The value of `C_(p) - C_(v)` is `1.09R` for a gas sample in state `A` and is `1.00R` in state `B`. Let `A_(A),T_(B)` denote the temperature and `p_(A)` and `P_(B)` denote the pressure of the states `A` and `B` respectively. Then

Options

A. `p_(A) lt p_(B)` and `T_(A) gt T_(B)`
B. `p_(A) gt p_(B)` and `T_(A) gt T_(B )`
C. `p_(A) = p_(B)` and `T_(A) lt T_(B)`
D. `p_(A) gt p_(B)` and `T_(A) lt T_(B)`

Explanation

To solve the problem, we need to analyze the given information regarding the specific heat capacities \(C_p\) and \(C_v\) for the gas in two different states, A and B.

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- For state A: \(C_p - C_v = 1.09R\)
- For state B: \(C_p - C_v = 1.00R\)
- We know that for an ideal gas, \(C_p - C_v = R\).

2. **Analyzing State B:**
- Since \(C_p - C_v = 1.00R\) in state B, this indicates that the gas behaves like an ideal gas in this state.
- Therefore, we can conclude that the temperature in state B is relatively high, allowing the gas to behave ideally.

3. **Analyzing State A:**
- In state A, \(C_p - C_v = 1.09R\), which is greater than \(1.00R\). This suggests that the gas in state A is not behaving like an ideal gas, likely due to intermolecular forces being significant at this state.

4. **Temperature Comparison:**
- Since state B behaves like an ideal gas, it must have a higher temperature than state A. Thus, we can conclude:
\[ T_B > T_A \]

5. **Pressure Comparison:**
- To behave like an ideal gas, the pressure in state B must be lower than in state A. This is because, for real gases to approximate ideal behavior, they typically need to be at lower pressures (where intermolecular forces are less significant). Therefore, we conclude:
\[ P_A > P_B \]

### Final Relations:
- From the analysis, we have:
- \(T_B > T_A\)
- \(P_A > P_B\)