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 `Cp // Cv = 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 at the given point on the P-V diagram, we can use the relationship between the slopes of isothermal and adiabatic processes.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Pressure, \( P = 1 \, \text{MPa} = 10^6 \, \text{Pa} \)
- Volume, \( V = 0.0025 \, \text{m}^3 \)
- Slope of the isothermal curve, \( \left( \frac{dP}{dV} \right)_{\text{isothermal}} = -400 \, \text{MPa/m}^3 = -400 \times 10^6 \, \text{Pa/m}^3 \)
- Ratio of specific heats, \( \frac{C_p}{C_v} = \gamma = 1.4 \)

2. **Use the Relationship Between Slopes:**
The relationship between the slope of the isothermal curve and the slope of the adiabatic curve is given by:
\[ \left( \frac{dP}{dV} \right)_{\text{adiabatic}} = \gamma \times \left( \frac{dP}{dV} \right)_{\text{isothermal}} \]

3. **Substitute the Values:**
Substitute the values into the equation:
\[ \left( \frac{dP}{dV} \right)_{\text{adiabatic}} = 1.4 \times \left( -400 \times 10^6 \, \text{Pa/m}^3 \right) \]

4. **Calculate the Slope of the Adiabatic Curve:**
\[ \left( \frac{dP}{dV} \right)_{\text{adiabatic}} = 1.4 \times -400 \times 10^6 \]
\[ = -560 \times 10^6 \, \text{Pa/m}^3 \]
\[ = -560 \, \text{MPa/m}^3 \]

5. **Final Result:**
The slope of the adiabatic curve at the given point is:
\[ \left( \frac{dP}{dV} \right)_{\text{adiabatic}} = -560 \, \text{MPa/m}^3 \]