For a reversible adiabatic ideal gas expansion `(dp)/(p)` is equal to

Options

A. `gamma(dV)/(V)`
B. `-gamma(dV)/(V)`
C. `((gamma)/(gamma-1))(dV)/(V)`
D. `(dV)/(V)`

Explanation

To solve the problem of finding the expression for \(\frac{dP}{P}\) during a reversible adiabatic expansion of an ideal gas, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Adiabatic Condition**: For an ideal gas undergoing a reversible adiabatic process, we have the relation:
\[ PV^\gamma = \text{constant} \]
where \(P\) is the pressure, \(V\) is the volume, and \(\gamma\) is the heat capacity ratio \(\left(\frac{C_p}{C_v}\right)\).

2. **Differentiate the Equation**: We differentiate the equation \(PV^\gamma = \text{constant}\) with respect to \(P\) and \(V\):
\[ d(PV^\gamma) = 0 \]
This implies:
\[ P d(V^\gamma) + V^\gamma dP = 0 \]

3. **Apply the Product Rule**: Using the product rule on \(V^\gamma\):
\[ d(V^\gamma) = \gamma V^{\gamma - 1} dV \]
Substituting this back into our differentiated equation gives:
\[ P \gamma V^{\gamma - 1} dV + V^\gamma dP = 0 \]

4. **Rearranging the Equation**: Rearranging the equation to isolate \(dP\):
\[ V^\gamma dP = -P \gamma V^{\gamma - 1} dV \]
Dividing both sides by \(PV^\gamma\) gives:
\[ \frac{dP}{P} = -\frac{\gamma dV}{V} \]

5. **Final Expression**: Thus, we arrive at the final expression:
\[ \frac{dP}{P} = -\gamma \frac{dV}{V} \]

### Summary of the Result:
The expression for the change in pressure relative to pressure during a reversible adiabatic expansion of an ideal gas is:
\[ \frac{dP}{P} = -\gamma \frac{dV}{V} \]