The U-tube device connected to the tank in Fig. 13-6 is...

The U-tube device connected to the tank in Fig. 13-6 is called a manometer. As you can see, the mercury in the tube stands higher in one side than the other. What is the pressure in the tank if atmospheric pressure is $76 \mathrm{~cm}$ of mercury? The density of mercury is $13.6 \mathrm{~g} / \mathrm{cm}^{3}$. from which $P=95 \mathrm{kPa}$. Or, more simply perhaps, we could note that the pressure in the tank is $5.0 \mathrm{~cm}$ of mercury lower than atmospheric. So the pressure is $71 \mathrm{~cm}$ of mercury, which is $94.6 \mathrm{kPa}$.

The U-tube device connected to the tank in Fig. 13-6 is called a manometer. As you can see, the mercury in the tube stands higher in one side than the other. What is the pressure in the tank if atmospheric pressure is $76 \mathrm{~cm}$ of mercury? The density of mercury is $13.6 \mathrm{~g} / \mathrm{cm}^{3}$.
from which $P=95 \mathrm{kPa}$.
Or, more simply perhaps, we could note that the pressure in the tank is $5.0 \mathrm{~cm}$ of mercury lower than atmospheric. So the pressure is $71 \mathrm{~cm}$ of mercury, which is $94.6 \mathrm{kPa}$.

Key Concepts

Manometer

A manometer is an instrument used to measure pressure by utilizing the height difference in a column of liquid. It typically consists of a U-shaped tube filled with a liquid (like mercury), where the difference in liquid levels directly reflects the pressure difference between the two ends of the tube.

Hydrostatic Pressure

Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It is determined by the fluid’s density, gravitational acceleration, and the height of the fluid column above the point in question. This concept is vital for understanding how differences in liquid column heights in a manometer relate to pressure differences.

Pressure Unit Conversion

Pressure unit conversion involves translating measurement units from one form, such as centimeters of mercury, to another, like kilopascals or atmospheres. This process requires an understanding of the relationships between the units and the properties of the measuring fluid, including its density and the effect of gravity.

Transcript

00:01 Good day, the topic is about monometry.

00:04 A monometer is a typical you tube device that will tell us the relative pressure of two fluids.

00:12 Consider a monometer here with one chamber containing a confined gas and the other end of the tube is connected to the atmosphere or is open to the atmosphere.

00:26 The tube is filled up with mercury and as we can see, the mercury level, is not the same for the two sides.

00:34 In particular, the level of mercury is closer to the compartment where there is the gas.

00:43 If the atmospheric pressure is equal to 76 centimeter hg and that the mercury density is 13 .6 grams per cubic centimeter, we wish to find the pressure exerted by the gas in the chamber.

00:58 So the diagram does so much information from the with regard to the difference in the pressure of the atmosphere and that of the gas.

01:07 You can imagine here as the pressure or as the fluid is lower at the side where it is at the side where it is connected or it is open to the atmosphere.

01:18 You can imagine that the atmosphere pushes more greatly on the mercury than the push of the gas.

01:27 So that means the pressure of the gas will be lower than the pressure of the gas will be lower than the pressure of the atmosphere by how much by whatever is this height is equivalent into pressure so let me say that again as we can see here that the side of the mercury is as if being pushed by the atmosphere so that it goes down over goes on to a certain height h then we can say that the pressure of the gas will be lower than the pressure of the atmosphere by the hydrostatic pressure equivalent to this height so we say the atm minus p gas is equal to the hydrostatic pressure due to this height that's then the hydrostatic pressure is solved as density of the fluid times g which is the acceleration due to gravity and the height so we have established here the formula that relates the pressures on the two sides of the monometer and all we need to do now is just to plug in into this equation argument.

02:45 So we will, we need to be consistent.

02:48 If we are going to solve this equation, we have to agree on a certain unit system.

02:55 And the easiest one is to agree on si.

02:58 So that means we will have to convert patm from centimeters hg into pascals...

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