Two ideal gases A & B are going through adiabatic process. Choose the correct option.

Options

A. both A & B are monoatomic
B. both A & B are diatomic
C. B is diatomic, A is monoatomic
D. B is monoatomic, A is diatomic

Answer

To solve the problem regarding the adiabatic processes of two ideal gases A and B, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Adiabatic Process**:
- An adiabatic process is one in which no heat is transferred to or from the system. For ideal gases, the relationship between pressure (P), volume (V), and the adiabatic index (γ) is given by the equation:
\[ PV^\gamma = K \]
where \( K \) is a constant.

2. **Graph Analysis**:
- The problem mentions a graph plotted between \( \ln P \) and \( \ln V \). The equation can be rearranged to express \( \ln P \) as:
\[ \ln P = -\gamma \ln V + \ln K \]
- This indicates that the graph will be a straight line with a slope of \( -\gamma \).

3. **Comparing Slopes**:
- If we have two gases A and B represented on the same graph, and if the slope for gas B is greater than the slope for gas A, we can conclude:
\[ -\gamma_B < -\gamma_A \quad \Rightarrow \quad \gamma_B > \gamma_A \]
- This means that the adiabatic index \( \gamma \) for gas B is greater than that for gas A.

4. **Identifying Gas Types**:
- For ideal gases, the adiabatic index \( \gamma \) is different for monoatomic and diatomic gases:
- For monoatomic gases (like Helium, Neon), \( \gamma \approx \frac{5}{3} \).
- For diatomic gases (like Oxygen, Nitrogen), \( \gamma \approx \frac{7}{5} \).
- Since \( \gamma_B > \gamma_A \), gas B must be monoatomic (higher \( \gamma \)) and gas A must be diatomic (lower \( \gamma \)).

5. **Conclusion**:
- Based on the analysis, we can conclude that gas B is monoatomic and gas A is diatomic. Therefore, the correct option is:
- **Option D**: B is monoatomic and A is diatomic.