Which of the following is correct?

Options

A. For `H_(2)andHe,zlt1,` and molar volume at STP is less than 22.4 L
B. For `H_(2)andHe,zlt1,` and molar volume at STP is greater than 22.4 L
C. For `H_(2)andHe ,zgt1,` and molar volume at STP is less than 22.4 L
D. For `H_(2)andHe,zgt1,` and molar volume at STP is greater than 22.4 L

Answer

To solve the question regarding the compressibility factor (Z) for H2 (hydrogen) and Helium, we can follow these steps:

### Step 1: Understand the Compressibility Factor (Z)
The compressibility factor (Z) is defined as:
\[ Z = \frac{V_{\text{real}}}{V_{\text{ideal}}} \]
where \( V_{\text{real}} \) is the molar volume of the gas under real conditions and \( V_{\text{ideal}} \) is the molar volume of the gas under ideal conditions.

### Step 2: Identify the Ideal Molar Volume
For an ideal gas at standard temperature and pressure (STP), the molar volume is approximately 22.4 liters.

### Step 3: Analyze the Given Information
According to the video transcript, it is stated that for both H2 and Helium, the compressibility factor Z is greater than 1. This implies:
\[ Z > 1 \]
Thus, we can conclude:
\[ V_{\text{real}} > V_{\text{ideal}} \]

### Step 4: Calculate the Real Molar Volume
Since \( V_{\text{ideal}} \) is 22.4 liters, and given that \( Z > 1 \), we can infer:
\[ V_{\text{real}} > 22.4 \, \text{liters} \]

### Step 5: Conclusion
From the above analysis, we can conclude that for H2 and Helium, the compressibility factor Z is greater than 1, and the molar volume at high temperature and pressure (HTP) is greater than 22.4 liters. Therefore, the correct option is:
- **Option 4:** For H2 and Helium, Z is greater than 1, and the molar volume at HTP is greater than 22.4 liters.

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