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Question Video: Solving for Gas Pressure in a Liquid Column Manometer Physics

The diagram shows a liquid column manometer connected at one end to a gas reservoir and at the opposite end to the atmosphere. The U-shaped tube contains mercury that has a density of 13,595 kg/m³. The top of the mercury column in contact with the atmosphere is vertically above the top of the mercury column in contact with the gas reservoir. The vertical distance between the column tops ℎ = 45.0 cm. Find the pressure of the gas in the reservoir. Use a value of 101.3 kPa for atmospheric pressure.

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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 U-shaped tube contains mercury that has a density of 13,595 kilograms per cubic meter. The top of the mercury column in contact with the atmosphere is vertically above the top of the mercury column in contact with the gas reservoir. The vertical distance between the column tops ℎ equals 45.0 centimeters. Find the pressure of the gas in the reservoir. Use a value of 101.3 kilopascals for atmospheric pressure.

In our diagram, we see the liquid column manometer with one end open to the atmosphere and the other end connected to this gas reservoir. The manometer is filled with mercury; that’s this gray fluid. The mercury we see is higher on the left-hand side of the manometer compared to the right-hand side. The height difference between these columns of mercury is given as 45.0 centimeters.

The fact that this height difference is not zero means that the pressure here at the top of the mercury column on the left is not the same as the pressure here at the top of the mercury column on the right. However, it is the case that the pressure along this dashed horizontal line at any point in the manometer is the same. If we consider a point on the right-hand side of the manometer at this elevation, the downward-acting pressure at that point, whatever it is, is due to the gas in the reservoir.

On the other side of our manometer, at the same elevation, we have the same total downward pressure. And this total pressure is due to the pressure of the atmosphere — we’ll call it 𝑃 sub atm — added to the pressure created by this column of mercury of height ℎ. The atomic symbol for mercury is Hg. So, we’ll call the pressure created from this height of mercury column 𝑃 sub Hg.

To see how all these pressures balance out, let’s first clear some space on screen. Considering the downward pressure acting here on the left side of our manometer, we know that this pressure is due to the sum of atmospheric pressure and the pressure created by our column of mercury of height ℎ. We’ve said that this downward pressure is equal to the downward pressure at the same elevation on the right-hand side. That downward pressure is due to the gas in the reservoir. So, we’ll call it 𝑃 sub gas. 𝑃 sub atm plus 𝑃 sub Hg equals 𝑃 sub gas.

Our goal here is to solve for the pressure of the gas. And in our problem statement, we are told the pressure of the atmosphere. However, we don’t yet know the pressure due to the column of mercury.

To help us solve for that pressure, though, we can recall the following mathematical expression. The pressure created by a fluid of density 𝜌 arranged in a column of height ℎ is equal to that density times the acceleration due to gravity 𝑔 times ℎ. This means that the pressure created by the column of mercury of height ℎ equals the density of mercury 𝜌 sub Hg times 𝑔, the acceleration due to gravity, times ℎ. In our problem statement, we were told that the density of mercury is 13,595 kilograms per cubic metre. Along with this, we know the height ℎ. And we can recall that the acceleration due to gravity is 9.8 meters per second squared.

So then, the pressure of the gas equals the pressure of the atmosphere plus the pressure of the mercury column. And the pressure due to the column of mercury equals the density of mercury times 𝑔 times the height of that column ℎ. Substituting in all are known values gives us this expression. Before we calculate the pressure of the gas though, we’ll want to make sure the units in both terms in this expression all agree. For that to be the case, there are two changes we’ll want to make.

First, we’ll want to convert our atmospheric pressure from units of kilopascals to units of pascals. One kilopascal is 1000 pascals. So, if we multiply 101.3 kilopascals by 1000, we get 101,300 pascals. In the next term in our expression, the one change we’ll want to make is to convert the units of this value, 45.0 centimeters, to meters. 100 centimeters is one meter. So to convert to meters, we’ll divide 45.0 centimeters by 100. That gives us 0.45 meters.

Notice then that in this second term, we have meters times meters divided by meters cubed. This means that two factors of meters cancel from numerator and denominator. We are left with overall units for this term being kilograms per meter-second squared. A kilogram per meter second squared is equal to a newton per meter squared, which is equal to a pascal.

Therefore, the units of the first term in this expression and the units of the second term agree. When we calculate piece of gas then, we’ll get an answer in units of pascals. Entering this expression on our calculator, we get a result of about 161,253 pascals.

Since some of the information given to us in our problem statement, specifically the height ℎ of 45.0 centimeters, had only three significant digits of precision, we’ll also want to report our final answer to only three significant digits. In that case, our answer becomes 161,000 pascals.

And then lastly, we can use this conversion that one kilopascal equals 1000 pascals to express our final answer in kilopascals. In this way we give our answer in the same units in which our atmospheric pressure was given originally. The pressure of the gas in the reservoir is 161 kilopascals.

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