this video is on the topic of gas pressure the learning objectives are on your screen so go ahead and pause the video and write those down in your notes to begin the topic of gas pressure uh it's useful to first discuss pressure in terms of something we might have heard of atmospheric pressure and atmospheric pressure is uh you know we define it as the force exerted by a column of air uh in the atmosphere um that force on one square inch of space that's one way to think about atmospheric pressure and the amount of uh of force exerted on a given space actually turns out to be about 14.7 pounds as a as a weight um of force on one square inch of uh space and so that's equal to one atmosphere of pressure so in general sense and as you'll see in a second we actually i already mentioned units of of atmospheres and i've also mentioned pounds per square inch so really there are a number of units of pressure and the the main thing to keep in mind is the general definition of pressure uh defined as capital p is going to be force over area and so pressure is is directly proportional to force and inversely proportional to area the pressure units i mentioned are right here i'm going to quickly breeze through these um definitely use this table um from uh uh use this table to help in in in your problem solving um uh practice so uh the pascal is the si unit um or the iu the recommended iupac unit um it is the only one here you'll notice that's defined in terms of si units of force and and sort of length so newton per meter squared um uh so the kilo pascal's just you know the the same prefix multiplier kilo so i'm not going to go into that psi pounds per square inch i just i already mentioned that that's at sea level which is about one atmosphere it's about 14.7 pounds per square inch um this is a really useful uh relationship here between atmospheres we have one atmosphere of pressure is the equivalent of 101 325 pascals and that's equal to 760 torr so um one of the common ones that i remember definitely is that one atmosphere is equal to 760 torr that one definitely is a very common um conversion we also have bar this is uh specific to not specific but commonly used in meteorology one bar is equal to exactly 100 000 pascals a millibar is just you know the prefix multi uh modifier milli um inches of mercury is one of my favorite ones just because it's literally the displacement of liquid mercury measured in inches um and that's typically um uh resulting from devices that have been designed to measure pressure barometers and manometers and we'll talk about those in a little bit one inch of mercury is equal to 3 386 pascals um the tour itself i mentioned tor up above that one atmosphere is equal to 760 taurus this is just the tour defined as in terms of just one tour it's equal to um one 760th atmospheres um i just find it easier to remember that one tour is equal to 700 um sorry that one atmosphere is equal to 760 torr millimeters of mercury um is is quite similar to inches of mercury and that it's the displacement of liquid mercury one millimeter of mercury's is is about one torr so the other way to say that is that 760 millimeters of mercury is equal to 760 torr and both of these are equal to one atmosphere um so interconverting these uh is pretty much dimensional analysis so i'll leave it to you to to to practice that we'll get practice in class what i want to jump into is how to measure um pressure so this is an example of a barometer this is one uh one type of instrument used to measure pressure and a barometer is typically used to measure atmospheric pressure and change its atmospheric pressure you can use a barometer that has liquid water in the bottom in this in this container that's open to atmosphere but you can also use liquid mercury that's in the container that's open to atmosphere the benefit of using mercury liquid mercury is that it's much uh more dense as a liquid with a much higher density than water so that means that if there is um if there are you know kind of big changes in pressure you uh the the the the vertical displacement up in the the tube uh of the barometer um won't be that high because it's such a dense liquid and i'll and i'll show i'll give you an expression for that accounts for liquid density in a second but you'll see that um water you know has much lower density it has it in order to measure pressure changes we actually have to have much higher displacements of water in the tube so for practicality reasons mercury makes a lot of sense for measuring changes in pressure we don't want to have have to make this gigantic barometer tube in which um just so that we can allow the water or some not very dense liquid to shoot up very uh and have very big displacements okay so it's strictly um practicality purposes certainly not for safety reasons to use mercury right um okay so the premise here is that this that there's a secondary container open to atmosphere and we have this closed end tube that is vacuum sealed on uh that's inverted so what happens is there's some equilibrium that's maintained um that's reached when atmospheric pressure is pushing down but if atmospheric pressure increases and it pushes down on that liquid surface um then the uh displacement uh in within this tube will go up and it will push if it's higher atmospheric pressure it will push the mercury up into the tube and we can measure that displacement in inches of mercury or millimeters of mercury whatever we want and also if the if the pressure drops back down below atmospheric pressure or there's just a decrease in atmospheric pressure we would see that vertical displacement go back down in the barometer the mathematical expression that i mentioned for a barometer is that we have um the pressure that we can measure is going to be equal to this um this is supposed to be a capital p is equal to h times uh this is not a p this is the greek letter rho i did not really it's kind of difficult to draw greek letters sometimes maybe i'll draw it more like this to emphasize that it's not a standard p times g h here is the height of the fluid and we can also think about the height displacement of the fluid rho here so what i'm saying row it's rho is a typical um greek letter used for density and this is where the density of the liquid comes into play and then g here is actually the acceleration due to gravity but if we use a a single liquid then that means that the density is a constant um the acceleration due to gravity is a constant so really the pressure is directly proportional then to the height of the liquid that we can measure in the barometer but a barometer is useful for atmospheric pressure what's more useful in a laboratory setting is to be able to actually measure the pressure of a sample for example not just the open atmosphere so this is the an example of a manometer and a manometer actually does not it's not opened atmosphere you have a sample uh reservoir that we can sort of in a controlled manner introduce to this um either a closed tube system with a with a youtube shape down here an inverted tube filled with liquid mercury there's also another design where we have an open atmosphere a open-ended tube where now we have to also account for the pressure of of the atmosphere in our calculations so the main