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## MCAT

### Course: MCAT > Unit 8

Lesson 8: Gas phase- Gas phase questions
- Absolute temperature and the kelvin scale
- Pressure and the simple mercury barometer
- Definition of an ideal gas, ideal gas law
- Derivation of gas constants using molar volume and STP
- Boyle's law
- Charles's law
- Avogadro's law
- The van der Waals equation
- Partial pressure

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# Boyle's law

Created by Ryan Scott Patton.

## Want to join the conversation?

- Why would Robert Boyles Wife be excited about the J tube?(6 votes)
- I think he's being sarcastic, and he actually means that she probably didn't like it.(40 votes)

- actually robert boyle didn't do the Experiment himself, he just took it from Henry Power. https://en.wikipedia.org/wiki/Henry_Power(9 votes)
- I still don't understand very well. So, do you have to find the inverse of your values in order to do the equation? And whats with the equation? So is it P1V1=P2V2, or V=m 1/p, or V=k(1/p), or P1V1=PfVf? PLEASE SOMEBODY EXPLAIN! :'((6 votes)
- Hello Nicole. I think the equation is P1V1 = P2V2. You can derive this from the Combined Gas Equation (P1V1/T1 = P2V2/T2). Since Boyle's law says it is at constant temperature, the temperatures cancel each other so you are left with P1V1 = P2V2 which is Boyle's Law.(3 votes)

- at6:12when he draw the second graph he took volume on the y axis but in the first graph he took pressure on the y axis .. why he did so?(3 votes)
- umm...also,can someone please explain what happened in the video?-im so sorry . i don't mean to be disrespectful or anything,but I'm kinda confused.(3 votes)
- Also, what is the difference between a Real Gas and an Ideal gas?(2 votes)
- What does Boyle's Law apparently state?(2 votes)
- pv=pla please explain for me about a which is always constant(2 votes)
- If a gas occupies 2.60 liters at a pressure of 1.00 atm, what will be its volume at a pressure of 3.50 arm?(2 votes)
- What's that thing called "atmospheric pressure?" The pressure of the atmosphere? Or something else? And what's a "J tube?"(1 vote)
- Thanks, Fathima. I'll make sure to remember that.(2 votes)

