Question

What is the density of water at a depth where the pressure is 80.0 atm, given that its density at the surface is 1.03×10^3kg⋅m^−3? Compressibility of water is 45.8×10^−11Pa^−1 and 1atm?

Answer 1

The density of water at a depth where the pressure is 80.0 atm is approximately 2.07 x 10^3 kg/m^3.

Step-by-step explanation:

To find the density of water at a depth where the pressure is 80.0 atm, we need to use the formula P = hpg, where P is the pressure, h is the depth, ρ is the density, and g is the acceleration due to gravity. The density of water at the surface is given as 1.03×10^3 kg/m^3. We can use this information to calculate the density at a depth of 80.0 atm.

First, we convert the pressure from atm to Pa using the conversion factor 1 atm = 101325 Pa. So the pressure is 80.0 atm * 101325 Pa/atm = 8.1 x 10^6 Pa.

Next, we can rearrange the formula P = hpg to solve for ρ:

ρ = P / (hg)

Substituting in the known values, we have:

ρ = (8.1 x 10^6 Pa) / (40.0 m * 9.8 m/s^2)

Calculating the expression, we find the density of water at a depth where the pressure is 80.0 atm to be approximately 2.07 x 10^3 kg/m^3.