What is the decrease in volume of one litre of water when a pressure of 10 atm is applied ? Given that the compressibility of water is ` 5 xx 10^(-10) m ^(2) N^(-1)`

Explanation

To find the decrease in volume of one liter of water when a pressure of 10 atm is applied, we will use the concept of compressibility and the bulk modulus of water.

### Step-by-Step Solution:

1. **Understand the given values:**
- Compressibility of water, \( \beta = 5 \times 10^{-10} \, \text{m}^2/\text{N} \)
- Pressure applied, \( P = 10 \, \text{atm} \)
- Volume of water, \( V = 1 \, \text{liter} = 1 \times 10^{-3} \, \text{m}^3 \)

2. **Convert pressure from atm to N/m²:**
- \( 1 \, \text{atm} = 1.01325 \times 10^5 \, \text{N/m}^2 \)
- Therefore, \( P = 10 \times 1.01325 \times 10^5 \, \text{N/m}^2 = 1.01325 \times 10^6 \, \text{N/m}^2 \)

3. **Calculate the bulk modulus \( K \):**
- The bulk modulus \( K \) is the reciprocal of compressibility:
\[ K = \frac{1}{\beta} = \frac{1}{5 \times 10^{-10}} = 2 \times 10^{9} \, \text{N/m}^2 \]

4. **Use the formula for change in volume \( \Delta V \):**
- The formula relating pressure, change in volume, and bulk modulus is:
\[ K = -\frac{P}{\Delta V / V} \]
Rearranging gives:
\[ \Delta V = \frac{P \cdot V}{K} \]

5. **Substitute the values into the formula:**
\[ \Delta V = \frac{(1.01325 \times 10^6) \cdot (1 \times 10^{-3})}{2 \times 10^{9}} \]

6. **Calculate \( \Delta V \):**
\[ \Delta V = \frac{1.01325 \times 10^3}{2 \times 10^{9}} = 5.06625 \times 10^{-7} \, \text{m}^3 \]

7. **Final answer:**
- The decrease in volume of one liter of water when a pressure of 10 atm is applied is approximately:
\[ \Delta V \approx 5.066 \times 10^{-7} \, \text{m}^3 \]