Question

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

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

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 × 10^{-6} m^3 Initial pressure, P1 = 1.1 atm × 101325 Pa/atm = 111457.5 Pa Final pressure, P2 = 1.2 atm × 101325 Pa/atm = 121590 Pa Initial volume, V1 = 8.16 mL × 10^{-6} m^3/mL = 8.16 x 10^{-6} m^3 Final volume, V2 = 7.48 mL × 10^{-6} m^3/mL= 7.48 x 10^{-6} m^3

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 ≈ P2 * V2 We can use this relationship to approximate the work done: W ≈ P1 * V1 - P2 * V2

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 ≈ (111457.5 Pa) * (8.16 x 10^{-6} m^3) - (121590 Pa) * (7.48 x 10^{-6} m^3) W ≈ 0.9092548 J - 0.9092624 J W ≈ -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).