Question

An air bubble of \(19.4 \mathrm{~cm}^{3}\) volume is at the bottom of a lake \(41.5 \mathrm{~m}\) deep where the temperature is \(3.80^{\circ} \mathrm{C}\). The bubble rises to the surface, which is at a temperature of \(22.6^{\circ} \mathrm{C}\). Take the temperature of the bubble to be the same as that of the surrounding water and find its volume just before it reaches the surface.

Short answer

The volume of the bubble just before it reaches the surface of the lake can be obtained by substituting the known values into the generalized gas law after converting units appropriately and performing the calculation.

Step-by-step solution

Convert temperatures from Celsius to Kelvin

Firstly, temperatures given needs to be converted from Celsius to Kelvin because absolute temperature should be used in gas laws. The formula for converting Celsius to Kelvin is \( T(K) = T(°C) + 273.15 \). So \( T1 = 3.80°C + 273.15 = 276.95K \) and \( T2 = 22.6°C + 273.15 = 295.75K \)

Calculate the pressure at the bottom of the lake

Next, the pressure at the bottom of the lake and at the surface (Patm) need to be determined. Pressure, P, at a depth h, in a liquid of density, ρ, is given by \( P = P_0 + ρgh \), where \(g\approx 9.81 m/s^2\), is acceleration due to gravity, ρ (density of water) = \(10^3 kg/m^3\), \(P_0 = 1 atm\), is the atmospheric pressure at sea level. Substituting the values: \( P = 1 atm + (10^3 kg/m^3) * 9.81 m/s^2 * 41.5 m \). Converting 1 atm to pascal(Pa), \( 1 atm = 1.013*10^5 Pa \), so \( P = 1.013*10^5 Pa + (10^3 kg/m^3) * 9.81 m/s^2 * 41.5 m \)

Apply the generalized gas law

With temperature and pressure calculated, these can now be substituted into the combined or generalized gas law given by \( P_1V_1/T_1 = P_2V_2/T_2 \), where \( P_1 and V_1 \) are initial pressure and volume, \( P_2 and V_2 \) are final pressure and volume, and \( T_1 and T_2 \) are initial and final absolute temperatures. Rearranging the formula for final volume: \( V_2 = (P_1V_1T_2)/(T_1P_2) \).

Plug in the values into generalized gas law to find the final volume

Now replace all variables in the generalized gas law with the values obtained and calculated: \( V_2 = (P_1V_1T_2)/(T_1P_2)= (1.013*10^5 Pa * 19.4 cm^3 * 295.75K)/(276.95K*P) \). Convert \(19.4 cm^3\) to \(m^3\), \(1cm^3 = 10^{-6} m^3\), so \( V_2 = (P_1V_1T_2)/(T_1P_2)= (1.013*10^5 Pa * 19.4*10^{-6}m^3 * 295.75K)/(276.95K*P) \).

Key concepts

Absolute Temperature

The concept of absolute temperature is crucial when working with gas laws. It refers to the measurement of temperature beginning from absolute zero, which is the point where particles have minimal thermal motion, and is the base of the Kelvin scale. Unlike Celsius or Fahrenheit, Kelvin does not have negative values, making it essential for scientific calculations.

In applying gas laws, temperatures must be converted to Kelvin to ensure accuracy. For instance, in the problem posed, the initial temperature of water at the bottom of the lake is converted from Celsius to Kelvin using the formula: \( T(K) = T(°C) + 273.15 \). These absolute temperatures allow us to determine how gas volume and pressure interact at different depths and temperatures.

Pressure and Depth Relationship

Understanding the relationship between pressure and depth is critical when studying fluid dynamics. Pressure increases linearly with depth in a fluid due to the weight of the fluid above. This is described by the formula: \( P = P_{0} + ρgh \), where \( P \) is the pressure at depth, \( P_0 \) is atmospheric or the surface pressure, \( ρ \) is the fluid density, \( g \) is the acceleration due to gravity, and \( h \) is the depth.

In the context of our problem, this means that the pressure experienced by the air bubble at the bottom of the lake is greater than the pressure at the surface, proportionate to the depth and density of the water above it. As the bubble rises, the surrounding pressure decreases, influencing the bubble's volume accordingly.

Gas Volume and Temperature Relationship

The volume of a gas is directly proportional to its temperature when pressure is held constant, a relationship known as Charles's Law. In mathematical terms, this is stated as \( V_1/T_1 = V_2/T_2 \), where \( V \) represents volume and \( T \) represents absolute temperature. As temperature increases, the kinetic energy of gas particles increases, causing them to move apart and expand the gas volume.

For the air bubble rising from the lake's bottom, this relation explains how the bubble expands. The water temperature—and consequently the bubble’s temperature—increases as it ascends. This increase in temperature results in an increased volume under the assumption that the bubble's pressure adjusts continuously to the surrounding water pressure.

Ideal Gas Law Applications

The ideal gas law provides a more comprehensive way to describe the state of a gas by including all four variables: pressure (\( P \)), volume (\( V \)), temperature (\( T \)), and the number of moles of gas (\( n \)). It is represented by the equation \( PV = nRT \), where \( R \) is the ideal gas constant. Although it assumes a hypothetical 'ideal' gas where particles do not interact, it nevertheless accurately describes the behavior of real gases under many conditions.

Its application allows us to solve problems like the air bubble in the lake by combining Boyle's Law, Charles's Law, and Avogadro's Law into a unified formula. In this way, scientists and engineers can predict how a gas will behave under varying conditions of temperature, pressure, volume, and quantity, which is critical for various fields such as meteorology, chemical engineering, and environmental science.