Compressibility factor (Z) for `H_2 (g)` at STP is

Options

A. `Z = 1`
B. `[Z = 1 +` `a`/`(V_m RT)]`
C. `[Z = 1 -` `a`/`(V_m RT)]`
D. `[``Z = 1` `+` `Pb`/`Rt``]`

Answer

To find the compressibility factor (Z) for hydrogen gas (H₂) at standard temperature and pressure (STP), we can follow these steps:

### Step 1: Understand the Compressibility Factor (Z)
The compressibility factor (Z) is defined as:
\[ Z = \frac{PV_m}{RT} \]
where:
- \( P \) = pressure,
- \( V_m \) = molar volume,
- \( R \) = universal gas constant,
- \( T \) = temperature.

### Step 2: Ideal Gas Behavior
For an ideal gas, the compressibility factor \( Z \) is equal to 1. This means that under ideal conditions, the gas behaves according to the ideal gas law:
\[ PV = nRT \]

### Step 3: Conditions at STP
At STP (Standard Temperature and Pressure):
- Temperature \( T = 273.15 \, K \)
- Pressure \( P = 1 \, atm \)

### Step 4: Molar Volume at STP
The molar volume \( V_m \) of an ideal gas at STP is approximately:
\[ V_m = 22.4 \, L/mol \]

### Step 5: Calculate Z for H₂
Now, substituting the values into the compressibility factor formula:
\[ Z = \frac{PV_m}{RT} \]
Substituting \( P = 1 \, atm \), \( V_m = 22.4 \, L/mol \), \( R = 0.0821 \, L \cdot atm/(K \cdot mol) \), and \( T = 273.15 \, K \):
\[ Z = \frac{(1 \, atm)(22.4 \, L/mol)}{(0.0821 \, L \cdot atm/(K \cdot mol))(273.15 \, K)} \]

### Step 6: Perform the Calculation
Calculating the denominator:
\[ 0.0821 \times 273.15 \approx 22.414 \]
Now substituting back into the equation for Z:
\[ Z = \frac{22.4}{22.414} \approx 0.999 \]

### Step 7: Conclusion
Since \( Z \) is very close to 1, it indicates that hydrogen gas behaves nearly ideally at STP. However, for hydrogen gas, there is a known positive deviation from ideal behavior due to intermolecular forces.

Thus, the compressibility factor \( Z \) for \( H_2(g) \) at STP is approximately:
\[ Z \approx 1.000 \]