Question

attached is the incorrect answer. Please answer with the correct solution and in Celsius

 An air bubble of radius 5.5 cm is released from the mouth of a diver. At the diver's depth the pressure is 1.20 atm . (We'll learn in chapter 13 why the diver's depth affects the pressure.) As it rises to the surface, the air inside the bubble does 8.938 J of work. The bubble rises so quickly to the surface that this process can be treated as being adiabatic. 

Part A 

If the bubble's temperature is initially the same as the diver's body (37 C), what is its temperature when it reaches the surface? Express your answer to three significant figures in degrees Celsius. 

Answer 1

P; = (101325 * 1.2) Pa = (121590 Pa) Vis grupa 4 a 0.055)3 = 6.96 *10-4 m3 Ti = 37+273 - (310K) MR = Pi Vi = 121590 ) (6.96X104 ) 310 nR=0.272 Wanr (Ti-Tf) y=.4 Y- 8.938- (0.272) (310-T6 ) .4-1) 3.575 - (310-T6) 0.272 13.144 = 310 - TF Tf : 310 - 13.144 To = 296.85k If = 296.85-273 Tf - 23.85° C K Hoping it may be ight.