For adiabatic reversible expansion of an ideal gas the expression relating pressure and volume of the gas is -

Options

A. (a) `P_(1)V_(1)=P_(2)V_(2)`
B. (b) `(P_(1)V_(1))/(T_(1))=(P_(2)V_(2))/(T_(2))`
C. (c) `P_(1)V_(1)^(gamma)=P_(2)V_(2)^(gamma)`
D. (d) `P=(1)/(V)`

Answer

To solve the question regarding the expression relating pressure and volume of an ideal gas during adiabatic reversible expansion, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Process**:
- An adiabatic process is one where no heat is exchanged with the surroundings. In the case of an ideal gas undergoing adiabatic reversible expansion, we need to find the relationship between pressure (P) and volume (V).

2. **Recall the Ideal Gas Law**:
- The ideal gas law is given by \( PV = nRT \), where \( n \) is the number of moles, \( R \) is the universal gas constant, and \( T \) is the temperature.

3. **Identify the Adiabatic Condition**:
- For an adiabatic process, the relationship between pressure and volume can be expressed using the equation:
\[ PV^\gamma = \text{constant} \]
- Here, \( \gamma \) (gamma) is the heat capacity ratio, defined as \( \gamma = \frac{C_p}{C_v} \), where \( C_p \) is the heat capacity at constant pressure and \( C_v \) is the heat capacity at constant volume.

4. **Analyze the Given Options**:
- The options provided are:
- (A) \( p_1 V_1 = p_2 V_2 \) (Isothermal process)
- (B) \( \frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2} \) (General gas law)
- (C) \( p_1 V_1^\gamma = p_2 V_2^\gamma \) (Adiabatic process)
- (D) \( p = \frac{1}{V} \) (Isothermal process)

5. **Select the Correct Expression**:
- Among the options, the correct expression for the adiabatic reversible expansion of an ideal gas is:
\[ p_1 V_1^\gamma = p_2 V_2^\gamma \]
- Therefore, the correct answer is option (C).

### Final Answer:
The expression relating pressure and volume of an ideal gas during adiabatic reversible expansion is:
\[ p_1 V_1^\gamma = p_2 V_2^\gamma \]