For a real gas `'G' Z gt1` as `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 `22.4L` 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 above

Explanation

To solve the question regarding the real gas 'G' with a compressibility factor \( Z > 1 \) at STP, we need to analyze the implications of this condition on the behavior of the gas.

### Step-by-Step Solution:

1. **Understanding the Compressibility Factor (Z)**:
- The compressibility factor \( Z \) is defined as:
\[ Z = \frac{V_m}{V_{real}} \]
where \( V_m \) is the molar volume of an ideal gas (22.4 L at STP) and \( V_{real} \) is the molar volume of the real gas.

2. **Given Condition**:
- We are given that \( Z > 1 \). This implies:
\[ \frac{V_m}{V_{real}} > 1 \implies V_m > V_{real} \]
- Rearranging gives us:
\[ V_{real} < V_m \]
- Since \( V_m = 22.4 \, \text{L} \), we conclude:
\[ V_{real} < 22.4 \, \text{L} \]

3. **Analyzing the Options**:
- The options provided are:
1. One mole of gas occupies 22.4 L at NTP.
2. One mole of gas occupies 22.4 L at pressure higher than that at STP.
3. One mole of gas occupies 22.4 L at pressure lower than that at STP.

4. **Relation Between Pressure and Volume**:
- According to Boyle's law, at constant temperature, pressure and volume are inversely related:
\[ P \propto \frac{1}{V} \]
- If the volume of the gas is less than 22.4 L (as we found), to achieve a volume of 22.4 L, the pressure must be increased.

5. **Conclusion**:
- Therefore, the correct statement is:
- **Option 2**: One mole of gas occupies 22.4 L at pressure higher than that at STP.

### Final Answer:
The correct option is **Option 2**: One mole of gas occupies 22.4 L at pressure higher than that at STP.