Q.What is the density of water at a depth where pressure is 80.0 atm, given that its density at the surface is $1.03 \times 10^{3}\ \text{kg m}^{-3}$?

The key idea is the Bulk Modulus Application, which describes how a fluid's volume (and thus density) changes under pressure.

1. The change in pressure $\Delta P$ from the surface to the depth is $80.0\ \text{atm} - 1.0\ \text{atm} = 79.0\ \text{atm}$. Converting to Pascals: $\Delta P = 79.0 \times 1.013 \times 10^5\ \text{Pa} = 7.999 \times 10^6\ \text{Pa}$.
2. The Bulk Modulus $B$ relates the pressure change to the fractional volume change: $B = -\frac{\Delta P}{\Delta V / V_0}$. For water, we use the standard value $B = 2.2 \times 10^9\ \text{Pa}$.
3. Since mass is conserved ($m = \rho V$), a decrease in volume corresponds to an increase in density. For small changes, the fractional change in volume is approximately the negative of the fractional change in density: $\frac{\Delta V}{V_0} \approx -\frac{\Delta \rho}{\rho_0}$. …

When a fluid like water is subjected to increased pressure, its volume decreases slightly, leading to a corresponding increase in its density. This resistance to compression is characterized by a material property called the Bulk Modulus ($B$). The Bulk Modulus essentially tells us how much pressure is needed to cause a certain fractional change in volume.

The higher the Bulk Modulus, the more resistant the material is to compression. Water has a relatively high Bulk Modulus, meaning it is not easily compressible, but a measurable change in density does occur under significant pressure.

Here's how we approach this problem:

1. We identify the initial density and the change in pressure.
2. We use the definition of Bulk Modulus, which relates the pressure change to the fractional volume change.
3. We then relate the fractional volume change to the fractional density change, as mass remains constant.
4. Finally, we calculate the new density.

Let's work through the steps.

1. Identify Given Values and Target

   • Initial density of water at the surface, $\rho_0 = 1.03 \times 10^3 \text{ kg m}^{-3}$.
   • Pressure at depth, $P = 80.0 \text{ atm}$.
   • Surface pressure, $P_0 = 1.0 \text{ atm}$.
   • We need to find the new density, $\rho$.

   The change in pressure ($\Delta P$) experienced by the water from the surface to the depth is the difference between the pressure at depth and the surface pressure:

   $\Delta P = P - P_0 = 80.0 \text{ atm} - 1.0 \text{ atm} = 79.0 \text{ atm}$.

   To use the Bulk Modulus, we must convert this pressure change to SI units (Pascals):

   $1 \text{ atm} = 1.013 \times 10^5 \text{ Pa}$.

   $\Delta P = 79.0 \times (1.013 \times 10^5 \text{ Pa}) = 7.999 \times 10^6 \text{ Pa}$.

2. Recall the Bulk Modulus Formula

   The Bulk Modulus ($B$) is defined as:

   $B = - \frac{\Delta P}{\Delta V / V_0}$

   Where:

   • $\Delta P$ is the change in pressure.
   • $\Delta V$ is the change in volume.
   • $V_0$ is the initial volume.
   • The negative sign indicates that an increase in pressure ($\Delta P > 0$) leads to a decrease in volume ($\Delta V < 0$), keeping $B$ positive.

3. Relate Volume Change to Density Change

   The mass ($m$) of the water remains constant. We know that mass is related to density ($\rho$) and volume ($V$) by $m = \rho V$.

   So, $m = \rho_0 V_0 = \rho V$.

   Taking the differential of $m = \rho V$ (assuming $m$ is constant):

   $d(\rho V) = 0$

   $\rho dV + V d\rho = 0$

   Dividing by $\rho V$:

   $\frac{dV}{V} + \frac{d\rho}{\rho} = 0$

   $\frac{dV}{V} = - \frac{d\rho}{\rho}$

   For small changes, we can approximate this as:

   $\frac{\Delta V}{V_0} \approx - \frac{\Delta \rho}{\rho_0}$

   Where $\Delta \rho = \rho - \rho_0$ is the change in density. …

Step 1: given surface density 1.03e3 kg/m^3, pressure at depth 80 atm, B_water=2.2e9 Pa. Step 2: convert pressure: ~8.10e6 Pa. Step 3: mass conserved => r …