The liquid in the open-tube manometer in Fig. 12.8a is mercury, y₁=3.00 cm,and y₂=7.00 cm. Atmospheric pressure is 980 millibars. What is (a) the absolute pressure at the bottom of the U-shaped tube; (b) the absolute pressure in the open tube at a depth of 4.00 cm below the free surface; (c) the absolute pressure of the gas in the container; (d) the gauge pressure of the gas in pascals?

Question

The liquid in the open-tube manometer in Fig. 12.8a is mercury, y₁=3.00 cm,and y₂=7.00 cm. Atmospheric pressure is 980 millibars. What is (a) the absolute pressure at the bottom of the U-shaped tube; (b) the absolute pressure in the open tube at a depth of 4.00 cm below the free surface; (c) the absolute pressure of the gas in the container; (d) the gauge pressure of the gas in pascals?

Accepted Answer

Hi, everyone in this problem, we are asked to actually look into an open to manometer which is used in at an atmospheric pressure of 0.7838 T M. And I am going to write that down as P A PM here. So we have B ATM which will equal to 0. ATM. And then the fluid used in the tube has a density, a row of 0.9 g per centimeter cube. And the MNO meter column hide is 30 centimeter here, as shown in this image, we are asked to actually first calculate the absolute pressure within the gas tank here and then second, the absolute pressure at the bottom And then next, the absolute pressure 15 cm above the bottom. And lastly the grade uh the gauge pressure off the gas tank. OK. So the way we want to tackle this problem is to by utilizing the fact that uh we know that the pressure in a mono meter in this case, is going to be the same in the bottom at the same level. So the pressure is going to be the same at the same level. So the left end of the tube here is going to be connected to the tank where a gas tank here where the pressure is P while the right end of the tube is open to the atmosphere where the atmospheric pressure is 0.78 to ATM the pressure exerted by a column of liquid of H. And density row is given by the hydrostatic pressure equation B well equals to row G H. And essentially we will have to add this row G H with uh the atmospheric pressure or whatever pressure is acting upon the system additionally to that. So first, you want to determine that the pressure at the bottom of the tube due to fluid in the left column here will equals two. So I'm just gonna indicate this point to P A. So I'm just gonna write down P A will equals to the pressure of the gas tank plus the hydrostatic pressure here which is the row G H. In this case, the P is going to be this pressure here. The row is row G and the H is going to be the 40 centimeter which in si is 0.4 m just like. So OK. Next we have point B and point B at point B pressure B will then have pretty much the same thing. But this time is P ATM which is the 0.783 ATM here plus row G H which will equals to P ATM plus row G. And this one, it's the whole height of the uh column here, which actually will be 70 cm or 0. m. So this is 0.7 m just like so, so because as I said previously, because this two pressures are measured at the same level. So this two pressure must be equal to one another. So we are going to equalize this two P A T A B plus row G, 0.4 m equals to P ATM plus row G 0.7 m just like. So, ok, so now we want to utilize this to actually calculate for our gas tank pressure here because we know all the other uh necessary information to solve for our gas tank. So our first part is to actually calculate the absolute pressure within the gas tank, which is the P. So the P will then be rearranging this equation here. P ATM plus row G 0.7 m minus 0.4 m just like. So, so the P will actually then be P ATM Plus row G 0.3 m. And I'm gonna input in all the values that we know. So first the 0.9 g per centimeter cube is essentially going to equals to 0.9 times 10 to the power of three kg per meter cube. This is the si unit and I'm gonna leave the ATM in ATM just so that it's easier for us to keep track righter than pascals because pascals are a bit more complicated. So this will then be P equals 0.7838 T M which is the atmospheric pressure plus the row which is 0. times 10 to the power of three kg per meter cube, the G which is 9.81 m per second squared. And lastly, we have the 0.3 m there just like so, but we want to remember that the right side here is all in si while our atmospheric pressure there is actually in ATM. So we want to convert whatever is on the right side into ATM by multiplying this right side by uh one ATM Over the pascal value which is essentially going to be 1.013 times 10 to the power of five pascals per one ATM. So multiplying this right side right here and adding that to uh the pressure of 0.783 E T M will have the absolute pressure within the gas tank P of 0.8 oh nine ATM just like. So OK, so the absolute pressure uh b within the gas tank is 0.8 oh nine ATM. And then now we can actually move on to calculate our second part of the problem. So in the second part of the problem, it is asking about the pressure at the bottom. So for the second part P bot P bot or P bottom is going to just be essentially we already kind of calculate or formulate our equation previously. So P bottom is essentially going to either be P A or PB. Both of them are going to give us the same value because this uh the pressure at the bottom at the same level is going to be the same. But I'm just gonna use P A in this case. So P bottom equals P A which P A well equals to the P which is this P that we just found, which is the um absolute uh pressure in the gas tank plus H row G row G multiplied by 0.4 m, which is the hydrostatic pressure. So this will then be a bot equals the B is 0.8 oh nine ATM Plus the row is 0.9 times standard of power of three kg Per meter Cube. And then the G is 9.81 m/s squared. The height is 0.4 m. And as always, you wanna multiply it by the one ATM over one point oh 13 times 10 to the power of five pascals to make sure that everything is in the same unit. So this will actually corresponds to P bot to be a value of 0.8438 T M just like. So, so the pressure in the bottom of the manometer is 0. ATM. OK. So that is our second part. Next. We want to move on to the third part, which the third part is actually asking us about the absolute pressure centimeters above the bottom. So I'm gonna write that down as P 15, P 15 is essentially going to just be pretty similar to this. But the 0.4 uh we are not looking at 0.4 anymore, the height because we're not looking at the bottom of the mono meter. We're looking at this point right here, which is 15 me uh 15 centimeters away from the bottom. So essentially the height here is going to be four T minus 15 Which then that will be 25cm just like. So OK, so, so P 15 is going to then be equals to the absolute pressure of the gas tank plus the hydrostatic pressure row G, 0.25 m or 25 centimeters. So this will actually correspond to uh the P is going to be the same because that is the absolute value of the gas ta which is 0.8 oh nine ad M plus the row is still the same 0.9 times 10 to the power of three kg per meter cube. The G is 9.81 m per second squared. And then the height is 0.25 m. And then lastly, we want to multiply everything by the uh conversion value that we have one E T M over one point oh 13 times 10 to the power of five pascals. And that will corresponds to P 15 of value 0.831 ATM. So the pressure, The absolute pressure 15 cm above the bottom is going to be 0.831 ATM. OK. Last, we want to move on to the last part of our problem which is the fourth part which is asking about the gauge pressure of the gas tank. So P gauge here is the gauge pressure of the gas tank. And we want to recall that P gauge is essentially just the pressure difference, which is just P minus B A PM. We know the, the P which is uh found from the first part of our problem which is the absolute value in the gas tank. And the P ATM is given which is 0.783 ATM. So this will give us a P gauge value of 0.0 26 ATM, which is just the pressure difference between the absolute value of the pressure in the gas tank and the atmosphere. So the pressure of the gas tank is going to be 0.8 oh nine ATM. The pressure in the bottom is 0.843 ATM. The pressure 15 centimeters away from the bottom is 0.81. And the gauge pressure instead of the gas tank is 0.0 26 ATM and all of that will actually correspond to option A. So option A is going to be the answer to our problem and that will be all for this particular practice problem. If you guys have any sort of confusion, still, please make sure to check out our other similar lesson videos and that will be all for this video right here. Thank you.

Verified video answer for a similar problem:

Key Concepts

Pressure Measurement

Hydrostatic Pressure

Manometer Functionality