## Video transcript

Voiceover: Robert Boyle
was an Irish scientist in the 1600s, and it's actually from his experiments that we get Boyle's Law, which actually preceded
the ideal gas equation, and we've already shown that. We're going to work backwards, and we'll use Boyle's Law to prove part of the ideal gas equation, and we'll get a little bit
of history along the way, which is always fun. Boyle was experimenting with gases, and he had a big J-tube set up in the entrance of his house, which I'm sure his wife
was thrilled about. He trapped some gas in that J-tube, and he filled the bottom of the tube, but with a little bit of mercury, which trapped that gas on the closed side because mercury is a pretty dense liquid, and gas has a hard time moving through it, so it trapped a little bit of
that gas on the other side. This left the open side
exposed to the atmosphere here, so that you have the pressure
of the gas on one side, and the pressure of the
atmosphere on the other. We know that they're pressing down with it with the same amount of pressure because as he started, the height of the mercury was the same on both sides. Things got really interesting when he added a bit more mercury because now that the two
levels didn't equalize, instead they were offset. What this meant was that the trapped gas pressure over here was greater than the atmospheric pressure, such that the gas pressure was equal to the atmospheric pressure plus the fluid pressure
of the height difference. You can think that the gas is pushing down on this part of the mercury with the same force, or with the same pressure
as the atmosphere pushing down over here, plus this bit of fluid
over here pushing down. Now, he added a bit more mercury, which compressed the gas even more, making its volume less, and he found that there
was an even greater offset, in the two fluid heights. He correctly took this to mean that the gas was exerting
even more pressure because now, the gas
pressure is equal to the atmospheric pressure plus even more fluid height. Robert Boyle plotted this data, and these are the values that he got in the middle of the 17th century. He plotted the volume in cubic inches, and he plotted the pressure
in inches of mercury, and he was measuring
that height difference for the pressure. Starting off, his volume
was 117.5 cubic inches, and his pressure was 12 inches of mercury. As he filled the J-tube with
a little bit more mercury, he had a volume of 87.2 and a pressure of 16 inches of mercury. As he continued to fill it, he got a volume of 70.7 with
a pressure of 20 inches, and continued for a volume of 58.8 and a pressure of 24 inches, and continued and got 44.2 and 32, and as he kept going,
he got 35.3 cubic inches and 40 inches of mercury, and then the last value was 29.1 inches of mercury after. That's the volume that he
had compressed it down to, and 48 inches of mercury for the pressure. What he did is he plotted this data, and he graphed pressure
as a function of volume. He had a graph with pressure
as a function of volume, and if we look at our pressure, about the highest it gets is 48 inches, so we'll make the top 50 and the middle part, we can say is 25 as a kind of benchmark, and if you look at volume, the highest we have is 117, so we'll make it 100, and we'll go a little bit over, and we can kind of fill in our graph. So, 50, 25, and 75
cubic inches for volume. We see that when our volume is 117.5, our pressure is 12, so that would be right about here, and we see that as our volume is 87.2, our pressure goes up a little bit to 16, and when we have our volume at 70.7, our pressure is about 20, so that would be about right there. When our volume is just about 60 here, we have a pressure of 24. Then as our volume goes to 44.2, we have 32-ish, that would be about right there, and then 35.3 for the volume is about 40 for the pressure. Right under a pressure of [48], our volume would be about 29, so about right there. What we have when we plot pressure as a function of volume is we have a hyperbola, and what we see is that as the volume drops by half from about 50 to 100, the pressure essentially doubles, and as we go from 50 to 25 for the volume, we go from 25 to 50 for the pressure, about so. We have an inverse relationship for the pressure and the volume, so if we graph volume, then as a function of
the inverse of pressure, we get this graph. We've got volume as a
function of the inverse of pressure, so we're going to need the inverse values of all of our pressure. One over 12, the inverse
of 12 would be 0.08. One over 16 for the inverse of 16 would be about 0.0625, and if we continue
finding the inverse values of these pressures, we would get 0.05 for 20, and then 24 would be 0.042, about. One over 32 would be 0.03125, and 40 would be 0.025, one divided by 40 is 0.025, and then one over 48 is 0.0208. We can populate our
graph with these values. About the highest inverse
pressure value we have is 0.08, and about the lowest is 0.02, so we can fill that in here. We're still working with
the same values for volume, so the highest is a little bit over 100, and then we can put in 50 and 25 and 75. When our volume is 117.5 cubic inches, the inverse pressure would be about 0.08. As we go down, 87.2 would be 0.0625, and 70.7 would be 0.05,
right in the middle here, and then 58.8, just
about 60 would be 0.042, and 44.2 would be 0.03125. 35.3 would be 0.025, and 29.1 would be 0.0208. This isn't a perfectly clean graph, but we do see that when we graph volume as a function of the inverse of pressure, we get a straight line. If we write this straight-line
graph as an equation, it would be y is equal to mx plus b. That's the equation for this graph where m is our slope,
and b is our y-intercept. Our y-intercept here is zero, so all we really need is y is equal to mx. Well, in our graph, our y value is our volume, and our x value is the
inverse of our pressure, so let's fill that in here. If we call our slope k instead of m, if we just use a different letter, then we'll get v is equal
to k times one over p. Multiplying both sides by p would give us pv is equal to k, or in other words, the
product of the volume and the pressure for a gas is a constant value, just like we see in
the ideal gas equation. Let's test this out by going back to those original values that Robert Boyle plotted. If we measure the product of the pressure and the volume here, we'll see that 117 times 12 is just about 1400, and 87 times 16 is just about 1400, and, in fact, all of
these volumes multiplied by the pressure, the product is always almost exactly 1400. One great application of this concept is that if the number of moles and the temperature of an
ideal gas are constant, then the initial product of p and v will equal the final product of p and v, so pf and vf for final. Let's try to use this in an example. If the pressure of a gas
in a 1.25 liter container is initially 0.872 atmospheres, what is the pressure if
the volume of the container is increased to 1.5 liters, assuming that the
temperature doesn't change? We know that if this
is a closed container, the number of particles
isn't going to change, so our moles are also constant. Let's use this idea that
p1v1 is equal to p2v2, and our initial pressure
is 0.872 atmospheres, and our initial volume is 1.25 liters. We're looking for the final pressure, when the final volume is 1.5 liters. The first thing that
we're going to need to do is divide both sides by 1.5 liters to isolate our final pressure. On this side, we completely
cancel out 1.5 liters, and on this side, we cancel
out our units of liters, and we get 0.872 times
1.25 divided by 1.5, then we'll retain our
unit of atmosphere here. That will give us our final pressure, which happens to be 0.727 atmospheres. Just as a final common sense check, this result follows Boyle's Law because we increased the volume from 1.25 to 1.5, and so we decreased the pressure from 0.872 to 0.727.