Change in pressure of a gas inside a tube

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1. Aug 30, 2015

lorele

1. The problem statement, all variables and given/known data
A U-shaped tube of uniform section "S" has one extreme closed and the other one open to the atmosphere. The tube contains mercury (M) in the central part, a gas (G) to the left that exerts a pressure "p" and a column of water (W) of height $$h_w$$ to the right. Then, a mass "m" of another fluid of unknown density is poured over the right side and, when equilibrium is reached, the level of mercury on the left has risen "$\Delta h$", and the pressure of G is now p'. Determine the increase of pressure p'-p of G according to the density of mercury ($\rho_m$), $\Delta h$, S and m.
The given solution is $p'-p={mg}/S - 2 \rho_m g \Delta h$ .
2. Relevant equations

$$p=\rho g h$$

3. The attempt at a solutionkrrk
$$p'=p_{water} + p_{mercury} + p_{air}=\rho_w g h_w + \rho_m g h_{mercury} + p_{air}$$
$$p=p_{water} + p'_{mercury} + p_{air} + p_{new}=\rho_w g h_w + \rho_m g (h_{mercury}+\Delta h) + p_{air} + {mg}/S$$
$$p'-p={mg}/S + \rho_m g \Delta h$$

2. Aug 31, 2015

3. Sep 1, 2015

lorele

Yes, I don't know how to get the actual answer.

4. Sep 1, 2015

Bystander

You'll want to re-examine your definitions and uses of "p ′ " and "p."

5. Sep 1, 2015

lorele

I can't edit the first post, so I made a new one. I swapped p' and p, but I think it's OK now. Still, I can't get the answer, and I don't know where I went wrong. Also, as the pressure increases with a decrease of height (that is, the change of height is actually negative), I changed the pressure of mercury in the second case to $$h-\Delta h$$, and I got somewhat closer to the solution, albeit without the 2, which I don't know where it comes from.

1. The problem statement, all variables and given/known data
A U-shaped tube of uniform section "S" has one extreme closed and the other one open to the atmosphere. The tube contains mercury (M) in the central part, a gas (G) to the left that exerts a pressure "p" and a column of water (W) of height $h_w$ to the right. Then, a mass "m" of another fluid of unknown density is poured over the right side and, when equilibrium is reached, the level of mercury on the left has risen "Δh", and the pressure of G is now p'. Determine the increase of pressure p'-p of G according to the density of mercury (ρm), Δh, S and m.
The given solution is p′−p=mg/S−2ρmgΔh.

2. Relevant equations
p=ρgh

3. The attempt at a solution

$$p=p_{water}+p_{mercury}+p_{air}=ρ_w g h_w+ρ_m g h_{mercury}+p_{air}$$
$$p'=p_{water}+p'_{mercury}+p_{air}+p_{new}=\rho_wgh_w+\rho_mg(h_{mercury}-\Delta h)+p_{air}+{mg}/S$$
$$p′−p=p_{water}-p_{water}+p_{air}-p_{air}+p'_{mercury}-p_{mercury}+mg/S=\rho_mg(h_{mercury}-\Delta h)-ρ_m g h_{mercury}+mg/S=mg/S-\rho_mg\Delta h$$

6. Sep 1, 2015

Bystander

... and, the level of mercury on the right has _________?

7. Sep 9, 2015

lorele

been reduced by $-\Delta h$ . OK, thanks!