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Chapter 10: Problem 91

A gas bubble with a volume of \(1.0 \mathrm{~mm}^{3}\) originates at the bottom of a lake where the pressure is \(3.0\) atm. Calculate its volume when the bubble reaches the surface of the lake where the pressure is 695 torr, assuming that the temperature doesn't change.

Short Answer

Expert verified
The volume of the gas bubble when it reaches the surface of the lake is \(3.28 \mathrm{~mm}^{3}\).

Step by step solution

01

List the given values

Initial Volume (Vi) = 1.0 mm鲁 Initial Pressure (Pi) = 3.0 atm Final Pressure (Pf) = 695 torr
02

Convert units

We need to convert the pressures to the same unit. Let's convert atm to torr: 1 atm = 760 torr 3.0 atm 脳 (760 torr/1 atm) = 2280 torr Initial Pressure (Pi) = 2280 torr Additionally, convert the volume to a larger unit like liters: 1.0 mm鲁 脳 (1 cm鲁 / 1000 mm鲁) 脳 (1 L / 1000 cm鲁) = 1.0 脳 10鈦烩伓 L Initial Volume (Vi) = \(1.0 \times 10^{-6}\) L
03

Apply Boyle's Law

Boyle's Law states that for an ideal gas at constant temperature, the product of the initial pressure and volume is equal to the product of the final pressure and volume: Pi 脳 Vi = Pf 脳 Vf Now, solve for the final volume (Vf): Vf = (Pi 脳 Vi) / Pf
04

Calculate the final volume

Plug in the given values and solve for Vf: Vf = (\(2280 \mathrm{~torr}\) 脳 \(1.0 \times 10^{-6} \mathrm{~L}\)) / \(695 \mathrm{~torr}\) Vf = \(2.28 \times 10^{-3} \mathrm{~L}\) / \(695 \mathrm{~torr}\) Vf = \(3.28 \times 10^{-6}\) L
05

Convert the final volume to mm鲁

To convert the final volume back to mm鲁, use the following conversion: Vf = \(3.28 \times 10^{-6}\) L 脳 (1000 cm鲁 / 1 L) 脳 (1000 mm鲁 / 1 cm鲁) = 3.28 mm鲁 The volume of the gas bubble when it reaches the surface of the lake is 3.28 mm鲁.

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Key Concepts

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

Pressure Conversion
One of the critical steps in gas calculations is converting pressure units. Pressure can be measured in various units such as atmospheres (atm) and torr. However, to perform accurate calculations, these units need to be consistent.
To convert pressure from atmospheres to torr or vice versa, remember that:
  • 1 atm is equivalent to 760 torr.
  • To convert from atm to torr, multiply by 760.
  • To convert from torr to atm, divide by 760.
In our original problem, the pressure of the gas bubble at the bottom of the lake is given as 3.0 atm. Using the conversion factor, you calculate this as 2280 torr. By ensuring that both initial and final pressures are in the same units (torr in this case), you can apply Boyle's Law more effectively.
Gas Laws
Gas laws provide vital relationships between pressure, volume, and temperature of gases. One of the famous gas laws is Boyle's Law. This principle describes how the pressure of a gas tends to increase as the volume of the container decreases, provided the temperature remains constant.
Boyle's Law is mathematically expressed as:
  • \[ P_i \times V_i = P_f \times V_f \]
Where:
  • \(P_i\) = Initial pressure
  • \(V_i\) = Initial volume
  • \(P_f\) = Final pressure
  • \(V_f\) = Final volume
This equation states that the product of the initial pressure and volume (on one side of the equation) is equal to the product of the final pressure and volume (on the other side). In the initial problem, you used Boyle's Law to calculate how the volume of the gas bubble changes as it ascends to the lake surface with reduced pressure.
Volume Calculation
Volume calculation is another critical aspect of gas problems. Sometimes, you need to convert between various volume units, such as milliliters, liters, or cubic millimeters.
In the original problem, the initial volume of the gas bubble is given as 1.0 mm鲁. Converting small units like mm鲁 into more standard or bigger units like liters makes handling numbers in calculations easier:
  • A cube with 10 mm on each side has a volume of 1 cm鲁.
  • 1000 cm鲁 equals 1 liter.
  • Thus, 1 mm鲁 is equivalent to \(1 \times 10^{-6}\) liters.
After calculating in liters, don't forget to convert the final volume back to the original units if needed. In this instance, the volume was eventually converted back to mm鲁 for the solution, providing a clear and comprehensible result. This step ensures that your answer matches the problem's original context.

