Question

Physics question. Please answer all parts. Included is an equation sheet for reference.

l) A scuba diver at a depth of 60 m and surrounding temperature of 5 C releases an air bubble of 10 cm a) Calculate the number of moles of the gas bubble. The pressure P at a given depth of water d is given by P pgd +P where p is the density of water, 1 g/cm and g is the acceleration due to gravity b) Calculate the volume of that air when it reaches the surface, which has a temperature of 25 C. c) The buoyancy force of a given bubble is the difference in weight between the material that the bubble displaces and the bubble itself. Given this fact, calculate the buoyancy force of the bubble when it is released and compare this to the buoyancy force of the bubble at the surface. d) Does a bubble have more buoyancy at the surface or at depth? What is the main reason for this? e) Sketch a P-V diagram of the bubble and label the start (point 1) and end (point 2) of its trip. f Use the diagram to estimate the work done by the bubble on its surroundings during its trip to the surface. Assume the path is a straight line. The work done is the area under the line. g) Using the fact that AE nRAT for a diatomic gas, int calculate the change in internal energy of the bubble as it rises to the surface. h) Use the answers to parts (f and (g) and the first law of thermodynamics to calculate the heat added to the bubble from the surrounding water. Does this result mean that the bubble cools the water off or heats it up?

Answer 1

PV = n RT P = P_atm + Pgh = (1.01 times 10^5) + (10^3 times 9.8 times 60) = 6.89 times 10^5 pa Moles = PV/RT = (6.89 times 10^5) (10 times 10^6)/(8.31) (273 + 5) = 2.98 times 10^-3 moles # V = nRT/P = (2.89 times 10^-3) (8.31) (273 + 25)/1.01 times 10^5 = 73.13 cm^3 # F_B1 = Pvg = 10^3 times 10 times 10^-6 times 9.8 = 0.098 N F_B2 = Pv'g = 10^3 times 73.13 times 10^-6 times 9.8 = 0.717 N