/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 50 Find the pressure at the bottom ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the pressure at the bottom of a graduated cylinder that is half full of mercury and half full of water. Assume the height of the cylinder is \(0.25 \mathrm{~m}\) and don't forget about atmospheric pressure!

Short Answer

Expert verified
The pressure at the bottom of the cylinder is \(2.70 \times 10^5 \, \mathrm{Pa}\).

Step by step solution

01

Define givens

Determine the given variables. We have half the cylinder filled with mercury (density of mercury = \(13.6 \times 10^3 \, \mathrm{kg/m^3}\)) covering a height of \(0.125\, \mathrm{m}\). The other half is filled with water (density of water = \(1000 \, \mathrm{kg/m^3}\)) also covering a height of \(0.125\, \mathrm{m}\). We are also given the atmospheric pressure which is approximately \(101.325\, \mathrm{kPa}\) or \(101325\, \mathrm{Pa}\).
02

Calculate pressure due to mercury

Pressure created by a fluid column is calculated by the formula: \( P = \rho g h \). Here, \( \rho \) represents fluid density, \( g \) represents gravitational acceleration, and \( h \) represents the height of the fluid column. \nFor mercury: \n\( P_{Hg} = (13.6 \times 10^3 \, \mathrm{kg/m^3}) \times (9.81 \, \mathrm{m/s^2}) \times (0.125 \, \mathrm{m}) = 1.67 \times 10^5 \, \mathrm{Pa} \)
03

Calculate pressure due to water

To calculate pressure created by water we also apply the same formula. \nFor water: \n\( P_{H_2O} = (1000 \, \mathrm{kg/m^3}) \times (9.81 \, \mathrm{m/s^2}) \times (0.125 \, \mathrm{m}) = 1.23 \times 10^3 \, \mathrm{Pa} \)
04

Combining pressures

To find the total pressure at the bottom of the cylinder, we combine the pressure caused by the water, the pressure caused by the mercury, and the atmospheric pressure. \nTotal Pressure: \n\( P_{Total} = P_{Hg} + P_{H_2O} + P_{Atmosphere} = (1.67 \times 10^5 \, \mathrm{Pa}) + (1. 23 \times 10^3 \, \mathrm{Pa}) + (101325 \, \mathrm{Pa}) = 2.70 \times 10^5 \, \mathrm{Pa} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Calculation
When it comes to understanding how pressure works within a container, like a graduated cylinder filled with different fluids, it's crucial to dive into the concept of pressure calculation. Pressure is defined as the force exerted per unit area. In the context of fluids, pressure at a point within a fluid can be calculated using the formula:\[ P = \rho g h \]Where:
  • \( P \) stands for pressure.
  • \( \rho \) is the density of the fluid.
  • \( g \) denotes the acceleration due to gravity, which is approximately \( 9.81 \, \mathrm{m/s^2} \) on Earth's surface.
  • \( h \) is the height of the fluid column above the point of pressure measurement.
Calculating the pressure due to a column of fluid involves multiplying these three components, telling us how much pressure is exerted at a certain depth due to the weight of the fluid above. Different fluids have different densities, which significantly affects the pressure each exerts for an equivalent height.
Density of Fluids
The density of a fluid plays a pivotal role in determining the pressure exerted by the fluid column. Fluid density is defined as mass per unit volume:\[ \rho = \frac{m}{V} \]Where:
  • \( \rho \) stands for density.
  • \( m \) represents mass.
  • \( V \) is the volume of the fluid.
In the context of our scenario with the graduated cylinder, mercury and water are two different fluids with distinct densities. Mercury's density is \( 13.6 \times 10^3 \, \mathrm{kg/m^3} \) which implies it is much heavier per unit volume compared to water's density of \( 1000 \, \mathrm{kg/m^3} \). The higher density of mercury results in it exerting more pressure than water even when both occupy the same height in the cylinder. That's why a greater pressure is calculated for mercury compared to water for the same height.
Atmospheric Pressure
Atmospheric pressure is the force exerted by the weight of the Earth's atmosphere. All objects located at or near the surface of the Earth experience this. On average, at sea level, the atmospheric pressure is about \( 101.325 \, \mathrm{kPa} \), which is equivalent to \( 101325 \, \mathrm{Pa} \).This pressure acts externally on the surface of any open container, such as a graduated cylinder. It is vital to include atmospheric pressure in total pressure calculations, especially when the container is open to the air. In the exercise solution, atmospheric pressure was integrated with the pressures exerted by both mercury and water. This incorporation is essential to collate all sources of pressure at the base of the cylinder to find the complete value.
Graduated Cylinder
A graduated cylinder is a common laboratory apparatus used to measure the volume of liquids with a fairly high degree of accuracy. They are usually narrow and tall. This design makes them perfect for observing the impact of different fluid heights on pressure, as seen in our scenario. Graduated cylinders are marked with lines to correspond to specific volumes, making them ideal for precise measurement of liquid volumes in experiments. In the exercise, the cylinder held equal amounts of mercury and water. By understanding each fluid's characteristics and combining them into the calculation, it enables the measuring of resulting pressure more effectively. The graduated cylinder allowed us to establish a clear height of the fluids, 0.125 meters each, helping us pinpoint the exact pressure exerted by both fluids at the bottom. Its usage simplifies the experimental setup required to determine the pressure and demonstrates fundamental fluid mechanics concepts.

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Most popular questions from this chapter

An elephant that has a mass of \(6000 \mathrm{~kg}\) evenly distributes her weight on all four feet. (a) If her feet are approximately circular and each has a diameter of \(50 \mathrm{~cm}\), estimate the pressure on each foot. (b) Compare the answer in part (a) with the pressure on each of your feet when you are standing up. Make some rough but reasonable assumptions about the area of your feet.

A diver is \(10.0 \mathrm{~m}\) below the surface of the ocean. The surface pressure is \(1 \mathrm{~atm}\). What is the absolute pressure and gauge pressure he experiences? The density of seawater is \(1.025 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\).

An ice cube floats in a glass of water so that the water level is exactly at the rim. After the ice cube melts, will all the water still be in the glass? Explain your answer.

If the gauge pressure is doubled, the absolute pressure will A. be halved. B. be doubled. C. be unchanged. D. be increased, but not necessarily doubled. E. be decreased, but not necessarily halved. SSM

Water is supplied to an outlet from a pumping station \(5.00 \mathrm{~km}\) away. From the pumping station to the outlet there is a net vertical rise of \(19 \mathrm{~m}\). Take the coefficient of viscosity of water to be \(0.0100\) poise. The pipe leading from the pumping station to the outelt is \(1.00 \mathrm{~cm}\) in diameter, and the gauge preassure in the pipe at the point where it exits the pumping station is \(520 \mathrm{kPa}\). At what volume flow rate does water flow from the outlet?

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