Which gas shows real behaviour?

Options

A. 16 g `O_(2)` at 1 atm & 273 K occupies 11.2 L
B. 1 g `H_(2)` in 0.5 L flask exerts pressure of 24.63 atm at 300 K
C. 1 mole `NH_(3)` at 300 K and 1 atm occupies volume 22.4 L
D. 5.6 L of `CO_(2)` at 1 atm & 273 K is equal to 11 g

Explanation

To determine which gas shows real behavior, we will analyze the compressibility factor (Z) for each gas using the formula:

\[ Z = \frac{PV}{nRT} \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = ideal gas constant (0.0821 L·atm/(K·mol))
- \( T \) = temperature in Kelvin

For an ideal gas, \( Z = 1 \). If \( Z \neq 1 \), the gas shows real behavior.

### Step-by-Step Solution:

1. **Analyze the first case: 16 grams of O2 at 1 atm and 273 K occupying 11.2 L.**
- Calculate the number of moles (\( n \)):
\[ n = \frac{16 \text{ g}}{32 \text{ g/mol}} = 0.5 \text{ moles} \]
- Substitute into the compressibility factor formula:
\[ Z = \frac{PV}{nRT} = \frac{1 \times 11.2}{0.5 \times 0.0821 \times 273} \]
- Calculate \( Z \):
\[ Z = \frac{11.2}{11.215} \approx 1 \]
- **Conclusion:** This gas behaves ideally.

2. **Analyze the second case: 1 gram of H2 in a 0.5 L flask at 24.63 atm and 300 K.**
- Calculate the number of moles (\( n \)):
\[ n = \frac{1 \text{ g}}{2 \text{ g/mol}} = 0.5 \text{ moles} \]
- Substitute into the compressibility factor formula:
\[ Z = \frac{PV}{nRT} = \frac{24.63 \times 0.5}{0.5 \times 0.0821 \times 300} \]
- Calculate \( Z \):
\[ Z = \frac{12.315}{12.315} = 1 \]
- **Conclusion:** This gas behaves ideally.

3. **Analyze the third case: 1 mole of NH3 at 300 K and 1 atm occupying 22.4 L.**
- Substitute into the compressibility factor formula:
\[ Z = \frac{PV}{nRT} = \frac{1 \times 22.4}{1 \times 0.0821 \times 300} \]
- Calculate \( Z \):
\[ Z = \frac{22.4}{24.63} \approx 0.91 \]
- **Conclusion:** This gas shows real behavior (since \( Z < 1 \)).

4. **Analyze the fourth case: 5.6 L of CO2 at 1 atm and 273 K (11 g).**
- Calculate the number of moles (\( n \)):
\[ n = \frac{11 \text{ g}}{44 \text{ g/mol}} = 0.25 \text{ moles} \]
- Substitute into the compressibility factor formula:
\[ Z = \frac{PV}{nRT} = \frac{1 \times 5.6}{0.25 \times 0.0821 \times 273} \]
- Calculate \( Z \):
\[ Z = \frac{5.6}{5.607} \approx 1 \]
- **Conclusion:** This gas behaves ideally.

### Final Answer:
The gas that shows real behavior is **NH3 (ammonia)** from the third case.