premise here is that if we introduce um to our manometer uh a sample of gas that has um any really measurable uh uh well gases will have measurable pressure but if it's a volatile liquid that has a substantial vapor pressure what will happen is that when we open up this valve here the pressure will force the the liquid mercury down right there's a greater pressure in the side of the manometer and that will cause this sort of vertical displacement on the other side of this inverted tube since this is closed to the atmosphere it's a closed system we don't have to account for the atmospheric pressure in our calculations the pressure of the gas is going to be equal to the height times rho times or height times density times the acceleration due to gravity so essentially that vertical displacement is directly proportional to the amount of pressure from our gas sample and i'll look at one practice problem in a second uh that deals with that in the open-ended cases so both of these on the right are open-ended cases but uh uh the what's the the situation over here is when the pressure of our sample is less than that of the atmospheric pressure okay and so uh what happens is since the other end is not vacuum sealed um it's open to the atmosphere then we also have to always account for the fact that there's atmospheric pressure pushing down on the open end of the tube so if we open up this valve to our sample gas and that sample gas is not pushing with one atmosphere of pressure or 760 millimeters of mercury or um 760 torr whatever unit you want to describe one atmosphere as then that means that um if it's less than one atmosphere the atmospheric pressure will push down the mer the liquid mercury and will actually measure a net negative displacement in the in the liquid height um in our uh manometer um and you can imagine a case it doesn't show it here but imagine a case where the gas sample pressure is exactly equal to one atmosphere then you would measure a net zero displacement in the liquid height okay that means that the sample pressure is equal to or the gas pressure is equal to that of atmospheric pressure now in the case where we have the gas sample when opened to the atmosphere actually is uh or open to the manometer is greater than one atmosphere of pressure um then that means that this will not only uh it has to overcome the force of the the atmospheric pressure but also it'll push just a little bit beyond that so if it was exactly one atmosphere we would measure no displacement if it's above one atmosphere it will push the liquid mercury down over here which causes a vertical displacement on the um inverted side of the of the manometer and then here you can see in these expressions for the for the for the first case that i described in the open-ended denominator of the gas pressure is actually equal to the atmospheric pressure it has to um account for that and then in this case um it's if we measure that negative displacement in height so it's minus height times density times acceleration due to gravity so that means that this would indicate a system or a gas pressure less than that of atmospheric pressure um and then the other manometer case where we have a gas sample with with greater than atmospheric pressure it um we have to say that the gas pressure is equal to one atmosphere because it overcame that one atmosphere the atmospheric pressure plus whatever that displacement height was so let's take a quick look um before uh the video uh before i end the video at two examples the first one is going to be the closed end situation so we do not have to account for atmospheric pressure in our calculations the question here that we want to answer is what is the pressure pressure in tor based off of this diagram that we can see here so what we can do is and again this is liquid mercury this is mercury liquid okay so uh to start we can uh use our starting value that we're reading directly off that manometer 26.4 um centimeters of mercury uh height displacement and essentially what we can do then is we can say that that is a unit of measure of pressure centimeters mercury although it's not a common one so we want to convert it to tor which is more common so we what we can do is actually say okay well if it's a centimeter of mercury that's just the length unit we can actually convert the length unit from one centimeter to 10 millimeters of mercury it's that is that straightforward and now that we're in millimeters of mercury that is a common pressure unit so now we can say use a conversion factor here one millimeter of mercury is um equal to one torr and so really this is two 26.4 times 10 so it's 264 torr of um for our gas pressure there okay and now in the um open end case we also now we have to consider the um uh atmospheric pressure um and i guess i'll point out too that the benefit of the closed end you notice that the gas pressure is 264 torr atmospheric pressure is 760 torque so that's not greater than atmospheric pressure so if this was the open-ended case we wouldn't get a vertical displacement in the liquid marker it would actually go down because atmospheric pressure is is is greater so speaking of open-ended cases let's look at this case here we can make a prediction that with this gas sample must have greater than one atmospheric pressure because look at this it was a vertical displacement of liquid mercury on the other side so that gas sample must have overcome the atmospheric pressure and then some so let's let's do a calculation to see if that's the case um what i'm going to do here is essentially uh jot down uh oh first let's see what do we want to solve for let's solve for this pressure in atmospheres pressure in atmospheres is it greater than one so what i'm going to do is actually take this step by step and say that we have 13.7 centimeters of mercury similar to before i'm going to convert that into millimeters of mercury first so this is one centimeter of mercury is equal to 10 millimeters of mercury and i can jump i can i can keep going from here but i want to actually just pause to now talk about what we just calculated this 137 milligrams of mercury is not the pressure of the gas sample that's the additional vertical height gained just right here but remember this is open to the atmosphere so this pressure 137 is what was on top of atmospheric pressure so really the pressure of the gas pressure of the gas is going to be equal to the uh pressure of the atmosphere plus whatever we just calculated that vertical displacement plus 137 millimeters of mercury so what this equals is 760 millimeters of mercury for atmospheric pressure which is approximating about one atmosphere um which is 760 ml millimeters of mercury plus 137 millimeters mercury and that will give us the pressure of the gas is actually 897 millimeters of mercury okay so that's that's from the equation that i gave you above so what is this in atmospheres we take 897 millimeters of mercury and we know that the conversion factor is 760 millimeters of mercury is equal to one atmosphere one atmosphere so uh what this gives us then is 1.18 atmospheres we did predict just based off of looking at the how the displacement occurred in the manometer that it's greater than one atmosphere and we have confirmed that in this example problem