One of the important approach to the study of real gases involves the analysis of parameter `Z` called the compressibility factor `Z = (PV_(m))/(RT)` where `P` is pressure, `V_(m)` is molar volume, `T` is absolute temperature and `R` is the universal gas constant. such a relation can also be expressed as `Z = ((V_(m)real)/(V_(m)ideal))` (where `V_(m ideal)` and `V_(m real)` are the molar volume for ideal and real gas respectively). Gas corresponding `Z lt 1` have attractive forces amoung constituent particles. As the pressure is lowered or temperature is increased the value of `Z` approaches 1. (reaching the ideal behaviour)
For a real gas `G Z gt 1` at `STP` Then for 'G': Which of the following is true:

Options

A. 1 mole of the gas occupies 22.4 L at `NTP`
B. 1 mole of the gas occupies 22.4 L at pressure higher than that at `STP` (keeping temperature constant)
C. 1 mole of the gas occupies 22.4 L at pressure lower than that at `STP` (keeping temperature constant)
D. None of the above

Answer

To solve the problem, we need to analyze the compressibility factor \( Z \) and its implications for a real gas at standard temperature and pressure (STP).

### Step-by-Step Solution:

1. **Understanding the Compressibility Factor \( Z \)**:
- The compressibility factor is defined as:
\[ Z = \frac{PV_m}{RT} \]
- Where:
- \( P \) = pressure
- \( V_m \) = molar volume
- \( R \) = universal gas constant
- \( T \) = absolute temperature

2. **Relation of \( Z \) with Ideal and Real Gases**:
- The compressibility factor can also be expressed as:
\[ Z = \frac{V_{m \text{ real}}}{V_{m \text{ ideal}}} \]
- For ideal gases, \( Z \) is equal to 1. For real gases, \( Z \) can be less than, equal to, or greater than 1 depending on the conditions.

3. **Behavior of Real Gases**:
- If \( Z < 1 \), it indicates that the gas has attractive forces among its particles.
- If \( Z > 1 \), it indicates that the gas behaves less ideally, suggesting repulsive forces dominate.

4. **Given Condition**:
- The problem states that for a real gas \( G \), \( Z > 1 \) at STP (Standard Temperature and Pressure).
- At STP, the pressure \( P = 1 \, \text{atm} \) and the molar volume of an ideal gas \( V_{m \text{ ideal}} = 22.4 \, \text{L} \).

5. **Analyzing the Condition \( Z > 1 \)**:
- From the definition of \( Z \):
\[ Z = \frac{PV_m}{RT} > 1 \]
- This implies:
\[ PV_m > RT \]

6. **Calculating Molar Volume**:
- At STP, we can substitute \( P = 1 \, \text{atm} \) and \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \) and \( T = 273 \, \text{K} \):
\[ RT = 0.0821 \times 273 = 22.4 \, \text{L atm} \]
- Therefore, for \( Z > 1 \):
\[ PV_m > 22.4 \, \text{L atm} \]

7. **Conclusion**:
- Since \( P = 1 \, \text{atm} \), we have:
\[ 1 \times V_m > 22.4 \implies V_m > 22.4 \, \text{L} \]
- This means that for the gas \( G \) at STP, the molar volume \( V_m \) must be greater than 22.4 L.

### Options Analysis:
- The options provided in the question can be evaluated based on our conclusion that \( V_m > 22.4 \, \text{L} \) at STP for gas \( G \).