To solve the problem regarding the compressibility factor (Z) for one mole of a gas under standard temperature and pressure (STP) conditions, we can follow these steps:
### Step-by-Step Solution:
1. **Understand the Compressibility Factor (Z)**:
The compressibility factor \( Z \) is defined as:
\[
Z = \frac{PV}{nRT}
\]
where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature
2. **Identify the Condition Given**:
The problem states that \( Z > 1 \) under STP conditions. This implies that the gas does not behave ideally.
3. **Interpret the Meaning of \( Z > 1 \)**:
- For an ideal gas, \( Z = 1 \).
- When \( Z > 1 \), it indicates that the gas occupies a volume greater than what is predicted by the ideal gas law. This suggests that the gas is less compressible than an ideal gas.
4. **Relate \( Z \) to Volume**:
Since \( Z \) is directly proportional to the volume when pressure and temperature are constant, if \( Z > 1 \), it implies:
\[
V > \frac{nRT}{P}
\]
Under STP (Standard Temperature and Pressure), \( n = 1 \) mole, \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \), \( T = 273.15 \, \text{K} \), and \( P = 1 \, \text{atm} \).
5. **Calculate the Ideal Volume at STP**:
The ideal volume for one mole of an ideal gas at STP is:
\[
V_{\text{ideal}} = \frac{nRT}{P} = \frac{(1 \, \text{mol})(0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1})(273.15 \, \text{K})}{1 \, \text{atm}} \approx 22.4 \, \text{L}
\]
6. **Conclusion**:
Since \( Z > 1 \), it implies that the actual volume of the gas is greater than 22.4 L. Therefore, the volume of the gas under these conditions must be greater than the ideal volume.
### Final Answer:
Thus, if the compressibility factor \( Z \) for one mole of a gas is more than one under STP conditions, it indicates that the volume of the gas is greater than 22.4 L.
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