Question

The increase in pressure required to decrease the volume of a water sample by 0.2percentage is \( P \times 10^5 \, \text{Nm}^{-2} \). Bulk modulus of water is \( 2.15 \times 10^9 \, \text{Nm}^{-2} \). The value of \( P \) is ________________.

Hint:

The bulk modulus describes the resistance of a substance to uniform compression. A smaller bulk modulus means it is easier to compress the substance, while a larger value means greater resistance.

Answer 1

Step 1: Identify Provided Data

The problem provides the following data:

• The volume of water decreases by 0.2% (percentage change is -0.2%).
  
• The bulk modulus of water is \( 2.15 \times 10^9 \, \text{Nm}^{-2} \).
  
• The required pressure change is represented as \( P \times 10^5 \, \text{Nm}^{-2} \); the objective is to determine \( P \).
  

Step 2: Bulk Modulus Formula

The bulk modulus \( B \) is defined by the relationship between pressure change \( \Delta P \) and fractional volume change \( \Delta V / V \) as follows:
\[ B = - \frac{\Delta P}{\frac{\Delta V}{V}} \]

where:
- \( B \) denotes the bulk modulus,
- \( \Delta P \) is the change in pressure,
- \( \frac{\Delta V}{V} \) represents the fractional change in volume.

Step 3: Input Values into Formula

The fractional volume change is 0.2%, equivalent to \( 0.2/100 = 0.002 \).
Using the given bulk modulus \( B = 2.15 \times 10^9 \, \text{Nm}^{-2} \), the equation becomes:
\[ 2.15 \times 10^9 = - \frac{\Delta P}{0.002} \] Solving for \( \Delta P \):
\[ \Delta P = - (2.15 \times 10^9) \times 0.002 = 4.3 \times 10^6 \, \text{Nm}^{-2} \]

Step 4: Equate Pressure Changes

The problem states the pressure change is \( P \times 10^5 \, \text{Nm}^{-2} \), so:
\[ P \times 10^5 = 4.3 \times 10^6 \] To find \( P \):
\[ P = \frac{4.3 \times 10^6}{10^5} = 43 \]

Conclusion

The determined value for \( P \) is 43.