Ethanol has a density of `659 kg//m^3` . If the speed of sound in ethanol is 1162 m/s, what is its adiabatic bulk modulus?

Options

A. `1.7 xx 10^8 N//m^2`
B. `7.7 xx 10^8 N//m^2`
C. `2.2 xx 10^8 N//m^2`
D. `8.9 xx 10^8 N//m^2`

Answer

To find the adiabatic bulk modulus of ethanol, we can use the relationship between the speed of sound in a medium, its density, and the bulk modulus. The formula we will use is:

\[ v = \sqrt{\frac{K}{\rho}} \]

Where:
- \( v \) is the speed of sound in the medium (in m/s),
- \( K \) is the adiabatic bulk modulus (in N/m²),
- \( \rho \) is the density of the medium (in kg/m³).

From this formula, we can rearrange it to solve for the bulk modulus \( K \):

\[ K = v^2 \cdot \rho \]

### Step-by-Step Solution:

1. **Identify the given values**:
- Speed of sound in ethanol, \( v = 1162 \, \text{m/s} \)
- Density of ethanol, \( \rho = 659 \, \text{kg/m}^3 \)

2. **Substitute the values into the formula**:
\[ K = (1162 \, \text{m/s})^2 \cdot (659 \, \text{kg/m}^3) \]

3. **Calculate \( v^2 \)**:
\[ v^2 = 1162^2 = 1350244 \, \text{m}^2/\text{s}^2 \]

4. **Multiply \( v^2 \) by the density \( \rho \)**:
\[ K = 1350244 \, \text{m}^2/\text{s}^2 \cdot 659 \, \text{kg/m}^3 \]

5. **Perform the multiplication**:
\[ K = 891153876 \, \text{N/m}^2 \]

6. **Convert to scientific notation**:
\[ K \approx 8.91 \times 10^8 \, \text{N/m}^2 \]

### Final Answer:
The adiabatic bulk modulus of ethanol is approximately \( 8.91 \times 10^8 \, \text{N/m}^2 \).

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