The compressibility factor of gases is less than unity at `STP`. Therefore,

Options

A. (a) `V_(m) gt 22.4` litres
B. (b) `V_(m) lt 22.4` litres
C. (c) `V_(m) = 22.4` litres
D. (d) `V_(m) = 44.8` litres

Explanation

To solve the question regarding the compressibility factor of gases at standard temperature and pressure (STP), we can follow these steps:

### Step-by-Step Solution:

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

2. **Given Condition**:
We know that at STP, the compressibility factor \( Z \) is less than 1:
\[ Z < 1 \]
This implies:
\[ \frac{PV_m}{nRT} < 1 \]

3. **Rearranging the Inequality**:
Rearranging the inequality gives us:
\[ PV_m < nRT \]

4. **Substituting Known Values at STP**:
At STP:
- \( P = 1 \, \text{atm} = 101.325 \, \text{kPa} \)
- \( R = 0.0821 \, \text{L atm K}^{-1} \text{mol}^{-1} \)
- \( T = 273 \, \text{K} \)
- For 1 mole of gas, \( n = 1 \)

5. **Calculating \( nRT \)**:
Substitute the values into the equation:
\[ nRT = 1 \times 0.0821 \times 273 \]
Calculating this gives:
\[ nRT \approx 22.414 \, \text{L} \]

6. **Finding the Molar Volume Condition**:
From the inequality \( PV_m < nRT \), we can conclude:
\[ PV_m < 22.414 \, \text{L} \]
Therefore, we can express this as:
\[ V_m < 22.414 \, \text{L} \]

7. **Conclusion**:
The condition that the compressibility factor \( Z < 1 \) at STP implies that the molar volume \( V_m \) of the gas must be less than 22.414 L.

### Final Answer:
Thus, the correct conclusion is that the molar volume \( V_m \) is less than 22.4 L at STP.