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an air bubble of radius 6.5 cm is released from the mouth of a diver. at the diver's depth the pressure is 1.15 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 10.93 j of work. the bubble rises so quickly to the surface that this process can be treated as being adiabatic. 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?

Answer :

The temperature of the diver's is 1403K when it reaches the surface if the process involved is adiabatic.

Firstly we need to find the radius of air bubble, as we apply the ideal gas equation.

Since, we know that an air bubble is spherical in size, so volume of sphere is =(4/3)πr³ where r is the radius of the air bubble.

Volume of air bubble=(4/3)πr³

                            =>V=(4/3)×π×(0.065)³

                             =>V=(4/3)×3.14×0.274625

                             =>V=3.44929/3

                              =>V=1.149m³

Since temperature given in degree, so in kelvin temperature is =C+273=37+273=310K

Now, we apply the ideal gas equation which is

=>PV=nRT

So,we firstly find value of nR,=>nR=(PV)/T

                                                =>nR= (1.15×1.149)/ 310

                                                =>nR=0.0042

Since, it is given that the process is adiabatic, so we can use formula of work

For air,we know that value of γ is 1.4

=>W=(nR/γ-1) ×(T₁ - T₂)

=>W=[0.0042/  1.4-1]×(T-310)

=>W=(0.0042/0.4) ×(T-310)

=>W=0.010×(T-310)

=>10.93=0.010×(T-310)

=>10.93/0.010=T-310

=>1093=T-310

=>T=1093+310

=>T=1403K

Hence, required temperature is 1403K.

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