can you tell me about the pressure inside this container?
boyle s law practice problems
Is it greater or less than the pressure in the first one? So in this container the volume is relatively high and here the volume is low, so as we decreased the volume, what happened? at the pressure now observe that in this container the particles are more dispersed and that is why they collide with the walls of the container less frequently here all the particles are grouped together there is not much space between the particles and that is why they collide more with the container It is often that the pressure of the gas is proportional to the number of collisions, therefore the pressure is higher in this container and lower in that container and this is the basic idea behind Boyle's law.
boyle s law practice problems

More Interesting Facts About,

boyle s law practice problems...

Pressure and volume are inversely related if you decrease the volume, the pressure will increase and in the same way, if the volume increases, the pressure will decrease, so the pressure of a gas inside a container given that if the temperature is maintained constant depends on the volume, if you decrease the volume at constant temperature, the pressure will increase if you increase the volume at constant pressure, I mean at a constant temperature, the pressure will decrease and that is Boyle's law. Now the equation associated with Boyle's law you need to know what it is. This equation p1 times v1 is equal to p2 times v2, so this is the equation you are going to use when solving volume and pressure

problems

.
boyle s law practice problems
Now you need to know the shape of the graph associated with Boyle's law, so let's say if we plot the pressure on the y-axis and the volume on the x-axis. Axis we know that as the volume increases, the pressure should decrease. Do you think the graph will look like this? It will not be a straight line, but it will be a curved line. As the volume increases, the pressure will thus decrease. so that is the correct way for this type of problem or this type of law. You may see this on a multiple choice

practice

test.
boyle s law practice problems
You may be given four or five graphs and need to know which one is associated with Boyle's law, so make sure you know. the shape of this graph now let's work on some

problems

number one a 2.5 liter container has a gas pressure of 4.6 atm if the volume is reduced to 1.6 liters what will be the new pressure inside the container well, let's do a list of what we have I'm not sure what just happened there, so v1 the first volume is 2.5 liters and the pressure that corresponds to that volume is four point six atm. Now the volume was reduced by one point to 1.6 liters, so that's the new v2 volume and we're on. looking for the new pressure p2 then all we have to do is use this equation p1 v1 is equal to p2 times v2 so p1 is 4.6 atm v1 is 2.5 liters our goal is to calculate the value of p2 and v2 is 1.6 liters so 4.6 times 2.5 that is 11.5 and that is equal to 1.6 times p2 so now to solve for p2 we have to divide both sides by 1.6 so p2 is 11.5 divided by 1.6 and that is approximately 7.19 atm, as we can see according to the law of the ball, You should always check to make sure this answer makes sense.
We decrease the volume from 2.5 to 1.6, so the pressure should increase from 4.6 to 7.19. Now let's move on to number two. The air inside a 3.5 liter flexible container has a pressure of 115 kpa. the volume of the container must be increased to decrease the pressure to 625 torr, so once again let's write down what we have: p1 is 115 kpa v1, that is 3.5 liters, now p2 is 625 torr and our goal is to calculate v2 now we need to realize is that p1 and p2 should have the same units now p does not always have to be an ATM it can be in tour it can be in kpa it can be in millimeters of mercury however these two units have to match, so if p1 is in kpa p2 should be in kpa if p2 is in tour then p1 should be in one it doesn't matter which is which as long as they match so I'm going to convert p1 to a tor value so we need to know that. the conversion between kpa and tor 101.3 kpa is equal to 1 atmosphere, which in turn is equal to 760 torr, so this is the conversion factor we need, so I'm going to put 101.3 kpa at the bottom so that the kpa units cancel out. and then I'm going to put 760 torr on top, so that's 115 times 760 divided by 101.3, so this is equal to 862.8 torr, so that's p2, I don't want to say p2 but p1, so I'm just going to write that here, so now these two. the units match so at this point we can now use the formula p1 v1 is equal to p2 v2 so p1 is 862.8 torr v1 is 3.5 liters p2 is 625 tour so now we are going to get v2 so notice that the units 2 are will cancel, which means v2 will be in liters, so to get the answer it will be 862.8 times 3.5 and that's 3019.8, so it's equal to 625 times v2.
Now let's divide both sides by 625, so v2 is 30 19.8 divided by 625, which is 4.83 liters and So that's the answer. Note that the volume increased from 3.5 to 4.83, causing a decrease in pressure from 862.8 to 625 torr. Now let's work on this last question. The volume of a gas at 17.5 psi decreases from 1.8 liters to 750 milliliters. is the new pressure of the gas and atm so now p1 is at 17.5 psi it is in pounds per square inch so when we get our answer for p2 it will be the same unit and then we can convert it to atm v1 is 1 .8 liters, but v2 is in milliliters, so we need to make sure that these two match, so I'm going to convert milliliters to liters.
Now the conversion factor is that a thousand milliliters is equal to 1 liter, so we just have to divide it by a thousand. 750 divided by a thousand is 0.75, so v2 is 0.75 liters, so now that v1 and v2 share the same unit, we can now use the equation, so p1 v1 is equal to p2 v2, so so p1 is 17.5 psi and v1 is 1.8 liters which we are solving for. p2 and v2 are 0.75 liters, so to calculate p2 it will be 17.5 times 1.8 and then we will divide that result by 0.75, so p2 is equal to 42 psi, the units of liters cancel out, so p2 has to be in psi, so that's our answer, however we need it. to change it to the correct unit, so what is the conversion factor from psi to atm? 14.7 psi is equal to one atm, that's the conversion factor we need to use, so starting with 42 psi we could say that one atm is equal to 14.7 psi and you need to write it in such a way that the psi cancels out, so which is just 42 divided by 14.7, so the new pressure is about 2.86 atm and that's the answer you get.

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