Hey folks, here is a question: A thread of mercury of lenght 15cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the lenght of the trapped air collumn is 20cm. Calculate the lenght of the air column when the tube is held. (i) Horizontally, (ii) Vertically with the open end underneath. (Atmospheric pressure= 76cm of mercury) In this question, am intrested in knowing what will happen when the tube is held horizontal and vertical as it will help in the calculation. What do they mean or it imply by refering the tube as having "uniform cross-sectional area"? Thank you as reply.
The initial state is telling you the volume of the air and the pressure - atmospheric pressure and the weight of the mercury. The gas will expand because the weight of the mercury will be removed. The gas will expand more (it will be at a vacuum). The pressure of the gas will be atmospheric minus the weight of the mercury. It simply makes the problem easier to calculate - you won't have to worry about the actual weight of the mercury - you can use the 76mm of Hg = atm. The volume can be calcualted usting PV=nRT and since you know the cross-section you are all set.
Remember that Boy's law states that the volume of given mass of gas is inversely proportional to the pressure , provided temperature remains constant. Can you relate your explaination to it?
Sorry - I will have to wait for tomorrow, probably, I have a presentation and an experiemnt that is starting right now. If everything goes perfectly i will get back today - and probably have a heart attack with surprise.Please Register or Log in to view the hidden image!
OK. Boyle's law says that PV=k or P=k/V which means that volume is inversely proportional to a change in pressure. The is makes sense in an everyday sort of way if you try to compress a cylinder of air the pressure goes up. That is essentially what the ideal gas law is saying; PV = nrt, for a constant number of atoms (n) at a constant temperature (T) PV= k = nrt. For your case when the capilary is turned so the water is in the bottom of the capillary the air pocket will expand. The increase in the volume is inversely proportional to the pressure - right? The decrease in the pressure means that the pressure will be less than atmospheric or a vacuum. The vacuum will prevent the further expansion of the volume. Not sure exactly what you are looking for in the molecular theory but try this idea. There are a certain number of molecules in the gas bubble the molecules have a constant velocity due to the temperature of the gas. The frequency of the collisions between the molecules and a surface unit area of the walls of the capillary is the pressure of the gas. When the capillary is turned so the water is down the volume increases which means the density of the gas is decreased so that there are fewer molecular collisions per unit area of the walls of the capillary so the pressure is lower. I hope I am not doing your homework.
This is the question: A thread of mercury of lenght 15cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the lenght of the trapped air collumn is 20cm. Calculate the lenght of the air column when the tube is held. (i) Horizontally, (ii) Vertically with the open end underneath. (Atmospheric pressure= 76cm of mercury) Quote: For your case when the capilary is turned so the water is in the bottom of the capillary the air pocket will expand : I never told you that water is in the bottom of the capillary tube. I said that, 'A thread of mercury of lenght 15cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end'. Quote: I hope I am not doing your homework?: You are not doing my home work, am only trying to understand what am not clear with. Ok!
A. If there was no mercury in the capillary what would the pressure of the air in the tube be? It would be at atmospheric pressure; assuming STP the pressure would be 101.325 kPa or14.696 psi. B. Now we will put 15 cm of Hg on into the tube. how much will the weight of the Hg increase the pressure? 76 cm = 14.696 psi. So the increase the in the pressure will be (15/76) X 14.696 psi. C. know we have our starting pressure: 14.696 psi + 2.900psi = 17.597 psi D. What would the volume be if we turned the capillary on it's side? What would be the forces on the gas? Essentially none, which means the pressure would equalize or the pressure in the capillary would go to 14.696 psi. E. So what would the volume go to? PV=nRT n is constant, T is constant (we can safely assume) SO PV = constant Therefore you can say PV(verticle) = PV(horizontal). Of couse P(hor) and P(vert) are different so lets calculate the volume change. (P(vert) X V(vert))/P(hor) = 17.597 psi x 20cm / 14.696psi = 23.95 cm 23.95 is not a volume - but the x-sectional area did not change so we are OK. This is done exactly the same way. I would take the starting point as the horizontal capillary but it does not matter which initial state you use. A Starting Pressure = 14.696psi Volume = 23.95 cm (this is not really a volume, the volume is acutally 23.95 cm x the constant x-section) B invert the capillary so we now have the weight expanding the gas and lowering the pressure. How much will the pressure decrease? 15/76 of 14.696psi. C You know have your intial pressure your final pressure and your starting volume so just use PV=PV and calculate your final volume. As a check it should of course be larger than the intial 23.95 cm. Always remember the physics that is going on and do not get lost in the math. I did this rather quickly since my boss is unkowingly paying me for this Please Register or Log in to view the hidden image! so hopefully there are no mistakes - don't believe there are. I didn't remember it was Hg I thought it was water we were talking about, but it makes no difference for the concept! edited to add I did make an error that I fixed.
