Video: Determining the Difference in Densities of Fluids from Differences in Pressure

The difference in pressures at a depth of 3.75 m for seawater and freshwater is 735 Pa. What is the difference in their densities?

04:51

Video Transcript

The difference in pressures at a depth of 3.75 meters for seawater and freshwater is 735 pascals. What is the difference in their densities?

So in this question, we’ve been given a difference in pressures. And at the depth of 3.75 meters, the difference in pressures is given for seawater and fresh water. And this difference is 735 pascals. We need to find the difference in their densities.

Now let’s say that the pressure for seawater is 𝑝 sub 𝑠 and the pressure for fresh water is 𝑝 sub 𝑓. We’ve been given the difference in pressures and the differences between the pressures of seawater and fresh water. This pressure difference is 735 pascals. Now both of these pressures were measured at a depth of 3.75 meters.

So let’s say that we’ve got a tank full of, let’s say, seawater. The pressure was measured at a depth of 3.75 meters. We can also describe this as the height below the surface of the water. And so we can live with this quantity β„Ž. Now since the pressure is measured at the same depth for both seawater and fresh water, we don’t need to separate heights. We don’t need β„Ž sub 𝑠 and β„Ž sub 𝑓. We just need β„Ž.

Finally, what we’re asked to do is to find the difference in the densities of the seawater and fresh water. In other words, we’re trying to find out 𝜌 sub 𝑠, which is the density of seawater, minus 𝜌 sub 𝑓 which is the density of fresh water. In order to solve this problem, we need to recall how we can find the pressure of a liquid.

The pressure of a liquid at a depth β„Ž is given by πœŒπ‘”β„Ž, where 𝜌 is the density of the liquid, 𝑔 is the gravitational field strength of the Earth, and β„Ž is the depth. So we can apply this to the quantities we’ve been given. For seawater, we’ve got 𝑝 sub 𝑠, the seawater pressure, is equal to 𝜌 sub 𝑠, the seawater density, times π‘”β„Ž. And we can do exactly the same thing for fresh water.

Now what we’re trying to do is to find out 𝜌 sub 𝑠 minus 𝜌 sub 𝑓. So let’s rearrange both of these equations so we can isolate 𝜌 sub 𝑠 in the first one and 𝜌 sub 𝑓 in the second one. What we can do is divide both sides of the equation by π‘”β„Ž in both cases. What that leaves us with is the following: 𝑝 sub 𝑠 over π‘”β„Ž is equal to two 𝜌 sub 𝑠 and 𝑝 sub 𝑓 over π‘”β„Ž is equal to 𝜌 sub 𝑓.

Now what we’re trying to find out is 𝜌 sub 𝑠 minus 𝜌 sub 𝑓. So all we need to do is do this minus this. And that leaves us with 𝑝 sub 𝑠 over π‘”β„Ž minus 𝑝 sub 𝑓 over π‘”β„Ž, because, as we saw earlier, 𝜌 sub 𝑠 is the same as 𝑝 sub 𝑠 over π‘”β„Ž. Therefore, we replace this with this. And 𝜌 sub 𝑓 is the same as 𝑝 sub 𝑓 over π‘”β„Ž. So we replace this with this.

Now what we’re trying to do is to find out the quantity on the left-hand side. So this is what we want to keep to the left-hand side. And then we can look at what we can do to the right-hand side. Well, we’ve got 𝑝 sub 𝑠 divided by π‘”β„Ž minus 𝑝 sub 𝑓 divided by π‘”β„Ž. So we can take out a common factor of one divided by π‘”β„Ž. But let’s make sure this is correct. We can use the distributive property of multiplication to go back to what we had earlier.

One divided by π‘”β„Ž times 𝑝𝑠 is the same as this term here. And one divided by π‘”β„Ž times 𝑝𝑓 is the same as this term here. Therefore, we have factorized correctly. Now the term in the brackets, 𝑝 sub 𝑠 minus 𝑝 sub 𝑓, we’ve already written down up here. That’s the difference in pressure. And we’ve been given that in the question. As for the value of 𝑔, well we know that the gravitational field strength of Earth is 9.8 meters per second squared. And we’ve already got β„Ž earlier on, 3.75 meters.

This means that we can substitute in our values to find the difference in densities. What that leaves us with is one divided by 𝑔, which is 9.8, times β„Ž, which is 3.75, all multiplied by 735, which is the difference in pressures. Evaluating this gives us a value of 20. But we still need to put in the units. To do this what we need to do is to look at what we’re trying to find out.

We’re trying to find out a difference in densities. That means that the right-hand side should have units of density. The units of density are mass divided by volume, because that’s what a density is. It’s the mass per unit volume. So the mass is kilograms, and the volume is meters cubed. So the units of mass divided by volume are kilograms per meters cubed. And hence, our final answer is that the difference in densities for seawater and fresh water is 20 kilograms per meters cubed.

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