During hibernation, an animal's metabolism slows down, and its body temperature lowers. For example, a California ground squirrel's body temperature lowers from \(40.0^{\circ} \mathrm{C}\) to \(10.0^{\circ} \mathrm{C}\) during hibernation. If we assume that the air in the squirrel's lungs is $75.0 \% \mathrm{N}_{2}\( and \)25.0 \% \mathrm{O}_{2},$ by how much will the rms speed of the air molecules in the lungs have decreased during hibernation?

The molar mass of the mixed gas in the lungs can be found by taking the weighted average of the molar masses of nitrogen and oxygen, according to their percentages in the lungs. Nitrogen has a molar mass of 28.02 g/mol and oxygen has a molar mass of 32.00 g/mol. Molar mass of reference_air: \(M = 0.75 × 28.02 + 0.25 × 32.00 = 21.015 + 8.000 = 29.015\) g/mol.

The rms speed of a gas is given by the equation: $$v_{rms} = \sqrt{\frac{3RT}{M}},$$ where \(v_{rms}\) is the rms speed, \(R\) is the gas constant, \(T\) is the temperature in Kelvin, and \(M\) is the molar mass of the gas in grams. We will calculate the rms speed for both initial and final temperatures. Initial temperature: \(T_1 = 40.0^{\circ}\mathrm{C} = 313.15\; \mathrm{K}\) Final temperature: \(T_2 = 10.0^{\circ}\mathrm{C} = 283.15\; \mathrm{K}\) Convert the molar mass into kg/mol by dividing by 1000: \(M = 29.015 / 1000 = 0.029015\) kg/mol. Use the gas constant \(R = 8.3145\) J/mol.K. We can now calculate the rms speeds: \(v_{rms1} = \sqrt{\frac{3 × 8.3145 × 313.15}{0.029015}} = 651.40\) m/s \(v_{rms2} = \sqrt{\frac{3 × 8.3145 × 283.15}{0.029015}} = 618.96\) m/s

Now, we will find the decrease in rms speed of the air molecules in the lungs during hibernation: Decrease in rms speed: \(v_{decrease} = v_{rms1} - v_{rms2} = 651.40 - 618.96 = 32.44\) m/s So, the rms speed of the air molecules in the lungs decreases by 32.44 m/s during hibernation.