On P-V coordinates, the slope of an isothermal curve of a gas at a pressure P = IMPa and volume V= 0.0025 `m^(3)` is equal to `-400 Mpa//m^(3)`. If `C_(p)//C_(v) =1.4` , the slope of the adiabatic curve passing through this point is :

Options

A. `-56 Mpa//m^(3)`
B. `-400 Mpa//m^(3)`
C. `-560 Mpa//m^(3)`
D. None of these

Explanation

To find the slope of the adiabatic curve passing through the given point on the P-V diagram, we can use the relationship between the slopes of the isothermal and adiabatic curves.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Slope of the isothermal curve (m_isothermal) = -400 MPa/m³
- Value of \( \frac{C_p}{C_v} \) (gamma, \( \gamma \)) = 1.4

2. **Use the Relationship Between Slopes:**
The slope of the adiabatic curve (m_adiabatic) is related to the slope of the isothermal curve by the formula:
\[ m_{\text{adiabatic}} = \gamma \times m_{\text{isothermal}} \]

3. **Substitute the Known Values:**
Substitute the values we have into the equation:
\[ m_{\text{adiabatic}} = 1.4 \times (-400 \, \text{MPa/m}^3) \]

4. **Calculate the Slope of the Adiabatic Curve:**
\[ m_{\text{adiabatic}} = -560 \, \text{MPa/m}^3 \]

5. **Conclusion:**
The slope of the adiabatic curve passing through the point where \( P = 1 \, \text{MPa} \) and \( V = 0.0025 \, \text{m}^3 \) is:
\[ m_{\text{adiabatic}} = -560 \, \text{MPa/m}^3 \]