Problem 169 - Chemistry

Question

(a) Show that the pressure exerted by a fluid \(P\) (in pascals) is given by \(P=h d g,\) where \(h\) is the column of the fluid in meters, \(d\) is the density in \(\mathrm{kg} / \mathrm{m}^{3},\) and \(g\) is the acceleration due to gravity \(\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)\) (Hint: See Appendix 1.) (b) The volume of an air bubble that starts at the bottom of a lake at \(5.24^{\circ} \mathrm{C}\) increases by a factor of 6 as it rises to the surface of water where the temperature is \(18.73^{\circ} \mathrm{C}\) and the air pressure is 0.973 atm. The density of the lake water is \(1.02 \mathrm{~g} / \mathrm{cm}^{3}\). Use the equation in (a) to determine the depth of the lake in meters.

Step-by-Step Solution

Answer

The depth of the lake is approximately 17.1 meters.

Step 1: Translate the fluid pressure formula

Given the formula \(P=h d g\), we know that \(P\) is the fluid pressure, \(h\) is the fluid column height, \(d\) is the density, and \(g\) is the acceleration due to gravity, measuring 9.81 \(m/s^{2}\).

Step 2: Understand the bubble situation

The air bubble starts deep in a lake and ascends to the surface. As it ascends, the volume of the air bubble increases by a factor of 6 due to pressure changes. The initial temperature at the bottom of the lake is 5.24 °C, and the final temperature at the surface is 18.73 °C. Additionally, the air pressure at the lake's surface is 0.973 atm, which is equivalent to 98513.25 Pa. Also, bear in mind that the density of the lake water is given in g/cm\(^3\), but the SI unit is Kg/m\(^3\). Converting \(1.02 g/cm^{3}\) to Kg/m\(^3\) gives 1020 Kg/m\(^3\).

Step 3: Apply Charles’s law

Charles's law describes how gases tend to expand when heated, that's why the bubble volume increases when it ascends to a warmer temperature - from \(273.15 + 5.24 = 278.39 K\) to \(273.15 + 18.73 = 291.88 K\). So, the initial volume \(V_1\) becomes six times larger at the end, increasing by a factor of \(\frac{V_2}{V_1} = \frac{T_2}{T_1}\), hence \(\frac{T_1}{T_2} = \frac{1}{6}\).

Step 4: Calculate lake’s depth

Knowing the pressure difference between the initial and final bubble can be represented by \(\Delta P = P_2 - P_1\). Given that \(P_1 = 0\) at the lake bottom, \(\Delta P = P_2 = h d g\). So, we can solve for \(h\), depth of the lake, by the equation \(h = \frac{P}{d g}\) where \(P = P_2 = \frac{P_0}{6}\) and \(P_0 = 98513.25 Pa\). Substituting these values in equation we find \(h = \frac{98513.25 Pa/6}{1020 Kg/m^{3} \times 9.81 m/s^{2}} = 17.1 m\).

Step 5: Final answer

So, the depth of the lake is about 17.1 m.

Key Concepts

Density

In the context of fluid pressure, density is an important factor. It refers to how much mass is contained within a given volume of a substance. Imagine fitting a group of particles into a space, the more packed they are, the denser the material. In mathematical terms, density is calculated as \(d = \frac{m}{V}\), where \(m\) is mass in kilograms and \(V\) is volume in cubic meters. This gives us units of kg/m³, used frequently to measure substances such as water or air.

When calculating fluid pressure, density plays a crucial role since it determines how heavy a column of fluid is. A higher density means more mass and, consequently, higher pressure at the base of the column. For instance, the density of water is about 1000 kg/m³, making it a reference point for most calculations involving aqueous solutions. In this exercise, the water in the lake has a density of 1020 kg/m³, which is slightly denser than pure water likely due to dissolved minerals.

Understanding density allows us to appreciate how it affects real-world phenomena, like how fish are able to float or sink in different water bodies.

Acceleration Due to Gravity

The acceleration due to gravity, denoted as \(g\), is a key component in calculating gravitational forces. On Earth, this acceleration is approximately 9.81 m/s². It is the rate at which objects accelerate towards the Earth's surface when in free fall, absent of other forces such as air resistance.

When computing fluid pressure as in the formula \(P = h d g\), \(g\) helps determine how much weight is exerted by the fluid column. A taller column of fluid will contribute to a higher pressure, and since the force due to Earth's gravity is consistent, it allows us to compute reliable fluid pressures.

• It emphasizes the direct proportionality between pressure and gravity: heavier gravitational pull results in higher pressure.
• By being a constant, \(g\) helps standardize calculations across different scenarios, making it easier to compare pressures from one environment to another.

Knowing \(g\) enables us to predict object behavior under different conditions, such as assessing the pressure a diver feels underwater, or the rise of an air bubble through liquid as seen in this exercise.

Charles's Law

Charles's law governs the relationship between the volume and temperature of a gas at constant pressure. It asserts that the volume of a gas is directly proportional to its absolute temperature, expressed in Kelvin. Simply put, if a gas is heated, it expands, and if cooled, it contracts.

The mathematical expression of Charles's law is \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\), where \(V_1\) and \(V_2\) are the initial and final volumes, and \(T_1\) and \(T_2\) are the initial and final temperatures, respectively. In practical terms, when a gas-filled bubble rises in warmer water, the gas inside expands as per Charles's law due to the increasing temperature. In this exercise, as the bubble ascends from cooler to warmer depths, its volume increases, indicating an application of this principle.

This law is essential in various applications, such as predicting the behavior of air in balloons as it rises through the atmosphere or during heating and cooling processes in scientific experiments. Understanding it helps predict changes in the environment and is crucial in science fields like meteorology and engineering.

• It shows the importance of temperature control in managing gas volumes.
• It highlights the practicalities of temperature and volume relations and why maintaining one affects the other.