U-Tube (Pv=nRT, Pascals) Problem

In summary, the problem involves a U-tube filled with mercury at a certain temperature and barometric pressure. The task is to determine the temperature at which the air column on the left side of the tube decreases by 10 cm. The solution involves using the ideal gas law and Pascal's principle to set up equations and solve for the final temperature. However, there is some confusion regarding the use of torr and pascals in the calculations.
  • #1
AKJ1
43
0
Challenge problem

1. Homework Statement


A U-tube of uniform cross section contains mercury. The initial temperature of the system is 301K and the barometric pressure is 749 torr. Suppose the left side of the tube is closed at the top and the temperature is raised to a temperature such that the air column on the left is 60cm high. Determine the temperature at which this occurs.

Initially, the mercury level is 50 cm on both sides. The height of each tube, as measured from the bottom, is 100 cm.

Homework Equations



Ideal gas law PV=nRT

Pascals principle P = P0+ρgh

The Attempt at a Solution



I know what my initial pressure and temperature is. In addition I can write my initial volume on the left tube as A(.50) .

My goal is to find the temperature when the the column of mercury on the left tube drops by 10cm (air column now 60cm).

So now I want to develop an expression to describe the final state of the left side of the tube.

Im assuming the number of moles will not change & R is a constant of course.

PliVli/(Ti) = PlfVlf/(Tf)

So now I have one equation and two unknowns ( I don't know the final pressure or temperature, the volume can be written in terms of AH). So I need to develop another expression. Here is where I can take full advantage of pascals principle.

When the mercury goes down on the left side & goes up on the right side, the pressure on the left side of the tube, at the "surface" of the mercury, must be equal to the pressure on the right tube at the same point.

The pressure on the right side of the tube can be written as

P = Pairhggh

Where, h= 0.20m . I choose 0.20m because when the left side goes down by .10m the right side goes up by .10m. Thus if I want to look at the pressure on the right side, I must acknowledge that the point is 0.20m below the surface of the liquid.

So now I can say Plf = P = Pairhggh

∴ I am now ready to solve for Tf .

Issue is, when I went to solve for Tf , I got an answer that did not correspond to one of the solutions.

here are the values I used

ρhg = 13500 kg/m^3
h = 0.20m
Pair = 101325 pa
Pli = 749 torr = 99858 pa
Vli[/SUB = A(0.50m)
Vlf[/SUB = A(0.40m)

And of course the initial temperature is 301 K.
 
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  • #2
AKJ1 said:
I went to solve for Tf , I got an answer
Could we see your work, please.
 
  • #3
Bystander said:
Could we see your work, please.

PliVli/(Ti) =PlfVlf/(Tf)

Tf =PlfVlfTi /(PliVli)

Note:
Plf = Pair + ρhggh = (101325)+(13500)(9.80)(.20) = 127785 pa

Subbing in values:Tf = (127785) (A(0.40))(301) / ((99858)(A(0.50)) = 308 K
 
  • #4
What is the final volume of the gas on the left?
 
  • #5
ehild said:
What is the final volume of the gas on the left?

I argue that the final volume on the left side will be, Vlf = A(0.40) .

The 0.40 is the height of the fluid on the left.

Is this a correct statement?
 
  • #6
Why did you use the height of the fluid when the gas law applies to gases?
 
  • #7
AKJ1 said:
initial temperature of the system is 301K and the barometric pressure is 749 torr
 
  • #8
ehild said:
Why did you use the height of the fluid when the gas law applies to gases?

Goodness me, I have no idea.

I now used (0.60) instead of (0.40) , I managed to get about 460K now.
 
  • #9

I don't follow, is the pressure given not an absolute pressure?
 
  • #10
AKJ1 said:
I don't follow
Yes you do.
AKJ1 said:
Pli = 749 torr = 99858 pa
 
  • #11
Bystander said:
Yes you do.

749 torr =/= barometric 749 torr ?
 
  • #12
You can measure the pressure in torr-s - the height in mm of a mercury column of equivalent pressure. So the 20 cm mercury column means additional 200 torr pressure. No need to convert to pascals.
The density of mercury depends on temperature, and so does its length, but assuming that only the air column is heated you can ignore it.
 
  • #13
ehild said:
You can measure the pressure in torr-s - the height in mm of a mercury column of equivalent pressure. So the 20 cm mercury column means additional 200 torr pressure. No need to convert to pascals.
The density of mercury depends on temperature, and so does its length, but assuming that only the air column is heated you can ignore it.

This is quite new to me, so excuse me if I sound a bit ignorant, but

1 cm of mercury column constitutes 10 torr pressure?

Therefore when I consider the 20cm mercury column, I should add 200 torr pressure. Consequently, I should account for the 40cm mercury column on the left tube?
 
  • #14
The 40 cm column in the left tube is below the air column, it does not add to the pressure. See attachment. The downward force at the red level has to be equal in both tubes, so the air pressure in the left tube is equal to the atmospheric pressure+ the pressure of the 20 cm mercury column.

mercurymanometer.JPG
 
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Likes AKJ1
  • #15
ehild said:
The 40 cm column in the left tube is below the air column, it does not add to the pressure. See attachment. The downward force at the red level has to be equal in both tubes, so the air pressure in the left tube is equal to the atmospheric pressure+ the pressure of the 20 cm mercury column.

View attachment 78923

Very clear explanation. Got it, thank you!
 
  • #16
You are welcome :).
 

Related to U-Tube (Pv=nRT, Pascals) Problem

1. What is the U-Tube (Pv=nRT, Pascals) Problem?

The U-Tube (Pv=nRT, Pascals) Problem is a physics problem that involves using the ideal gas law, PV=nRT (where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature) and Pascal's law (which states that pressure in a closed system is the same everywhere) to solve for an unknown variable in a system involving a U-shaped tube filled with gas.

2. How do I solve the U-Tube (Pv=nRT, Pascals) Problem?

To solve the U-Tube (Pv=nRT, Pascals) Problem, you will need to set up an equation using the ideal gas law and Pascal's law. This will involve identifying the known variables and plugging them into the equation to solve for the unknown variable. It may also involve rearranging the equation to isolate the unknown variable.

3. What are some common mistakes when solving the U-Tube (Pv=nRT, Pascals) Problem?

Some common mistakes when solving the U-Tube (Pv=nRT, Pascals) Problem include not properly identifying the known variables, not converting units to match, and not correctly applying the ideal gas law and Pascal's law. It is important to double check your work and make sure all steps are clearly shown in order to avoid mistakes.

4. Are there any tips for solving the U-Tube (Pv=nRT, Pascals) Problem?

One tip for solving the U-Tube (Pv=nRT, Pascals) Problem is to draw a diagram of the U-shaped tube and label all known variables. This can help visualize the problem and make it easier to identify which variables are known and which are unknown. It is also helpful to double check all units and make sure they are consistent throughout the problem.

5. How can I use the U-Tube (Pv=nRT, Pascals) Problem in real-life situations?

The U-Tube (Pv=nRT, Pascals) Problem can be applied to real-life situations in which gas is involved, such as in the study of meteorology, scuba diving, and chemical reactions. It can also be used to understand the behavior of gases in closed systems, such as in a car engine or a hot air balloon.

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