The compressibillity of a gas is greater than unity at 1 atm and 273 K. Therefore :

Options

A. `V_(m)gt22.4L`
B. `V_(m)lt22.4L`
C. `V_(m)=22.4L`
D. `V_(m)=44.8L`

Explanation

To solve the question regarding the compressibility of a gas being greater than unity at 1 atm and 273 K, we can follow these steps:

### Step 1: Understand Compressibility Factor
The compressibility factor (Z) is defined as the ratio of the molar volume of a real gas (Vm real) to the molar volume of an ideal gas (Vm ideal) under the same conditions. Mathematically, it is expressed as:
\[ Z = \frac{V_{m \, \text{real}}}{V_{m \, \text{ideal}}} \]

### Step 2: Identify Conditions
The question states that the compressibility of the gas is greater than unity (Z > 1) at 1 atm and 273 K. This implies that the volume of the real gas is greater than that of the ideal gas under the same conditions.

### Step 3: Determine Ideal Gas Volume
At standard temperature and pressure (STP), which is 0°C (273 K) and 1 atm, the molar volume of an ideal gas (Vm ideal) is approximately 22.4 liters.

### Step 4: Relate Real Gas Volume to Ideal Gas Volume
Since we know Z > 1, we can express this relationship as:
\[ V_{m \, \text{real}} > V_{m \, \text{ideal}} \]
Thus:
\[ V_{m \, \text{real}} > 22.4 \, \text{liters} \]

### Conclusion
From the above steps, we conclude that if the compressibility factor of a gas is greater than unity at 1 atm and 273 K, the volume of the real gas must be greater than 22.4 liters.