Which of the following choices match (es) with `CO_(2)` at P = 1 atm and T = 273 K?

Options

A. `Zne1`
B. Attractive forces are dominant
C. PV = nRT
D. P(V - nb) = nRT

Answer

To solve the question regarding the behavior of carbon dioxide (CO₂) at a pressure of 1 atm and a temperature of 273 K, we can follow these steps:

### Step-by-Step Solution

1. **Identify the Conditions**:
- The question provides a pressure of 1 atm and a temperature of 273 K. These conditions correspond to standard temperature and pressure (STP).

2. **Understand Gas Behavior at STP**:
- At STP, gases can behave ideally or as real gases. The behavior of real gases deviates from ideal gas behavior due to intermolecular forces and the volume occupied by gas molecules.

3. **Determine the Nature of CO₂**:
- Carbon dioxide (CO₂) is a non-polar molecule with a linear structure. Its dipole moment is zero, indicating that it does not have a significant dipole-dipole interaction.

4. **Analyze the Intermolecular Forces**:
- Since CO₂ is a non-polar molecule, the intermolecular forces present are primarily London dispersion forces. However, due to its higher molecular weight compared to lighter gases, the attractive forces (dispersion forces) will dominate over repulsive forces.

5. **Evaluate the Compressibility Factor (Z)**:
- The compressibility factor (Z) is defined as Z = PV/nRT. For an ideal gas, Z = 1. However, for real gases, Z can be greater or less than 1 depending on the intermolecular forces.
- Since CO₂ is a real gas at these conditions, Z will not equal 1, confirming that it behaves as a real gas.

6. **Conclusion**:
- Based on the analysis, we can conclude that CO₂ at 1 atm and 273 K behaves as a real gas, and the attractive forces dominate due to its molecular weight.

### Final Answer
The correct choices that match with CO₂ at P = 1 atm and T = 273 K are the options that indicate it behaves as a real gas with Z not equal to 1.