If the slope for isotherm is X and the slope for adiabat is Y then

Options

A. X=`gammaY`
B. `Y=gammaX`
C. X=Y
D. `X^(gamma)=Y`

Answer

To solve the problem regarding the slopes of isotherms and adiabats in thermodynamics, we will derive the relationship between the slopes \(X\) (for isotherm) and \(Y\) (for adiabat).

### Step-by-Step Solution:

1. **Understanding Isotherm and Adiabat**:
- An isotherm is a curve on a pressure-volume (P-V) diagram where the temperature remains constant.
- An adiabat is a curve on a P-V diagram where no heat is exchanged with the surroundings.

2. **Differentiating the Isotherm Equation**:
- For an isothermal process, we can use the ideal gas law:
\[ PV = nRT \]
- Differentiating this with respect to volume \(V\) gives:
\[ PdV + VdP = 0 \]
- Rearranging this, we find:
\[ \frac{dP}{dV} = -\frac{P}{V} \]
- The slope of the isotherm, denoted as \(X\), can be represented as:
\[ X = -\frac{P}{V} \]

3. **Differentiating the Adiabat Equation**:
- For an adiabatic process, the relationship can be expressed as:
\[ PV^\gamma = \text{constant} \]
- Differentiating this gives:
\[ PdV + VdP \cdot \gamma = 0 \]
- Rearranging this, we find:
\[ \frac{dP}{dV} = -\frac{\gamma P}{V} \]
- The slope of the adiabat, denoted as \(Y\), can be represented as:
\[ Y = -\frac{\gamma P}{V} \]

4. **Relating the Slopes \(X\) and \(Y\)**:
- Now, we can relate the slopes \(X\) and \(Y\):
\[ \frac{X}{Y} = \frac{-\frac{P}{V}}{-\frac{\gamma P}{V}} = \frac{1}{\gamma} \]
- This implies:
\[ \gamma X = Y \]

5. **Conclusion**:
- The relationship between the slopes of the isotherm and the adiabat is given by:
\[ Y = \gamma X \]