Question: A thread of mercury of lenght 15cm is used to trap some air in a capillary tube with uniform cross-sectional area and closed at one end. With the tube vertical and the open end uppermost, the lenght of the trapped air collumn is 20cm. Calculate the lenght of the air column when the tube is held. (i) Horizontally, (ii) Vertically with the open end underneath. (Atmospheric pressure= 76cm of mercury) It should become clear by now that am looking for one thing which is no other thing but how to derive the pressures that we be used for the calculation. I already know the formular to use which is no other but p1v1=p2v2, where p and v are pressure and volume respectively. Quote: If there was no mercury in the capillary what would the pressure of the air in the tube be? It would be at atmospheric pressure; assuming STP the pressure would be 101.325 kPa or14.696 psi. B. Now we will put 15 cm of Hg on into the tube. how much will the weight of the Hg increase the pressure? 76 cm = 14.696 psi. So the increase the in the pressure will be (15/76) X 14.696 psi. C. know we have our starting pressure: 14.696 psi + 2.900psi = 17.597 psi I want to understand what you mean this way; if there is no mecury in the tube, atmospheric pressure (76cm) will be exerted on the tube. Therefore, if a mecury thread of lenght 15cm is use to trap air in the tube, then the pressure will increase becuase of the atmospheric pressure (i.e 15cm+ 76cm) and this occured when the tube was vertical and the open end uppermost. When the tube was held vertical and the open end uppermost, the lenght of the trapped air collumn we were made to understand was 20cm. Now that the the tube is held. (i) Horizontally, we are required to calculate the lenght the new air collum which equals the volume of air. This means that as the tube is held horizontal with one end open, then the air collumn which was initially 20cm when it was held vertical with uppermost end open, will either increase or decrease depending on wether the pressure which increased to 91cm (15cm+76cm) due to the atmospheric pressure being exerted on the mecury and the mecury in turn exert on air trapped below it; it depends on wether the pressure exceeds far more than 91cm or decreases far below it when the tube was held horizontal. That above is the physics behind it which I don't understand. Please can you help me out? (ii) Vertically with the open end underneath: Now that the tube is held vertically again as in the first instance but this time with the open end underneath: Will the pressure increase or decrease? If the pressure increases or decreases then why and how did it happen? That is the physics behind it which I don't understand. Can you please help me out?
It is clear what you are looking for and I gave it to you. Yes I don't think you are trying here, I told you the answer and told you why. Maybe this will help. Imagine you are in the bottom of a 5 m tube and you are given 2 option: Option 1: The tube is verticle and and there will be 5 - 100 kg bags of sand dropped into the tube. Option 2: The tube is horizontal and there will be 5 - 100 kg bags of sand put into the mouth of the tube. I think you would pick option 2 because in option 1 there would be a much greater pressure on your head - got it? Maybe this will help. Get a bicycle pump. Imagine that the end of pump is plugged so no air comes out. The pump is your simulated capillary - the handle end is the mercury end of the capillary and the bottom of the pump is the closed end of the capillary. We can use a 10 pound weight to simulate the mercury. Lay the pump on it's side on the ground and attach a 10 pound weight to the handle, will the handle move in, move out or not move? Hold the pump vertically with the handle on top, attach a 10 pound weight to the handle what direction will the handle move? Hold the pump vertically so the handle is on the bottom, attach a 10 pound weight to the handle, which direction will the pump handle move? If you cannot figure out which direction the handle will move in these scenarios or cannot see how this relates to your experiment then I would have to suggest that you should pursue a career in business, art or music and not science. The other possibility is that you need a better translator.:shrug:
Quote: Imagine you are in the bottom of a 5 m tube and you are given 2 option: Option 1: The tube is verticle and and there will be 5 - 100 kg bags of sand dropped into the tube. Option 2: The tube is horizontal and there will be 5 - 100 kg bags of sand put into the mouth of the tube. I think you would pick option 2 because in option 1 there would be a much greater pressure on your head - got it? I think I understand your explaination this way; when the tube is vertical with tube open at the uppermost end, the atmospheric pressure will be exerted higher on the mecury thread but when the tube is horizontal the atmospheric pressure will be exerted less on the mecury thread since the tube is lying horizontal and the atmospheric pressure has been distributed all round the tube along it full lenght. So the atmospheric pressure is partly exerted on the tube and on the air column. Quote: Get a bicycle pump. Imagine that the end of pump is plugged so no air comes out. The pump is your simulated capillary - the handle end is the mercury end of the capillary and the bottom of the pump is the closed end of the capillary. We can use a 10 pound weight to simulate the mercury. Lay the pump on it's side on the ground and attach a 10 pound weight to the handle, will the handle move in, move out or not move? The handle will not move in because the end is plugged so that no air is coming out. The reason been that the air that is suppose to escape through the plugged end and give the handle chance to move in cannot because the passage way has been blocked. Therefore, as long as the air remains inside, it will give the handle no space to move in. Thus the person using this pump will exprience the air resistance and he will finding it difficult using the pump. Quote: Hold the pump vertically with the handle on top, attach a 10 pound weight to the handle what direction will the handle move? The handle will move towards the direction of the 10 pound weight on it because the end is still plugged. Quote: Hold the pump vertically so the handle is on the bottom, attach a 10 pound weight to the handle, which direction will the pump handle move? The handle will move towards the 10 pound weight because the 10 pound weight act like some kind of gravity pulling on the handle downward. Does that mean that the when the tube is turned so that the open end is underneath a kind of gravity is been exerted on the thread of mecury and the air collumn. If that is the case, how do I explain that. That is how I can figure it out.
