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: Understand the Given Information

The problem provides the following information:

• The percentage change in volume of water is 0.2% (i.e., the volume decreases by 0.2%).
  
• The bulk modulus of water is given as \( 2.15 \times 10^9 \, \text{Nm}^{-2} \).
  
• The pressure change required to decrease the volume by this percentage is given by \( P \times 10^5 \, \text{Nm}^{-2} \), and we are asked to find the value of \( P \).
  

Step 2: Formula for Bulk Modulus

The bulk modulus \( B \) is related to the change in pressure \( \Delta P \) and the corresponding fractional change in volume \( \Delta V / V \) by the formula:
\[ B = - \frac{\Delta P}{\frac{\Delta V}{V}} \]

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

Step 3: Substitute the Given Values

We are given the fractional change in volume as 0.2%, which is equal to \( 0.2/100 = 0.002 \).
The bulk modulus \( B \) is given as \( 2.15 \times 10^9 \, \text{Nm}^{-2} \), so we can write the equation as:
\[ 2.15 \times 10^9 = - \frac{\Delta P}{0.002} \] Solving for \( \Delta P \), we get:
\[ \Delta P = - (2.15 \times 10^9) \times 0.002 = 4.3 \times 10^6 \, \text{Nm}^{-2} \]

Step 4: Relating the Pressure Change to \( P \)

We are given that the pressure change is \( P \times 10^5 \, \text{Nm}^{-2} \), so:
\[ P \times 10^5 = 4.3 \times 10^6 \] Solving for \( P \), we get:
\[ P = \frac{4.3 \times 10^6}{10^5} = 43 \]

Conclusion

The value of \( P \) is 43.

Answer 2

Step 1: Understand the formula for Bulk Modulus.
The bulk modulus \( B \) is defined as the ratio of the change in pressure \( \Delta P \) to the relative change in volume \( \frac{\Delta V}{V} \), given by the equation:
\[ B = \frac{\Delta P}{\frac{\Delta V}{V}}. \] Here, \( \Delta P \) is the change in pressure, and \( \frac{\Delta V}{V} \) is the fractional change in volume.

Step 2: Apply the given data.
We are given the following information:
- The bulk modulus of water \( B = 2.15 \times 10^9 \, \text{Nm}^{-2} \).
- The fractional change in volume is \( \frac{\Delta V}{V} = -0.2\% = -0.002 \).
We are asked to find the change in pressure \( \Delta P \), which is given by:
\[ \Delta P = B \times \frac{\Delta V}{V}. \] Step 3: Calculate the change in pressure.
Substitute the known values into the equation:
\[ \Delta P = 2.15 \times 10^9 \times (-0.002) = -4.3 \times 10^6 \, \text{Nm}^{-2}. \] The magnitude of the change in pressure is \( 4.3 \times 10^6 \, \text{Nm}^{-2} \). The value of \( P \) is 43 when expressed in units of \( P \times 10^5 \, \text{Nm}^{-2} \).

Final Answer:
\[ \boxed{43}. \]