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 $3.5 \mathrm{J} /\left(\mathrm{g} \cdot^{\circ} \mathrm{C}\right)$.

We will use the ideal gas law \(P V = n R T\) to find the new volume. First, we need to convert the temperatures from Celsius to Kelvin: \(T_1 = 20.0^{\circ} C + 273.15 = 293.15\,\mathrm{K}\) \(T_2 = 22.0^{\circ} C + 273.15 = 295.15\,\mathrm{K}\) As pressure and the amount of air (n) are constants, we can set up the equation: \(\frac{V_1}{T_1} = \frac{V_2}{T_2}\) Plugging in the values: \(\frac{8.16\,\mathrm{mL}}{293.15\,\mathrm{K}} = \frac{V_2}{295.15\,\mathrm{K}}\) Now solve for \(V_2\): \(V_2 = \frac{8.16\,\mathrm{mL} \cdot 295.15\,\mathrm{K}}{293.15\,\mathrm{K}}\) \(V_2 = 8.24\, \mathrm{mL}\)

Since the fish is swimming horizontally, the pressure stays constant at 1.1 atm. The work done can be calculated using the equation: \(W = P \Delta V\) First, we need to convert the pressure to Pascals: \(P = 1.1 \,\mathrm{atm} \times \frac{101325 \,\mathrm{Pa}}{1\, \mathrm{atm}} = 111457.5\, \mathrm{Pa}\) Now, find \(\Delta V\) and convert it to cubic meters: \(\Delta V = V_2 - V_1 = 8.24\,\mathrm{mL} - 8.16\,\mathrm{mL} = 0.08\, \mathrm{mL} = 0.08 \times 10^{-6}\,\mathrm{m^3}\) Finally, calculate the work done: \(W = (111457.5\,\mathrm{Pa})(0.08 \times 10^{-6}\,\mathrm{m^3}) = 8.92\,\mathrm{J}\)

For a diatomic ideal gas, the constant pressure heat capacity (C_p) is given by: \(C_p = \frac{7}{2}R\) Using the first law of thermodynamics, we can find the heat gained (Q) by the system: \(Q = \Delta U + W\) Since \(\Delta U = n C_p \Delta T\), we can write: \(Q = n C_p \Delta T + W\) We have already calculated the work done, W = 8.92 J, and we have: \(\Delta T = T_2 - T_1 = 295.15\,\mathrm{K} - 293.15\,\mathrm{K} = 2\,\mathrm{K}\) Now, we need to find the number of moles (n) using the initial conditions (V1 and 1.1 atm) and the ideal gas constant R: \(n = \frac{P_1 V_1}{R T_1} = \frac{(1.1 \cdot 101325)\, \mathrm{Pa} \cdot 8.16 \times 10^{-6}\,\mathrm{m^3}}{8.314\,\mathrm{J \,mol^{-1}K^{-1}} \cdot 293.15 \,\mathrm{K}} = 3.20 \times 10^{-4} \,\mathrm{mol}\) Calculate the heat gained by the air in the swim bladder: \(Q = (3.20 \times 10^{-4}\,\mathrm{mol}) \times \left(\frac{7}{2} \times 8.314\,\mathrm{J \,mol^{-1}K^{-1}}\right) \times (2\,\mathrm{K}) + 8.92\,\mathrm{J}\) \(Q = 10.33\,\mathrm{J}\)

The fish loses an amount of heat equal to the heat gained by the air in the swim bladder (Q). We can use the specific heat equation, \(Q = mc\Delta T\), to find the temperature decrease of the fish. Rearrange the equation to solve for \(\Delta T\) and plug in the given values: \(\Delta T = \frac{Q}{mc} = \frac{10.33\, \mathrm{J}}{(5.00 \mathrm{g})\left(3.5\,\mathrm{J}/(\mathrm{g}\cdot^{\circ}\mathrm{C})\right)} = 0.591\,^{\circ}\mathrm{C}\) The temperature of the fish will decrease by \(0.591^{\circ} \mathrm{C}\) after losing the heat.