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Chapter 15: Problem 60

If the pressure on a fish increases from 1.1 to 1.2 atm, its swim bladder decreases in volume from \(8.16 \mathrm{mL}\) to \(7.48 \mathrm{mL}\) while the temperature of the air inside remains constant. How much work is done on the air in the bladder?

Short Answer

Expert verified
The temperature remains constant during this process. Answer: The approximate work done on the air in the swim bladder of the fish is -7.6 x 10^{-6} J.

Step by step solution

01

Convert units to SI

Firstly, we need to convert the given volumes and pressures into the International System of Units (SI). The pressure values must be converted from atmospheres (atm) to Pascals (Pa), and the volume values must be converted from milliliters (mL) to cubic meters (m^3). 1 atm = 101325 Pa 1 mL = 1.0 脳 10^{-6} m^3 Initial pressure, P1 = 1.1 atm 脳 101325 Pa/atm = 111457.5 Pa Final pressure, P2 = 1.2 atm 脳 101325 Pa/atm = 121590 Pa Initial volume, V1 = 8.16 mL 脳 10^{-6} m^3/mL = 8.16 x 10^{-6} m^3 Final volume, V2 = 7.48 mL 脳 10^{-6} m^3/mL= 7.48 x 10^{-6} m^3
02

Calculate the work done

The work done in an isothermal process (constant temperature) can be calculated using the following formula: W = nRT ln(V2 / V1) where W is the work done, n is the number of moles, R is the gas constant, T is the temperature, and ln(V2 / V1) is the natural logarithm of the ratio of the final volume to the initial volume. However, we are not given the number of moles and temperature in this problem. But we notice that the product of the initial pressure and volume is almost equal to the product of the final pressure and volume: P1 * V1 鈮� P2 * V2 We can use this relationship to approximate the work done: W 鈮� P1 * V1 - P2 * V2
03

Calculate the approximate work done

Plug in the values of initial and final pressures and volumes into the formula and calculate the work done: W 鈮� (111457.5 Pa) * (8.16 x 10^{-6} m^3) - (121590 Pa) * (7.48 x 10^{-6} m^3) W 鈮� 0.9092548 J - 0.9092624 J W 鈮� -7.6 x 10^{-6} J So the approximate work done on the air in the bladder is -7.6 x 10^{-6} J. Note that the negative sign indicates that the work was done on the air (compression), not done by the air (expansion).

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Most popular questions from this chapter

A \(0.500-\mathrm{kg}\) block of iron at \(60.0^{\circ} \mathrm{C}\) is placed in contact with a \(0.500-\mathrm{kg}\) block of iron at $20.0^{\circ} \mathrm{C} .\( (a) The blocks soon come to a common temperature of \)40.0^{\circ} \mathrm{C} .$ Estimate the entropy change of the universe when this occurs. [Hint: Assume that all the heat flow occurs at an average temperature for each block.] (b) Estimate the entropy change of the universe if, instead, the temperature of the hotter block increased to \(80.0^{\circ} \mathrm{C}\) while the temperature of the colder block decreased to \(0.0^{\circ} \mathrm{C}\) [Hint: The answer is negative, indicating that the process is impossible. \(]\)
List these in order of increasing entropy: (a) 0.01 mol of \(\mathrm{N}_{2}\) gas in a \(1-\mathrm{L}\) container at \(0^{\circ} \mathrm{C} ;\) (b) 0.01 mol of \(\mathrm{N}_{2}\) gas in a 2 -L container at \(0^{\circ} \mathrm{C} ;\) (c) 0.01 mol of liquid \(\mathrm{N}_{2}\).
A fish at a pressure of 1.1 atm has its swim bladder inflated to an initial volume of \(8.16 \mathrm{mL}\). If the fish starts swimming horizontally, its temperature increases from \(20.0^{\circ} \mathrm{C}\) to $22.0^{\circ} \mathrm{C}$ as a result of the exertion. (a) since the fish is still at the same pressure, how much work is done by the air in the swim bladder? [Hint: First find the new volume from the temperature change. \(]\) (b) How much heat is gained by the air in the swim bladder? Assume air to be a diatomic ideal gas. (c) If this quantity of heat is lost by the fish, by how much will its temperature decrease? The fish has a mass of \(5.00 \mathrm{g}\) and its specific heat is about $3.5 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)$.
The motor that drives a reversible refrigerator produces \(148 \mathrm{W}\) of useful power. The hot and cold temperatures of the heat reservoirs are \(20.0^{\circ} \mathrm{C}\) and \(-5.0^{\circ} \mathrm{C} .\) What is the maximum amount of ice it can produce in \(2.0 \mathrm{h}\) from water that is initially at \(8.0^{\circ} \mathrm{C} ?\)
An inventor proposes a heat engine to propel a ship, using the temperature difference between the water at the surface and the water \(10 \mathrm{m}\) below the surface as the two reservoirs. If these temperatures are \(15.0^{\circ} \mathrm{C}\) and \(10.0^{\circ} \mathrm{C},\) respectively, what is the maximum possible efficiency of the engine?
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