A holding tank at sea level (atmospheric pressure 1.01 x 10^{5} Pa) containing water has a pressure of 2 x 10^{5} Pa at 10 m depth. What is the pressure in an equivalent holding tank containing mercury (specific gravity = 13) at 1 m depth in Denver, CO (atmospheric pressure 8.4 x 10^{4} Pa)?

a) 2 x 10^{5} Pa

b) 1 x 10^{5} Pa

c) 1.4 x 10^{6} Pa

d) 8.4 x 10^{4} Pa

Formula for hydrostatic pressure in a fluid:

P_{tot} = P_{0} + ρgh

**Explanation**

Here, we can simply plug in the values provided:

P_{tot} = (8.4 x 10^{4}) + (13,000 kg/m^{3} x 10m/s^{2} x 1m) = 2.14 x 10^{5} Pa

a) 2 x 10^{5} Pa correct, matches our calculation

b) 1 x 10^{5} Pa, incorrect, this is the same pressure at sea level for 10m of water, ignores fluid change and atmospheric pressure change

c) 1.4 x 10^{6} Pa, incorrect, this answer fails to change the fluid depth to 1m for the mercury rather than the 10m depth used for water

d) 8.4 x 10^{4} Pa, incorrect, this answer ignores the fluid contribution to pressure and only uses the atmospheric contribution*Next Step Test Preparation provides one-on-one MCAT tutor programs nationwide.*

Hi! Can you share the equation for specific gravity please? Thank you!

specific gravity (unitless)= density of the substance/density of water

Density of water= 1g/cm^3 or 1000kg/m^3 or 1kg/L

Hello I am also not understanding how we got to the answer mathematically. Could you elaborate on that please.

Thank you

Absolutely! First, note the formula given in the question stem: Ptot = P0 + ρgh. This is a good formula to understand. Here, Ptot refers to the total pressure, which we’re solving for. P0 (sometimes written as Patm) refers to the atmospheric pressure, which was given here as 8.4 x 10^4 Pa. Finally, ρgh is the gauge pressure, or the pressure that is due to the fluid (and NOT the atmosphere above it). In this formula, ρ represents the density of the fluid, g is 10 m/s^2, and h is the relevant depth (here, 1 meter).

Now, I think the part that is most confusing here is the density (ρ). The question doesn’t give the density of mercury, only its specific gravity. But do you remember the definition of specific gravity? The specific gravity of a substance is equal to its density divided by the density of water (1000 kg/m^3 – memorize this value!). If 13 = (density of mercury) / (1000 kg/m^3), then (density of mercury) = 13,000 kg/m^3.

Finally, we can plug in our values! 8.4 x 10^4 Pa + (13,000 kg/m^3)(10 m/s^2)(1 m) = 8.4 x 10^4 Pa + 130,000 = 8.4 x 10^4 Pa + 1.3 x 10^5 Pa. Note that these values have different exponents, so we need to convert one to match the other. I’d change 8.4 x 10^4 Pa into 0.84 x 10^5 Pa, giving us a final answer of 0.84 x 10^5 Pa + 1.3 x 10^5 Pa = 2.14 x 10^5 Pa.

Good luck with your studies!

When you have a question like this one, where they give you all the useless information in the beginning, how do you tackle that? Because I spent at least 1 minute trying to figure out what the question was talking about but at the end I didn’t even need to use the water tank information.

Hi Gaby, very good question. First of all, it’s extremely helpful to know your physics formulas and units extremely thoroughly. Here, if we see that the question is asking for pressure and we immediately recall the equation given in this explanation [Ptot = Patm + (rho)gh], then we will know that we only need one value for atmospheric pressure, one value for fluid density (rho), and one value for height. Since the question ends by asking about the mercury tank, we then know that we can disregard the water values.

Alternatively, when we’re given lots of information, think carefully about how the pieces of information relate to each other. Here, the water tank and the mercury tank are at different atmospheric pressures, the fluids inside them have different specific gravities, etc. In fact, the only relationship we can see between them is that the question describes them as “equivalent” holding tanks – likely implying that they’re the same shape/volume. But the shape/volume/size of a holding tank doesn’t determine the pressure at different points in the tank – only the values in the equation given above have any impact on that. So, we can conclude that the water tank information must not be helpful here.

Hello,

For this question, why is the value for height in the equation not inputted as -1? Given that height normally refers to height *above* the ground, shouldn’t depth refer to height below ground?

Obviously for this specific question that would produce a negative answer, but in other cases how would we logically think to *add* a height value when it’s described as depth?

Thanks!

Hi there, great question! This tends to be a point of confusion for fluids problems in particular. From the given equation for hydrostatic pressure (Ptot = P0 + ρgh), it’s easy to think that h refers to the “height” of the object, like you mentioned. However, it’s best to instead think of “h” as the height of the fluid above the object, which is what actually produces this pressure. You may also see “h” described in that equation as the “height of the fluid column.” Since there is 1 m of fluid above the object, we plug in 1, not -1.

(In general, this confusion only applies to fluids, since this is the only realm of MCAT physics that routinely involves depth below a surface. For typical projectile motion problems, for example, it’s perfectly fine to plug in a negative value for h if we are dealing with a situation below ground.)