Question

Problem: For K=2.2×109 Pa for the bulk modulus of elasticity of water, what pressure is required to reduce its volume by 0.5%?

Answer 1

Answer on Question #53003, Physics / Mechanics | Kinematics | Dynamics

For $K = 2.2 \times 10^{9} \mathrm{Pa}$ for the bulk modulus of elasticity of water, what pressure is required to reduce its volume by $0.5\%$?

Solution:

We have the following given data.

Bulk modulus of elasticity of water $(K) = 2.2 \times 10^{9} \mathrm{Pa} = 2.2 \mathrm{Gpa}$

Reduction in volume $\left(\frac{\Delta V}{V}\right) = -0.5\% = -0.005$

We need to determine the pressure $(\Delta P)$

We know that the modulus of elasticity is equal

$$K = - \frac {\Delta P}{\frac {\Delta V}{V}}$$

Thus, we can substitute the given values according to the condition of the task into the noted above formula.

$$2.2 \times 10^{9} = - \frac {\Delta P}{-0.005}$$

Now, we can simplify the obtained equation in order to find the value of pressure which is required to reduce its volume by $0.5\%$.

$$\Delta P = (2.2 \times 10^{9}) \cdot 0.005 = 11000 \mathrm{KPa}$$ $$\Delta P = 11000 \mathrm{KPa}$$

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