THERMAL CYU 14

Physical Concept: The condition “without any swimming effort” implies that the diver is neutrally buoyant. This occurs when the average density of the diver equals the density of the surrounding water, or equivalently, when the buoyant force equals the gravitational force.

Let $V_{max}$ be the volume of air in the lungs at the surface (where $P = p_0$) and $V_{body}$ be the total volume of the diver at the surface. The volume of the flesh and bones is constant, but the volume of air in the lungs compresses as pressure increases with depth $h$.

Volume of flesh ($V_f$): $$V_f = V_{body} – V_{max} = 82\,\text{L} – 4.0\,\text{L} = 78\,\text{L} = 78 \times 10^{-3}\,\text{m}^3$$

At depth $h$, the pressure is $P_h = p_0 + \rho g h$. Assuming the air in the lungs compresses isothermally (or using the Boyle’s law approximation for the contained gas), the volume of air $V_{air}’$ at depth $h$ is given by:

$$p_0 V_{max} = P_h V_{air}’ \implies V_{air}’ = \frac{p_0 V_{max}}{p_0 + \rho g h}$$

For neutral buoyancy, the total weight of the diver ($mg$) must equal the weight of the displaced water ($F_B$). The total volume at depth $h$ is $V_f + V_{air}’$.

$$mg = \rho (V_f + V_{air}’) g$$ $$m = \rho \left( V_f + \frac{p_0 V_{max}}{p_0 + \rho g h} \right)$$

Substituting the values:

• $m = 80\,\text{kg}$
• $\rho = 1000\,\text{kg/m}^3$
• $V_f = 78 \times 10^{-3}\,\text{m}^3$
• $p_0 = 10^5\,\text{Pa}$
• $V_{max} = 4.0 \times 10^{-3}\,\text{m}^3$

$$80 = 1000 \left( 78 \times 10^{-3} + \frac{10^5 (4.0 \times 10^{-3})}{10^5 + 10^4 h} \right)$$ $$\frac{80}{1000} = 0.078 + \frac{400}{10^5 + 10^4 h}$$ $$0.080 – 0.078 = \frac{400}{10^4(10 + h)}$$ $$0.002 = \frac{400}{10000(10 + h)} = \frac{0.04}{10 + h}$$ $$10 + h = \frac{0.04}{0.002} = 20$$ $$h = 10\,\text{m}$$