Question Video: Finding Fluid Density in a Liquid Column Manometer | Nagwa Question Video: Finding Fluid Density in a Liquid Column Manometer | Nagwa

Question Video: Finding Fluid Density in a Liquid Column Manometer Physics • Second Year of Secondary School

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The diagram shows a liquid column manometer connected at one end to a gas reservoir and at the opposite end to the atmosphere. The pressure of the gas reservoir ๐_(gas) = 112.5 kPa, and the atmosphere pressure ๐_(atm) = 101.3 kPa. The U-shaped tube contains an oil. The top of the oil column in contact with the atmosphere is vertically above the top of the oil column in contact with the gas reservoir. The vertical distance between the column tops โ = 95.15 cm. What is the density of the oil?

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Video Transcript

The diagram shows a liquid column manometer connected at one end to a gas reservoir and at the opposite end to the atmosphere. The pressure of the gas reservoir ๐ sub gas equals 112.5 kilopascals, and the atmosphere pressure ๐ sub atm equals 101.3 kilopascals. The U-shaped tube contains an oil. The top of the oil column in contact with the atmosphere is vertically above the top of the oil column in contact with the gas reservoir. The vertical distance between the column tops โ equals 95.15 centimeters. What is the density of the oil?

Our diagram shows us a manometer with a U-shaped tube that is open at one end to the atmosphere and at the other end is connected with a gas reservoir. An oil is inside of the tube. And the height of the oil column on the left side of the tube is greater than the height of the oil column on the right side by a distance โ. Knowing both the pressure of the gas and the reservoir and the pressure of the atmosphere, we want to solve for this oilโs density.

Recording the information weโre given about the pressure of the atmosphere and the pressure of the gas as well as the height difference between the oil columns on either side of the U-shaped tube, letโs now clear some space on screen and begin by thinking about what it means that the height of the oil column on one side of this tube is different from the height on the other side.

The first thing we can note is that the pressure along this pink dashed line at any point inside the tube is the same. In other words, the pressure here in the tube is the same as the pressure here. Since our system is in equilibrium, that means the pressure at any given point is the same in all directions. Therefore, the total downward acting pressure on the right side of our tube along this pink line is the same as the total downward acting pressure on the left side of the tube along that line.

On the right-hand side, that downward pressure is entirely due to the pressure of the gas and the reservoir. On the left, that downward pressure is due to the pressure of the atmosphere added to the pressure created by this height of oil in the column. That is, this much oil, because it weighs something, creates a pressure at the bottom of the column of height โ.

In general, the pressure created by a fluid of density ๐ arranged so that itโs in a column of height โ is equal to ๐ times the acceleration due to gravity ๐ times โ. We can write then that the pressure due to a height โ of oil in our column is equal to the density of the oil times ๐ times โ.

Letโs now recall how this pressure of the oil relates to other pressures in the manometer. Weโve said that the total downward acting pressure at this point is the same as that at this point. Mathematically, that means that the pressure due to the atmosphere added to the pressure due to the oil of a column of height โ is equal to the pressure of the gas in the reservoir. We know this is true because our system is in equilibrium; the column of oil is neither moving upward in this direction or upward in this direction.

We can rewrite this equation using our expression for the pressure created by the column of oil. We recall that itโs the density of the oil ๐ sub oil that we want to solve for. We can begin doing this by subtracting the atmospheric pressure from both sides. Since on the left-hand side, we both add and subtract this pressure, these two terms cancel one another out. As a next step, we can divide both sides of the equation by ๐ times โ. Doing this makes both ๐ and โ cancel on the left.

We arrive at this simplified expression for the density of our oil. When we get to our final answer, weโll want our oil density to be expressed in units of kilograms per cubic meter. As we calculate ๐ sub oil, letโs ensure that we are indeed getting these units. Note that in our numerator, we have two pressures, both of which are given in units of kilopascals. One kilopascal is equal to 1000 pascals. And therefore, to convert these pressures into units of pascals, we can multiply each one by 1000. ๐ sub atm, 101.3 kilopascals multiplied by 1000 is 101300 pascals. Likewise, when we multiply ๐ sub gas by 1000, we get 112500 pascals.

A pascal, we can recall, is equal to a newton over a meter squared. Going one step further, a newton is equal to a kilogram meter per second squared. If we combine these two conversions, then we can write that a pascal is equal to a kilogram meter per second squared, thatโs a newton, divided by meters squared. If we then multiply both numerator and denominator of this fraction by one over meters squared, then in the denominator, the unit of meters squared cancels out. In the numerator, one factor of meters cancels from top and bottom. So we can write the units of pascals simply as kilograms per meter second squared.

Having worked through these units, letโs now enter in the known values for ๐ sub gas and ๐ sub atm in units of pascals, which we have converted to kilograms per meter second squared. This gives us 112500 kilograms per meter second squared minus 101300 kilograms per meter second squared. If we factor out the units from both of these two terms, we get this result. And if we subtract 101300 from 112500, thatโs equal to 11200. This then is the pressure difference between the pressure of the gas and the pressure of the atmosphere. All this is divided by ๐, which is 9.8 meters per second squared, times โ, which is 95.15 centimeters.

Before we substitute in this value for โ though, letโs recall that 100 centimeters is equal to one meter. And therefore, we can convert our height โ to meters by moving the decimal point two spots to the left. This is equivalent to dividing by 100. Our height โ then is 0.9515 meters. And notice that now in our denominator, we have meters times meters, or meters squared, all divided by seconds squared. Factoring out these units, in our denominator, we have 9.8 times 0.9515 meters squared per second squared.

Notice then that one over second squared appears in both the overall denominator and overall numerator of this fraction. Therefore, those units cancel out. We have then overall units of kilograms per meter divided by meters squared. If we multiply numerator and denominator by one over meters squared, then the units in the denominator cancel out entirely. And in the numerator, meters times meters squared is meters cubed.

Weโve confirmed then that the units weโre calculating are kilograms per cubic meter, the units of density we wanted. When we calculate this fraction, we get a result of 1201 kilograms per cubic meter. Here, weโve kept four significant figures to be consistent with the precision of the information given to us in this problem. The density of oil in the manometer is 1201 kilograms per cubic meter.

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