Change in pressure of a gas inside a tube

[lorele]

Homework Statement

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. Homework Equations
$$p=\rho g h$$

The Attempt at a Solution

krrk
$$p'=p_{water} + p_{mercury} + p_{air}=\rho_w g h_w + \rho_m g h_{mercury} + p_{air}$$
$$p=p_{water} + p'_{mercury} + p_{air} + p_{new}=<b>\rho_w g h_w + \rho_m g (h_{mercury}+\Delta h) + p_{air} + {mg}/S </b>$$
$$p'-p={mg}/S + \rho_m g \Delta h$$[/B]

 

[Bystander]

Your question presumably has to do with the difference between your answer and the given answer?

 

[lorele]

Your question presumably has to do with the difference between your answer and the given answer?

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

 

[Bystander]

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

 

[lorele]

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

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.

Homework Statement

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.

Homework Equations

p=ρgh

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

 

[Bystander]

closer to the solution, albeit without the 2, which I don't know where it comes from

level of mercury on the left has risen "Δh"

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

 

[lorele]

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

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