Question

A cylindrical drum of radius 0.5 m is used to hold 400 liters of petroleum ether (density = .68 g/mL or $680 \textrm{ kg/m}^3$).
(Note: $1 \textrm{ liter} = 0.001 \textrm{ m}^3$)

- Determine the amount of pressure applied to the walls of the drum if the petroleum ether fills the drum to its top.
- Determine the amount of pressure applied to the floor of the drum if the petroleum ether fills the drum to its top.
- If the drum were redesigned to hold 800 liters of petroleum ether:
- How would the pressure on the walls change? Would it increase, decrease, or stay the same?
- How would the pressure on the floor change? Would it increase, decrease, or stay the same?

Final Answer

- $1.70 \times 10^3 \textrm{ Pa}$
- $3.40 \times 10^3 \textrm{ Pa}$
- i) average pressure increases with increases in volume. ii) Pressure at the floor of the container increases with volume.

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Video Transcript

This is College Physics Answers with Shaun Dychko.
The density of petroleum ether is 0.68 grams per milliliter or 680 kilograms per cubic meter and it's in a cylindrical container of radius 0.5 meters and the container has a volume of 400 liters which we'll convert into cubic meters by multiplying by one cubic meter for every 1000 liters. So the pressure in this cylinder is the first thing we have to calculate. The maximum pressure at the floor is going to be the pressure due to this column of petroleum ether and that pressure is density times g times the total height of this cylinder. The answer to our question however, what is the pressure in the cylinder, we'll have to take an average of the pressures at all the different heights. Now since the pressure changes linearly with height, the average will just be the simple average of the pressure at the top plus the pressure at the bottom divided by two. The pressure at the top is zero so then that means we're just taking the pressure at the bottom and dividing it by two to get our average. So, we'll calculate the height knowing the volume and the radius. So the volume is the cross-sectional area of the cylinder which is pi

*r*squared times*h*its height. We'll divide both sides by pi*r*squared to solve for*h*. So*h*is volume divided by pi times radius squared. We'll substitute that in for*h*here. So the maximum pressure at the bottom is density times g times volume divided by pi times radius squared. So that's 680 kilograms per cubic meter times 9.81 Newtons per kilogram times 0.4 meters cubed, yep, then divide that by pi times 0.5 meters radius squared. That gives 3.4 times ten to the three pascals which we then divide by two to get the average pressure which is 1.7 times ten to the three pascals. Now part B, we'll just copy this maximum down here, that's the pressure at the floor. Then in part C, re-do it knowing that -- or just say what would happen if the volume increased. Well the average pressure is maximum pressure over two and maximum pressure is*rho g V*over pi*r*squared and divide all that by two. As volume increases, the average pressure will increase because we're increasing the numerator here. Well, it's the same story down here. The maximum pressure at the floor is just this without the two in the denominator and as*V*increases so does*P max*at the floor.