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Chapter 10: Problem 98

Consider a lake that is about \(40 \mathrm{~m}\) deep. A gas bubble with a diameter of \(1.0 \mathrm{~mm}\) originates at the bottom of a lake where the pressure is \(405.3 \mathrm{kPa}\). Calculate its volume when the bubble reaches the surface of the lake where the pressure is 98 \(\mathrm{kPa}\), assuming that the temperature does not change.

Short Answer

Expert verified
The volume of the gas bubble when it reaches the surface of the lake is approximately \(2.15 \times 10^{-3} \mathrm{cm^3}\).

Step by step solution

01

Identify the Ideal Gas Law

The Ideal Gas Law can be expressed as: \(PV = nRT\) where P is the pressure, V is the volume, n is the amount of gas (in moles), R is the universal gas constant, and T is the temperature. Since the temperature and the amount of gas do not change in this problem, the Ideal Gas Law can be simplified to: \(\frac{P_1V_1}{P_2V_2}=1\) where the subscripts 1 and 2 refer to the bottom and surface of the lake, respectively.
02

Calculate the initial volume of the bubble

We are given the diameter of the bubble at the bottom of the lake as 1.0 mm. We will first calculate the radius: Radius = Diameter / 2 = \(1.0 \mathrm{mm} / 2 = 0.5 \mathrm{mm}\) Now, convert the radius to meters: Radius = \(0.5 \times 10^{-3} \mathrm{m}\) The volume of a sphere can be calculated using the formula: Volume = \(\frac{4}{3}\pi r^3\) Substituting the radius value, we get: \(V_1 = \frac{4}{3}\pi (0.5 \times 10^{-3})^3 = 5.24 \times 10^{-10} \mathrm{m^3}\)
03

Calculate the volume of the bubble at the surface

We will now apply the simplified Ideal Gas Law equation to find the volume at the surface \(V_2\): \(\frac{P_1V_1}{P_2V_2}=1\) Rearrange the formula to isolate \(V_2\): \(V_2 = \frac{P_1V_1}{P_2}\) Substitute the given pressures and initial volume: \(V_2 = \frac{405.3 \times 10^3 \mathrm{Pa} \times 5.24 \times 10^{-10} \mathrm{m^3}}{98 \times 10^3 \mathrm{Pa}}\) Now, calculate the value of \(V_2\): \(V_2 = 2.15 \times 10^{-9} \mathrm{m^3}\)
04

Convert the volume to cubic centimeters

The volume at the surface is given in cubic meters, but it is often more convenient to express it in cubic centimeters. To convert the volume from cubic meters to cubic centimeters, use the conversion factor: 1 \(\mathrm{m^3} = 10^6 \mathrm{cm^3}\) \(V_2 = 2.15 \times 10^{-9} \mathrm{m^3} \times 10^6 \mathrm{cm^3/m^3} = 2.15 \times 10^{-3} \mathrm{cm^3}\) The volume of the gas bubble when it reaches the surface of the lake is approximately 2.15 x 10^{-3} cubic centimeters.

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Most popular questions from this chapter

(a) What are the mole fractions of each component in a mixture of $15.08 \mathrm{~g}\( of \)\mathrm{O}_{2}, 8.17 \mathrm{~g}\( of \)\mathrm{N}_{2},$ and \(2.64 \mathrm{~g}\) of \(\mathrm{H}_{2}\) (b) What is the partial pressure in atm of each component of this mixture if it is held in a \(15.50-\mathrm{L}\) vessel at \(15^{\circ} \mathrm{C} ?\)

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required \(105 \mathrm{~s}\) for $1.0 \mathrm{~L}$ of the gas to effuse. Under identical experimental conditions it required \(31 \mathrm{~s}\) for \(1.0 \mathrm{~L}\) of \(\mathrm{O}_{2}\) gas to effuse. Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of $1.0 \mathrm{~L} ;$ in other words, rate is the amount that diffuses over the time it takes to diffuse.)

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is \(210.0 \mathrm{~L}\) that contains \(\mathrm{O}_{2}\) gas at a pressure of \(16,500 \mathrm{kPa}\) at \(23^{\circ} \mathrm{C}\). (a) What mass of \(\mathrm{O}_{2}\) does the tank contain? (b) What volume would the gas occupy at STP? (c) At what temperature would the pressure in the tank equal $15.2 \mathrm{MPa} ?\( (d) What would be the pressure of the gas, in \)\mathrm{kPa}$, if it were transferred to a container at \(24^{\circ} \mathrm{C}\) whose volume is \(55.0 \mathrm{~L} ?\)

Suppose you have a fixed amount of an ideal gas at a constant volume. If the pressure of the gas is doubled while the volume is held constant, what happens to its temperature? [Section 10.4\(]\)

Imagine that the reaction $2 \mathrm{CO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)$ occurs in a container that has a piston that moves to maintain a constant pressure when the reaction occurs at constant temperature. Which of the following statements describes how the volume of the container changes due to the reaction: (a) the volume increases by \(50 \%,\) (b) the volume increases by \(33 \%\), (c) the volume remains constant, (d) the volume decreases by \(33 \%,(\mathbf{e})\) the volume decreases by $50 \%\(. [Sections 10.3 and 10.4\)]$
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