Question

An air bubble of radius 7.0 cm is released from the mouth of a diver. At the diver's depth the pressure is 1.30 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 28.27 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 1

The equation for adiabatic process is given by

$PV^{\gamma}=Constant, \ or\ P^{1-\gamma}T^{\gamma}=Constant$

In this problem, P₁ = 1.35 atm, P₂ = 1 atm, T₁ = 37 ^oC = 310 K, T₂ = ?, The γ = 1.4 for air, which is predominantly a diatomic gas. Therefore, from the above equation you can write

$P_{1}^{1-\gamma}T_{1}^{\gamma}=P_{2}^{1-\gamma}T_{2}^{\gamma}$

$\therefore T_{2}=T_{1}\left (\frac{P_{1}}{P_{2}} \right )^{\frac{1-\gamma}{\gamma }}=310\times \left ( 1.35 \right )^{\frac{1-1.4}{1.4 }}=284.5\ K=11.5\ ^{o}C$