### 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 below the top of the mercury column in contact with the gas reservoir. The vertical distance between the column tops โ equals 25 centimeters. Find the pressure of the gas in the reservoir. Use a value of ๐ one equals 101.3 kilopascals for atmospheric pressure.

In this example, we have a liquid column manometer. Thatโs this U-shaped tube here. One end of the manometer is open to the atmosphere, and one end is connected to a gas reservoir at some pressure. The liquid in the manometer is mercury. And we can see that the height of the mercury column on the left is not equal to the height of the mercury column on the right. This tells us that there is a pressure difference between the atmospheric pressure acting down on the mercury here and the pressure of the gas in the reservoir acting on the mercury here. We want to solve for the pressure of the gas in the reservoir, and weโll label that ๐ sub gas.

Letโs recall that the density of mercury, weโll label it ๐ sub Hg, since Hg is the atomic symbol for mercury, is given as 13,595 kilograms per cubic meter. That density ends up being important because, as weโve seen, there is a pressure difference between the pressure of the atmosphere and the pressure of the gas.

If we call that difference in pressure on either side of the manometer ฮ๐, then looking at our diagram, we can recognize that that difference in pressure is equal to the height of the mercury in this column right here. We were told in our problem statement that this height โ is 25 centimeters. So then, the pressure thatโs created by a 25-centimeter-tall column of mercury is equal to the difference in pressure ฮ๐ between atmospheric pressure and the pressure of our gas.

In general, for a fluid with a density ๐ having a height โ, the pressure created by that fluid is ๐ times ๐ times โ. We can write then that ฮ๐ equals the density of mercury multiplied by the acceleration due to gravity times โ, our height of 25 centimeters. Letโs label atmospheric pressure ๐ sub atm, even though itโs called ๐ sub one in our problem statement, just to make it easy to remember what it represents. Then, by seeing which side of our manometer has a taller column of mercury, we can figure out which is greater, ๐ sub atm or ๐ sub gas.

Because of the additional 25 centimeters of mercury on the right-hand side of our manometer, we know since that pressure is transmitted all through the mercury to act at this interface with the atmosphere that the pressure due to the atmosphere, ๐ sub atm, is holding up, we could say, more mercury than the pressure due to the gas in the reservoir. Compared to the pressure of the gas in the reservoir, atmospheric pressure is supporting an additional 25 centimeters of mercury. Therefore, ๐ sub atm is greater than the pressure of the gas.

This difference is what weโve called ฮ๐. Since ฮ๐ equals the difference in these pressures, and it also equals the pressure due to this height of mercury of 25 centimeters, we can combine these equations and write that ๐ sub atm minus ๐ sub gas equals the density of mercury times ๐ times โ.

Now, letโs recall that itโs the pressure of the gas that we want to solve for. Rearranging this equation, if we add ๐ sub gas to both sides and subtract ๐ times ๐ times โ from both sides, then we get this result, where ๐ sub gas is the subject of the equation.

Letโs now fill in what we know from the right-hand side of this expression. In our problem statement, weโre told that atmospheric pressure is 101.3 kilopascals. We know the density of mercury. We know that ๐, the acceleration due to gravity, is 9.8 meters per second squared. And weโve also been told that the height โ is 25 centimeters. At this point, letโs clear some space on screen and consider the units in this whole expression.

In the first term, we have units of kilopascals. 1000 pascals equals one kilopascal, and a pascal itself is equal to a newton of force divided by a square meter of area. So, to convert 101.3 kilopascals to more recognizable units, we can move the decimal place one, two, three spots to the right, giving us 101,300 pascals. And noting as we have that a pascal is a newton per square meter, we can write this as 101,300 newtons per meter squared.

Considering then the rest of our expression, letโs first change these 25 centimeters into a distance in meters. Since 100 centimeters equals one meter, we can shift the decimal point two spots to the left. That shows us that 25 centimeters is 0.25 meters. In this term, we note that we have units of meters cubed in the denominator and units of meters times meters, or meters squared, in the numerator. If we simplify these units of meters on the right and then group the numbers and the units separately in this term, we see that, for this term, weโll get units of kilograms per meter second squared.

What we want to do here though is check that these units are the same as these units so that we can truly combine these terms through subtraction. Now, a newton is equal to a kilogram meter per second squared. Therefore, we can write that a newton per meter squared equals a kilogram meter per second squared per meter squared. Note that now in this numerator, we have units of meters and in the denominator units of meters squared. Therefore, the units of meters simplify. So we just have one factor of meters in the denominator.

And notice now that the units of this first term are exactly equal to those of the second term. All this means that when we do combine these two numbers through subtraction, weโre performing a physically valid operation.

Entering this expression on our calculator, we get 67,992.25 kilograms per meter second squared. And letโs remember that a kilogram per meter second squared equals a newton per meter squared and that a newton per meter squared is equal to a pascal. We can write our final answer in units of pascals then. But since our atmospheric pressure was originally given to us in units of kilopascals, we can convert this number from pascals to kilopascals by moving the decimal point three spots to the left. This is the pressure of the gas in the reservoir.

But before we box it as our final answer, letโs round it to the correct number of significant figures. In our problem statement, we were told that the height โ of the column of mercury on the right-hand side that exceeds the height on the left is 25 centimeters. This value has two significant figures, so our final answer will have two as well.

To see what this final value will be, we can look at the first digit after the decimal. Since this value is greater than or equal to five, we round up, to get a result of 68 kilopascals. This then is the pressure of the gas in the gas reservoir. We see that itโs less than atmospheric pressure, which agrees with the scenario shown in our diagram.