Question

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°C to 14.6°C during hibernation. If we assume that the air in the squirrel’s lungs is 75.0% N2 and 25.0% O2, by how much will the rms speed of the air molecules in the lungs have decreased during hibernation? Boltzmann's constant is 1.38 × 10−23 J/K. Molar mass of Nitrogen is 14.00674 u and Oxygen is 15.9994 u.

answer must be in m/s.

Answer 1

The found as in a rms speed is va (BKT Vm where k - m - - T ! Boltzmann constant mass of molecule Temperature Molar mass of Molar mass of MA oxygen = 164 (02) Mitrogen = 144 (Ne) e mass of moleule, m = (0.75x3x14) + (0.95**16) * Au whore Av is Avogadro's No. txt - conversion factor for u to kg -luc 1.6707= ?? kg • m= 4.814x16 26 kg L AV Viral Vimittel Are 13x138x15 23 4.8148 (14.64873.15) : 13x138x153x (hot 213:15) 4.814xloeb TAG= -21.49 m C Negative sign means debtcase) e rms speed decreases by /21049 m/s) RAS