91Ó°ÊÓ

A steel cargo drum has a height of \(880 \mathrm{~mm}\) and a diameter of \(610 \mathrm{~mm}\). With its top removed it has a mass of \(17.3 \mathrm{~kg}\). The drum is turned upside down at the surface of the North Atlantic and is pulled downward into the ocean by a robotic submarine. On this day the surface temperature is \(23.0^{\circ} \mathrm{C}\) and the surface air pressure is \(p_{0}=101 \mathrm{kPa}\). The water temperature decreases linearly with depth to \(3.0^{\circ} \mathrm{C}\) at \(1000 \mathrm{~m}\) below the surface. As the drum moves downward in the ocean, the air inside the drum is compressed, reducing the upward buoyant force. (a) At what depth \(y_{\text {neutral }}\) is the barrel neutrally buoyant? (Hint: The pressure in the drum is equal to the sea pressure, which at depth \(y\) is \(p_{0}+\rho g y\) where \(\rho=1025 \mathrm{~kg} / \mathrm{m}^{3}\) is the density of seawater. The temperature at depth can be determined using the information above. Together with the ideal- gas law, you can derive a formula for the volume of air at depth \(y,\) and therefore a formula for the upward buoyant force as a function of depth.) (b) What is the volume of the air in the drum at depth \(y_{\text {neutral }} ?\)

Step by step solution

01

Find the initial volume of the drum

We know that the steel drum is a cylinder with a height of \(880 \, mm\) and a diameter of \(610 \, mm\). Let's convert these dimensions to meters and calculate the volume \(V_0\) of the drum using the formula for volume of a cylinder, \(V_0= \pi r^2 h\), where \(r\) is the radius and \(h\) is the height of the drum.

02

Get the equation for the force on the drum

The total force on the drum, when it is submerged in the water, is given by the force of gravity (weight of the drum) minus the buoyancy force. This can be formulated as \(F = m_{\text{drum}}g - \rho V g\), where \(m_{\text{drum}} = 17.3 \, kg\) is the mass of the drum, \(\rho = 1025 \, kg/m^3\) is the density of seawater, \(V\) is the volume of the drum, and \(g\) is the acceleration due to gravity. The drum will be neutrally buoyant when this force equals zero, thus we can express this condition as \(m_{\text{drum}}g = \rho V g\).

03

Calculate the volume of the air at depth

From the ideal gas law \(PV=nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is number of moles (constant), \(R\) is ideal gas constant and \(T\) is absolute temperature, we can obtain an expression for volume \(V = \frac{nRT}{P}\). As the temperature decreases linearly with depth, we can derive the temperature at depth \(y\) as \(T(y) = 23 - 0.02y\) (converting to Kelvin by adding 273). Given \(P_0 = 101 \, kPa = 101000 \, Pa\) as the surface pressure and \(\rho = 1025 \, kg/m^3\) as the density of the seawater, the pressure \(P(y)\) at a given depth \(y\) is \(P(y) = P_0 + \rho g y\). Substituting these expressions and \(nRT/P_0\) for \(V_0\) into our equation describing neutral buoyancy, we end up with an equation for \(y_{\text{neutral}}\).

04

Calculate the volume of the air at depth \(y_{\text{neutral}}\)

We can substitute the calculated depth \(y_{\text{neutral}}\) into our pressure and temperature equations to find these quantities at this depth. Then, substituting these values into the ideal gas law \(V = \frac{nRT}{P}\) allows us to calculate the change in volume of the air, when the drum is at depth \(y_{\text{neutral}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Buoyant Force

The buoyant force is a crucial concept when it comes to understanding how objects float, sink, or remain suspended in a fluid. It's the upward force exerted by a fluid that opposes the weight of an object immersed in the fluid. Think of it like the fluid is trying to 'lift' the object. Archimedes' principle tells us that the buoyant force on an object submerged in a fluid is equal to the weight of the fluid that the object displaces. So, the denser the fluid, the greater the buoyant force.

Applying this to the steel drum in our exercise, when the drum is turned upside down and submerged in seawater, it experiences a buoyant force equal to the weight of the seawater it displaces. As the drum is pulled deeper, the air inside it compresses. This changes the volume of the drum and therefore changes the buoyant force as well, since the amount of displaced water (and thus its weight) changes. The drum achieves neutral buoyancy when the downward force of gravity on the drum equals the upward buoyant force.

The Ideal Gas Law In Action

The ideal gas law is an important equation in physics and chemistry that relates the pressure (P), volume (V), and temperature (T) of a gas to the number of moles (n) and the ideal gas constant (R): \( PV = nRT \) . It's a valuable tool for understanding how gases will behave under different conditions.

When applied to our exercise, the ideal gas law helps us determine the volume of the compressed air in the submerged drum at various depths. As the air is compressed with increasing pressure from the surrounding water, its volume decreases if its temperature changes or the number of air moles remains the same. This change in volume affects the buoyant force on the drum.

Seawater Density Insights

The density of seawater is another significant factor in this context and can vary with location, depth, and temperature. In our exercise, a standard density of \(1025 \text{kg/m}^3\) is used to represent seawater. This density is crucial in calculating the pressure exerted by the ocean at different depths, as well as the buoyant force. It's important to note that an object will float if it is less dense than the surrounding fluid, and sink if it is more dense.

Therefore, understanding the density of seawater enables us to predict and calculate at what point the drum in our scenario will have neutral buoyancy, as the weight of water displaced by the submerged drum at any point (contributing to the buoyant force) is directly tied to this density.

Pressure With Depth Relationship

As one descends through a fluid such as water, pressure increases. This relationship is described by the equation \( P = P_0 + \rho gh \), where \( P \) is the pressure at depth, \( P_0 \) is the surface pressure, \( \rho \) is the density of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the depth below the surface. The increase in pressure with depth is a result of the weight of the fluid above pressing down.

In our exercise, the pressure inside the drum at depth \( y \) is equal to the surrounding seawater pressure, which increases as the drum is pulled deeper. This increasing pressure significantly reduces the volume of the air inside the drum, which in turn affects the drum's buoyancy. This principle helps us determine the depth at which the drum becomes neutrally buoyant by understanding how pressure increases with depth and influences the properties of the entrapped air.