The cross-sectional area of the U-tube shown below is uniform and is equal to one square centimeter (1.00cm^2). One end is open to the atmosphere. Atmosphere pressure may be taken as P1=1.01 * 10^5 Pa. The other end of the U-tube is connected to a pipe in which a constant gas pressure of P2=1.25 atmospheres is maintained. The U-tube contains a liquid which has a density of 1.0*10^4kg/m^3.
QUESTION: What the second liquid still in the arm that is open to the atmosphere, the pressure p2 is now reduced until it is equal to atmospheric pressure. What now is the difference (if any) between the liquid levels in the two arms of the U-tube?
To determine the difference in the liquid levels in the two arms of the U-tube, we need to consider the balance between the pressure exerted by the liquid column and the atmospheric pressure.
Given:
- Cross-sectional area of the U-tube: 1 cm^2
- Atmospheric pressure (P1): 1.01 × 10^5 Pa
- Gas pressure at the connected pipe (P2): 1.25 atmospheres
- Density of the liquid in the U-tube: 1.0 × 10^4 kg/m^3
When the pressure at the open end of the U-tube is reduced to atmospheric pressure (P1), the liquid levels in the two arms of the U-tube will equalize. The reason is as follows:
At the initial state, the pressure difference between the two arms of the U-tube is the difference between the gas pressure (P2) and the atmospheric pressure (P1). Using the conversion 1 atmosphere = 1.01 × 10^5 Pa, we have P2 = 1.25 × 1.01 × 10^5 Pa = 1.2625 × 10^5 Pa.
The pressure difference (ΔP) between the two arms is ΔP = P2 - P1 = 1.2625 × 10^5 Pa - 1.01 × 10^5 Pa = 0.2525 × 10^5 Pa.
By applying Pascal's law, we know that the pressure difference between the two arms will cause a difference in the heights of the liquid columns, given that the cross-sectional area is uniform.
The pressure difference ΔP is related to the height difference Δh by the formula ΔP = ρgh, where ρ is the density of the liquid and g is the acceleration due to gravity.
Rearranging the equation to solve for Δh: Δh = ΔP / (ρg).
Substituting the given values, we get:
Δh = (0.2525 × 10^5 Pa) / (1.0 × 10^4 kg/m^3 × 9.8 m/s^2)
Δh ≈ 0.2576 meters
Therefore, when the gas pressure is reduced to atmospheric pressure, the difference in the liquid levels in the two arms of the U-tube will be approximately 0.2576 meters.