In an adiabatic change, the pressure p and temperature T of a diatomic gas are related by the relation `ppropT^alpha`, where `alpha` equals

Options

A. (a) 1.67
B. (b) 0.4
C. (c) 0.6
D. (d) 3.5

Explanation

To solve the problem of finding the relation between pressure \( p \) and temperature \( T \) for a diatomic gas undergoing an adiabatic change, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Adiabatic Law**:
The adiabatic process for an ideal gas is described by the equation:
\[ PV^\gamma = \text{constant} \]
where \( \gamma \) (gamma) is the heat capacity ratio \( C_p/C_v \).

2. **Use the Ideal Gas Law**:
The ideal gas equation is given by:
\[ PV = nRT \]
From this, we can express volume \( V \) as:
\[ V = \frac{nRT}{P} \]

3. **Substitute Volume in the Adiabatic Law**:
Substitute the expression for \( V \) into the adiabatic law:
\[ P \left(\frac{nRT}{P}\right)^\gamma = \text{constant} \]

4. **Simplify the Equation**:
This can be rearranged to:
\[ P \cdot \frac{(nRT)^\gamma}{P^\gamma} = \text{constant} \]
Simplifying gives:
\[ P^{1 - \gamma} (nR)^\gamma T^\gamma = \text{constant} \]

5. **Isolate Pressure**:
Rearranging the equation, we find:
\[ P^{1 - \gamma} \propto \frac{1}{T^\gamma} \]
This implies:
\[ P \propto T^{-\frac{\gamma}{1 - \gamma}} \]

6. **Identify Alpha**:
From the proportionality, we can identify \( \alpha \):
\[ \alpha = -\frac{\gamma}{1 - \gamma} \]

7. **Determine Gamma for Diatomic Gas**:
For a diatomic gas, \( \gamma = \frac{7}{5} \). Substitute this value into the expression for \( \alpha \):
\[ \alpha = -\frac{\frac{7}{5}}{1 - \frac{7}{5}} = -\frac{\frac{7}{5}}{-\frac{2}{5}} = \frac{7}{2} \]

8. **Final Result**:
Therefore, the value of \( \alpha \) is:
\[ \alpha = 3.5 \]

### Conclusion:
The relation between pressure \( p \) and temperature \( T \) for a diatomic gas undergoing an adiabatic change is given by:
\[ p \propto T^{3.5} \]