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This is a doubt on question 5.3 from the 5th chapter of the book Heat and Thermodynamics by Zemansky and Dittman. The problem statement is as follows:

Mercury is poured into the open end of a j-shaped tube, closed at the short end, trapping air in that end. How much mercury can be poured in before it overflows? The length of the tube is given as 1.5 meters, effects due to curvature of the bottom can be neglected, and 1 atm = 76 cm Hg.

I get that at some point after pouring in mercury, the pressure of air trapped inside is so much that you can't further add mercury, thus leading to spilling. I also tried equating the product of the initial pressure and volume to the final pressure and volume of the air column. I took $P_i=76 cm Hg$ and $V_i=(150cm)(Area)$ and equated it to the final product(assuming isothermal conditions hold), taking the length of the air column as x cm, and final pressure as the sum of atmospheric and the pressure due to mercury column corresponding to the same horizontal level as the air column inside. But, the answer I am getting is 60 cm, which cannot be true as the shorter arm is given to be only 50 cm in length, and the air is trapped in the short end.

Could someone explain where am I wrong?

Edit: I am attaching a diagram as well. Here, x is the length of the air column and the shaded region is mercury filled in the tube. Initially, this tube was empty, open to atmosphere.

J-tube with mercury

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  • $\begingroup$ Please post or link a picture so that we can see dimensions. For one thing, this answer could depend on how much air is trapped. I really would need to see a picture. $\endgroup$
    – Poisson Aerohead
    Commented Mar 1, 2023 at 3:39
  • $\begingroup$ No diagrams have been given in the text, but I will edit the post with a diagram I made. $\endgroup$
    – V Govind
    Commented Mar 1, 2023 at 4:31
  • $\begingroup$ As it stands, you drop this surprise with the last sentence that the short end is 50 cm, so clearly geometry was given. $\endgroup$
    – Poisson Aerohead
    Commented Mar 1, 2023 at 4:43
  • $\begingroup$ I should have mentioned that the longer arm is 1 m and the short one is 0.5 m at the beginning itself, my bad. But yeah, no diagrams were given in the book, $\endgroup$
    – V Govind
    Commented Mar 1, 2023 at 4:51

1 Answer 1

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I am not sure about the answer that you got, but judging by your picture and your reasoning for it being wrong, it sounds to me like you are not considering the full solution set. The arrangement in your picture will occur if the mercury compresses the air until it rounds the J hook. If that does not happen, the mercury portion will cut off somewhere on the right straight length before the hook and the uncolored portion representing air will extend through the full hook to the closed off segment on the left.

This is actually where I found a solution. I found that the mercury ends 26cm from the bottom, so that the air fills the full closed side on the left and 26cm of the right. I have not checked the answer too closely though.

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  • $\begingroup$ In the solution set given to us, the final length of the mercury column is given as 125 cm. I was trying to get to this answer. But your argument also makes sense. $\endgroup$
    – V Govind
    Commented Mar 1, 2023 at 6:16
  • $\begingroup$ If that means the Hg goes down the full 100 cm and back up the other side 25 cm, then the total pressure head change is 75 cm, which adds to the 76 cm at the open atmosphere to about double it. The air trapped in the tube goes from 150 cm length to 25 cm length (the area is constant), so its volume is one sixth the starting volume. Thus its pressure is six times atmospheric pressure. Unless I misunderstand the set up, that solution makes no sense. Note if your professor mixed up about it going around the hook, it agrees with my solution. I am at about ~25cm on the other side of the hook. $\endgroup$
    – Poisson Aerohead
    Commented Mar 1, 2023 at 6:23
  • $\begingroup$ Are you saying that this solution makes no sense, because the area is the same, and both sides having unequal pressures imply that equilibrium of forces cannot be achieved? $\endgroup$
    – V Govind
    Commented Mar 1, 2023 at 7:01
  • $\begingroup$ This is a static equilibrium problem. The pressure of the trapped air must equal the pressure of the mercury at that surface. The Hg is just done with a Bernoulli equation with no velocity. The air, I assumed, was the same temperature, so that P1V1 = P2V2. If I understand your setup, the answer you are telling me has the air go from 1.5 m to 0.25 m. Since the area is constant, the volume goes down as the length, to 1/6. So the pressure is 6 times the starting pressure. Hg at atmospheric is a 72 cm column, so adding another 75 cm is about twice atmospheric. Am I misunderstanding the setup? $\endgroup$
    – Poisson Aerohead
    Commented Mar 1, 2023 at 7:12
  • $\begingroup$ No no, this was just the setup I was also imagining. I understand it now. $\endgroup$
    – V Govind
    Commented Mar 1, 2023 at 9:21

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