Final answer:
The pressure at the bottom of the Marianas Trench is approximately 1.08 × 10¸ Pa, and the percent decrease in the volume of seawater due to this pressure is roughly 5.045%. This indicates that the assumption of constant density is not entirely valid, and the actual pressure would be slightly higher. Option (C) is correct.
Step-by-step explanation:
To calculate the pressure at the bottom of the Marianas Trench, we use the formula for pressure due to a fluid column, which is P = ρgh, where ρ (rho) is the density of seawater, g is the acceleration due to gravity, and h is the height of the fluid column (depth of the ocean).
In this case, ρ = 1030 kg/m³ (the average density of seawater), g = 9.81 m/s² (acceleration due to Earth's gravity), and h = 11000 m (depth of the trench).
Substituting the values, we get:
P = 1030 kg/m³ * 9.81 m/s² * 11000 m
P = 1.11 * 10¸ Pa
To calculate the percent decrease in volume, we use the formula for bulk modulus B = -P / (ΔV/V), rearranging it to find (ΔV/V) and then multiplying by 100% to convert it to a percentage. If B (bulk modulus of seawater) is similar to that of water which is approximately 2.2 × 10¹ Pa, we get:
(ΔV/V) = -P/B
(ΔV/V) = -(1.11 * 10¸ Pa) / (2.2 * 10¹ Pa)
(ΔV/V) = -0.05045
Percent decrease in volume = 0.05045 * 100%
Percent decrease in volume ≈ 5.045%
This percentage represents a small, but noticeable, compression of seawater at the trench's depth. Therefore, the assumption of constant density is not entirely valid; however, for most calculations, it can be an acceptable approximation. As the volume of seawater slightly decreases under such high pressure, its density increases. The actual pressure would be somewhat greater than the calculated value, because as the volume decreases, the density increases, leading to a higher pressure at the deeper parts.