No. I have explained it mathematically, I have explained per your example and given a different example. I have to believe you are just messing with us. Or because there would be no force on the handle, your still messing with us. No. It is not possible you think that or you could not operate a toaster without injuring yourself. Some sort of gravity? Really? I have spoon fed you the answer and you still ask me how do you explain it? As I have realized - sure I am thick - you have been just messing with people. Well it doesn't matter it was kind of fun to do the little exercise.:shrug:
I meant no insult or abuse on anybody. Please don't be offended. What exactly are you saying about my post? Tell me your mind in clear terms. What do you think about my last reply? What do you mean by "am messing up with you?"
Your post shows that you have no idea about the concept of force, gravity or pressure I can't help you any more than I have. It is very depressing. You apparently do not have even the most basic of understanding. I told you how to do the problem in post #2 I actually did the problem for you in post #10 I gave you another example to help visualize what was going on in post #12. Then you ask me again what is going on in the problem. I assume you are messing with me because I cannot believe you are that clueless on the problem. And if you are then give up, and learn the basic concepts of pressure and force before trying to tackle a problem like this. Good luck.
You have actually done your best. Is left for me to do my own best by studing through all your replies thouroghly and see how good I make of them. I repeat once again, never be offended, am only trying to understand what I don't know. Thank you and remain blessed.
This is my working from what me and Origin has discussed so far Using the equation P[sub]1[/sub]V[sub]1[/sub] = P[sub]2[/sub]V[sub]2[/sub] Where P[sub]1[/sub] = initial pressure, V[sub]1[/sub] = initial volume, P[sub]2[/sub] = final pressure, and V[sub]2[/sub] = final volume. But the lenght, l of the air column is proportional to the volume of air in the tube. Also the lenght of the mecury column is proportional to Pressure, P which the mecury exert on the air below it. As the tube is vertical with open end uppermost, atmospheric pressure = 76cm (according to the data given) is being exerted on the mecury column. Thus, the total pressure = atmosperic pressure + pressure of mecury (76cm + 15cm) = 91cm = P[sub]1[/sub] Using the established fact in solving for V[sub]2[/sub] when the tube is held horizontal with one end open: V[sub]2[/sub] = P[sub]1[/sub]V[sub]1[/sub]/P[sub]2[/sub] = 91*20/76 23.95cm[sup]3[/sup] correct to d.p When the tube is vertical with the open end underneath, the atmospheric pressure on the mecury column will get reduced Thus P[sub]2[/sub] = atomspheric pressure - mecury pressure (76 - 15) =61cm V[sub]2[/sub] = P[sub]1[/sub]V[sub]1[/sub]/P[sub]2[/sub] = 91*20/61 = 29.84cm[sup]3[/sup] correct to d.p I hope Origin is watching this.
This is my working from what me and Origin has discussed so far Using the equation P[sub]1[/sub]V[sub]1[/sub] = P[sub]2[/sub]V[sub]2[/sub] Where P[sub]1[/sub] = initial pressure, V[sub]1[/sub] = initial volume, P[sub]2[/sub] = final pressure, and V[sub]2[/sub] = final volume. But the lenght, l of the air column is proportional to the volume of air in the tube. Also the lenght of the mecury column is proportional to Pressure, P which the mecury exert on the air below it. As the tube is vertical with open end uppermost, atmospheric pressure = 76cm (according to the data given) is being exerted on the mecury column. Thus, the total pressure = atmosperic pressure + pressure of mecury (76cm + 15cm) = 91cm = P[sub]1[/sub] Using the established fact in solving for V[sub]2[/sub] when the tube is held horizontal with one end open: V[sub]2[/sub] = P[sub]1[/sub]V[sub]1[/sub]/P[sub]2[/sub] = 91*20/76 23.95cm[sup]3[/sup] correct to d.p When the tube is vertical with the open end underneath, the atmospheric pressure on the mecury column will get reduced Thus P[sub]2[/sub] = atomspheric pressure - mecury pressure (76 - 15) =61cm V[sub]2[/sub] = P[sub]1[/sub]V[sub]1[/sub]/P[sub]2[/sub] = 91*20/61 = 29.84cm[sup]3[/sup] correct to d.p I hope Origin is watching this. What do Origin has to say about this?
