The increase in pressure required to decrease the 200 L volume of a liquid by 0.008 % in kPa is (Bulk modulus of the liquid = 2100 M Pa is )

Options

A. 8.4
B. 84
C. 92.4
D. 168

Answer

To solve the problem, we need to calculate the increase in pressure required to decrease the volume of a liquid by 0.008%. We will use the formula related to the bulk modulus of the liquid.

### Step-by-Step Solution:

1. **Understand the given values:**
- Volume of the liquid, \( V = 200 \, \text{L} \)
- Change in volume, \( \Delta V = 0.008\% \) of \( V \)
- Bulk modulus of the liquid, \( B = 2100 \, \text{MPa} = 2100 \times 10^6 \, \text{Pa} \)

2. **Convert the percentage change in volume to a decimal:**
\[ \Delta V = 0.008\% = \frac{0.008}{100} = 0.00008 \]

3. **Calculate the actual change in volume (\( \Delta V \)):**
\[ \Delta V = 0.00008 \times V = 0.00008 \times 200 \, \text{L} = 0.016 \, \text{L} \]
(Note: 1 L = \( 10^{-3} \, \text{m}^3 \), so \( 0.016 \, \text{L} = 0.016 \times 10^{-3} \, \text{m}^3 = 1.6 \times 10^{-5} \, \text{m}^3 \))

4. **Use the bulk modulus formula:**
The bulk modulus \( B \) is defined as:
\[ B = -\frac{\Delta P}{\frac{\Delta V}{V}} \]
Rearranging gives us:
\[ \Delta P = -B \cdot \frac{\Delta V}{V} \]

5. **Substituting the values into the equation:**
\[ \Delta P = -2100 \times 10^6 \, \text{Pa} \cdot \left( \frac{1.6 \times 10^{-5}}{200 \times 10^{-3}} \right) \]

6. **Calculate \( \frac{\Delta V}{V} \):**
\[ \frac{\Delta V}{V} = \frac{1.6 \times 10^{-5}}{0.2} = 8 \times 10^{-5} \]

7. **Now calculate \( \Delta P \):**
\[ \Delta P = -2100 \times 10^6 \cdot 8 \times 10^{-5} \]
\[ \Delta P = -168000 \, \text{Pa} = -168 \, \text{kPa} \]

8. **Final answer:**
The increase in pressure required to decrease the volume of the liquid by 0.008% is approximately:
\[ \Delta P \approx 168 \, \text{kPa} \]