# An 0.25*L air bubble is released by a diver at a pressure of 2.4*atm, and a temperature of 288*K. What volume will it express at the surface with T=300*K?

Jan 18, 2016

$\frac{{P}_{1} {V}_{1}}{T} _ 1 = \frac{{P}_{2} {V}_{2}}{T} _ 2$; we use the combined gas law and absolute temperature.

#### Explanation:

So ${V}_{2} = \frac{{P}_{1} {V}_{1} {T}_{2}}{{T}_{1} {P}_{2}}$

$=$ $\frac{2.4 \cdot \cancel{a t m} \times 0.250 \cdot L \times 300 \cdot \cancel{K}}{288 \cdot \cancel{K} \times 1.0 \cdot \cancel{a t m}}$ $=$ ??L

It is fairly clearly that the volume of the bubble WILL expand, both on the basis of the change in pressure, and on the basis of the change in temperature.

Note that the golden rule of scuba diving as opposed to free diving is $\text{NEVER HOLD YOUR BREATH}$. Why should holding your breath be an issue for a scuba diver?