For a p-V plot, the slope of an adiabatic curve = x × slope of isothermal at the same point. Here x is :

Options

A. (a)`R/(lamda - 1)`
B. (b)`1/lamda`
C. (c)`lamda`
D. (d)`1/(lamda-1)`

Explanation

To solve the problem, we need to determine the relationship between the slopes of the adiabatic and isothermal curves on a p-V plot.

### Step-by-Step Solution:

1. **Understand the Isothermal Process**:
- For an isothermal process, the temperature (T) remains constant. The equation for an isothermal process is given by:
\[ PV = nRT \]
- Taking the differential of this equation gives us:
\[ d(PV) = 0 \implies PdV + VdP = 0 \]
- Rearranging this, we find:
\[ \frac{dP}{dV} = -\frac{P}{V} \]
- This indicates that the slope of the isothermal curve at any point is:
\[ \text{slope}_{\text{isothermal}} = -\frac{P}{V} \]

2. **Understand the Adiabatic Process**:
- For an adiabatic process, there is no heat exchange (Q = 0). The relationship for an adiabatic process is given by:
\[ PV^\gamma = \text{constant} \]
- Here, \(\gamma\) (gamma) is the adiabatic index, which is the ratio of specific heats (\(C_p/C_v\)).
- Taking the differential of this equation gives us:
\[ d(PV^\gamma) = 0 \implies Pd(V^\gamma) + V^\gamma dP = 0 \]
- Using the product rule, we can express \(d(V^\gamma)\) as:
\[ \gamma PV^{\gamma-1}dV + V^\gamma dP = 0 \]
- Rearranging gives:
\[ dP = -\frac{\gamma P}{V} dV \]
- Thus, the slope of the adiabatic curve at any point is:
\[ \text{slope}_{\text{adiabatic}} = -\frac{\gamma P}{V} \]

3. **Relate the Slopes**:
- Now, we can relate the slopes of the adiabatic and isothermal curves:
\[ \text{slope}_{\text{adiabatic}} = \gamma \times \text{slope}_{\text{isothermal}} \]
- Therefore, we can express this as:
\[ \text{slope}_{\text{adiabatic}} = \gamma \left(-\frac{P}{V}\right) \]

4. **Conclusion**:
- From the relationship derived, we see that:
\[ x = \gamma \]
- Hence, the value of \(x\) is equal to the adiabatic index \(\gamma\).

### Final Answer:
The value of \(x\) is \(\gamma\).