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Chapter 9: Problem 64

An air bubble of 1.0 -mm radius is rising in a container with vegetable oil of specific gravity 0.85 and viscosity \(0.12 \mathrm{Pa} \cdot \mathrm{s} .\) The container of oil and the air bubble are at \(20^{\circ} \mathrm{C} .\) What is its terminal velocity?

Short Answer

Expert verified
Based on the given information and using Stoke's Law, the terminal velocity of the air bubble in vegetable oil was calculated to be approximately 0.0044 m/s.

Step by step solution

01

Calculate the density of vegetable oil

Since specific gravity is defined as the ratio of the oil's density to the density of water, we can use this information to find the density of the vegetable oil. The density of water at \(20^{\circ}\mathrm{C}\) is approximately \(1000 \mathrm{kg/m^3}\). Hence, the density of the vegetable oil can be calculated as follows: \(Density_{oil} = Specific\; gravity\;*\;Density_{water}\) \(Density_{oil} = 0.85 * 1000 \mathrm{kg/m^3}= 850 \mathrm{kg/m^3}\) .
02

Calculate the gravitational force acting on the air bubble

The mass of the air bubble can be found using the density of air, which at \(20^{\circ}\mathrm{C}\) is approximately \(1.20 \mathrm{kg/m^3}\). The volume of the air bubble is given by the formula for the volume of a sphere: \(V_{bubble} = \frac{4}{3} * \pi * r^3\) where \(r=1.0\;\mathrm{mm} = 1.0*10^{-3}\;\mathrm{m}\). Now, the mass of the air bubble can be found by: \(Mass_{bubble} = Density_{air} * V_{bubble}\) And the gravitational force acting on the air bubble is: \(F_{gravity} = Mass_{bubble} * g\) where \(g = 9.81\;\mathrm{m/s^2}\) (acceleration due to gravity).
03

Calculate the buoyant force acting on the air bubble

To find the buoyant force acting on the air bubble, we need to use the Archimedes' principle, which states that the upward buoyant force is equal to the weight of the fluid displaced by the object. The buoyant force is given by: \(F_{buoyancy} = V_{bubble} * Density_{oil} * g\)
04

Apply Stoke's Law to calculate the terminal velocity

Stoke's Law states that the terminal velocity of a small, spherical object moving through a viscous fluid is given by the following formula: \(V_{terminal} = \frac{2}{9}*\frac{r^{2}(Density_{oil} - Density_{air})g}{\eta}\) , where \(\eta = 0.12 \,\mathrm{Pa\cdot s}\) is the viscosity of the vegetable oil. Now, substitute the values found in Steps 1, 2, and 3 into the Stoke's Law formula, and calculate the terminal velocity of the air bubble.
05

Calculate the terminal velocity

Using the values calculated in the previous steps, we have: \(V_{terminal} = \frac{2}{9}*\frac{(1.0*10^{-3}\;\mathrm{m})^{2}(850\;\mathrm{kg/m^3} - 1.20\;\mathrm{kg/m^3})9.81\;\mathrm{m/s^2}}{0.12\;\mathrm{Pa\cdot s}}\) After calculating the above expression, we obtain the terminal velocity of the air bubble: \(V_{terminal} \approx 0.0044\;\mathrm{m/s}\)

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