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Most popular questions from this chapter

Chlorine dioxide gas \(\left(\mathrm{ClO}_{2}\right)\) is used as a commercial bleaching agent. It bleaches materials by oxidizing them. In the course of these reactions, the \(\mathrm{ClO}_{2}\) is itself reduced. (a) What is the Lewis structure for \(\mathrm{ClO}_{2}\) ? (b) Why do you think that \(\mathrm{ClO}_{2}\) is reduced so readily? (c) When a \(\mathrm{ClO}_{2}\) molecule gains an electron, the chlorite ion, \(\mathrm{ClO}_{2}^{-}\), forms. Draw the Lewis structure for \(\mathrm{ClO}_{2}^{-}\). (d) Predict the \(\mathrm{O}-\mathrm{Cl}-\mathrm{O}\) bond angle in the \(\mathrm{ClO}_{2}^{-}\) ion. (e) One method of preparing \(\mathrm{ClO}_{2}\) is by the reaction of chlorine and sodium chlorite: $$ \mathrm{Cl}_{2}(\mathrm{~g})+2 \mathrm{NaClO}_{2}(\mathrm{~s}) \longrightarrow 2 \mathrm{ClO}_{2}(\mathrm{~g})+2 \mathrm{NaCl}(\mathrm{s}) $$ If you allow \(10.0 \mathrm{~g}\) of \(\mathrm{NaClO}_{2}\) to react with \(2.00 \mathrm{~L}\) of chlorine gas at a pressure of \(1.50 \mathrm{~atm}\) at \(21{ }^{\circ} \mathrm{C}\), how many grams of \(\mathrm{ClO}_{2}\) can be prepared?

After the large eruption of Mount St. Helens in 1980 , gas samples from the volcano were taken by sampling the downwind gas plume. The unfiltered gas samples were passed over a gold-coated wire coil to absorb mercury (Hg) present in the gas. The mercury was recovered from the coil by heating it, and then analyzed. \(\underline{\text { In one }}\) particular set of experiments scientists found a mercury vapor level of \(1800 \mathrm{ng}\) of \(\mathrm{Hg}\) per cubic meter in the plume, at a gas temperature of \(10^{\circ} \mathrm{C}\). Calculate (a) the partial pressure of \(\mathrm{Hg}\) vapor in the plume, \((\mathrm{b})\) the number of \(\mathrm{Hg}\) atoms per cubic meter in the gas, \((\mathrm{c})\) the total mass of Hg emitted per day by the volcano if the daily plume volume was \(1600 \mathrm{~km}^{3}\).

Newton had an incorrect theory of gases in which he assumed that all gas molecules repel one another and the walls of their container. Thus, the molecules of a gas are statically and uniformly distributed, trying to get as far apart as possible from one another and the vessel walls. This repulsion gives rise to pressure. Explain why Charles's law argues for the kinetic- molecular theory and against Newton's model.

(a) How high in meters must a column of water be to exert a pressure equal to that of a \(760-\mathrm{mm}\) column of mercury? The density of water is \(1.0 \mathrm{~g} / \mathrm{mL}\), whereas that of mercury is \(13.6 \mathrm{~g} / \mathrm{mL}\). (b) What is the pressure in atmospheres on the body of a diver if he is \(39 \mathrm{ft}\) below the surface of the water when atmospheric pressure at the surface is \(0.97\) atm?

Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to markets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and thus has a boiling point at atmospheric pressure of \(-164^{\circ} \mathrm{C}\). One possible strategy is to oxidize the methane to methanol, \(\mathrm{CH}_{3} \mathrm{OH}\), which has a boiling point of \(65^{\circ} \mathrm{C}\) and can therefore be shipped more readily. Suppose that \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane at atmospheric pressure and \(25^{\circ} \mathrm{C}\) are oxidized to methanol. (a) What volume of methanol is formed if the density of \(\mathrm{CH}_{3} \mathrm{OH}\) is \(0.791 \mathrm{~g} / \mathrm{mL} ?\) (b) Write balanced chemical equations for the oxidations of methane and methanol to \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(l)\). Calculate the total enthalpy change for complete combustion of the \(10.7 \times 10^{9} \mathrm{ft}^{3}\) of methane described above and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of \(0.466 \mathrm{~g} / \mathrm{mL} ;\) the density of methanol at \(25^{\circ} \mathrm{C}\) is \(0.791 \mathrm{~g} / \mathrm{mL}\). Compare the enthalpy change upon combustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume?
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