# Change in pressure of a gas inside a tube

• lorele
In summary: I missed that! In summary, the given solution for the increase of pressure p'-p of gas G in a U-shaped tube containing mercury, water, and another fluid of unknown density is p'-p=mg/S-2ρmgΔh, where Δh is the change in the level of mercury on the left side of the 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
$$p'=p_{water} + p_{mercury} + p_{air}=\rho_w g h_w + \rho_m g h_{mercury} + p_{air}$$
$$p=p_{water} + p'_{mercury} + p_{air} + p_{new}=\rho_w g h_w + \rho_m g (h_{mercury}+\Delta h) + p_{air} + {mg}/S$$
$$p'-p={mg}/S + \rho_m g \Delta h$$
[/B]

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

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

Bystander said:
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

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

lorele said:
closer to the solution, albeit without the 2, which I don't know where it comes from
lorele said:
level of mercury on the left has risen "Δh"
... and, the level of mercury on the right has _________?

Bystander said:
... and, the level of mercury on the right has _________?
been reduced by $-\Delta h$ . OK, thanks!

## 1. What is the relationship between temperature and pressure in a gas inside a tube?

The relationship between temperature and pressure in a gas inside a tube is known as the ideal gas law, which states that pressure and temperature are directly proportional. This means that as temperature increases, pressure also increases, and vice versa.

## 2. How does the volume of a gas inside a tube affect its pressure?

According to Boyle's Law, the volume and pressure of a gas inside a tube are inversely proportional. This means that as the volume of the gas decreases, the pressure increases, and vice versa. So, if the volume of the tube is decreased, the gas particles will collide more frequently with the walls of the tube, resulting in an increase in pressure.

## 3. What happens to the pressure of a gas inside a tube when the number of gas particles is increased?

When the number of gas particles inside a tube is increased, the pressure of the gas also increases. This is because there are more gas particles colliding with the walls of the tube, resulting in a higher pressure. This relationship is known as Gay-Lussac's Law.

## 4. How does the type of gas affect the pressure inside a tube?

The type of gas inside a tube can affect the pressure in several ways. First, different gases have different molecular weights, which can impact how the gas particles collide with the walls of the tube. Additionally, the intermolecular forces between gas particles can also affect the pressure. For example, gases with stronger intermolecular forces, such as water vapor, will have higher pressures compared to gases with weaker intermolecular forces, such as helium.

## 5. What factors can cause a change in pressure of a gas inside a tube?

The pressure of a gas inside a tube can be affected by various factors, including temperature, volume, and the number and type of gas particles. Other factors that can cause a change in pressure include external forces, such as compression or expansion of the tube, and changes in altitude or atmospheric pressure. Additionally, chemical reactions or phase changes within the gas can also cause a change in pressure.

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