During adiabatic process, pressure P and density equation is:

Options

A. `P rho ^(gamma)`= constant
B. `P rho^(- gamma)` = constant
C. `P^(gamma) rho^(gamma+1)` = constant
D. ` P ^(1/gamma) rho^(gamma)` = constant

Explanation

To derive the equation relating pressure (P) and density (ρ) during an adiabatic process, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Adiabatic Process**:
- In an adiabatic process, there is no heat exchange between the system and its surroundings. Therefore, the heat transfer \( Q = 0 \).

2. **Using the General Equation for Adiabatic Processes**:
- The general equation for an adiabatic process is given by:
\[ P V^\gamma = \text{constant} \]
where \( P \) is the pressure, \( V \) is the volume, and \( \gamma \) (gamma) is the adiabatic index (ratio of specific heats).

3. **Relating Volume to Density**:
- We know that density \( \rho \) is defined as:
\[ \rho = \frac{m}{V} \]
where \( m \) is the mass of the gas. Rearranging this gives:
\[ V = \frac{m}{\rho} \]

4. **Substituting Volume in the Adiabatic Equation**:
- Substitute \( V \) in the adiabatic equation:
\[ P \left(\frac{m}{\rho}\right)^\gamma = \text{constant} \]

5. **Simplifying the Equation**:
- This can be rewritten as:
\[ P \cdot \frac{m^\gamma}{\rho^\gamma} = \text{constant} \]
- Since \( m^\gamma \) is a constant for a given mass of gas, we can denote it as a new constant \( C \):
\[ P \cdot \frac{1}{\rho^\gamma} = C \]

6. **Final Relation**:
- Rearranging gives us:
\[ P = C \cdot \rho^\gamma \]
- This shows that pressure is proportional to the density raised to the power of gamma during an adiabatic process.

### Final Equation:
\[ P \propto \rho^\gamma \]