The increasing order of effusion among the gases `H_(2), O_(2), NH_(3)` and `CO_(2)` is

Options

A. `H_2, CO_2, NH_3, O_2`
B. `H_2, NH_3, O_2, CO_2`
C. `H_2,O_2, NH_3 , CO_2`
D. `CO_2 , O_2 , NH_3 , H_2`

Answer

To determine the increasing order of effusion among the gases \( H_2 \), \( O_2 \), \( NH_3 \), and \( CO_2 \), we will use Graham's law of effusion. This law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

### Step-by-Step Solution:

1. **Identify the Molar Masses of the Gases:**
- \( H_2 \): The molar mass is \( 2 \, \text{g/mol} \).
- \( O_2 \): The molar mass is \( 16 \times 2 = 32 \, \text{g/mol} \).
- \( NH_3 \): The molar mass is \( 14 + (3 \times 1) = 17 \, \text{g/mol} \).
- \( CO_2 \): The molar mass is \( 12 + (16 \times 2) = 44 \, \text{g/mol} \).

2. **List the Molar Masses in Order:**
- The molar masses we have calculated are:
- \( H_2 = 2 \, \text{g/mol} \)
- \( NH_3 = 17 \, \text{g/mol} \)
- \( O_2 = 32 \, \text{g/mol} \)
- \( CO_2 = 44 \, \text{g/mol} \)

3. **Determine the Order of Molar Masses:**
- Arranging these from lowest to highest:
- \( H_2 < NH_3 < O_2 < CO_2 \)

4. **Apply Graham's Law:**
- According to Graham's law, the rate of effusion is inversely proportional to the square root of the molar mass. Therefore, the gas with the lowest molar mass will effuse the fastest, and the gas with the highest molar mass will effuse the slowest.

5. **Determine the Increasing Order of Effusion:**
- Since \( CO_2 \) has the highest molar mass and \( H_2 \) has the lowest, the increasing order of effusion will be:
- \( CO_2 < O_2 < NH_3 < H_2 \)

### Final Answer:
The increasing order of effusion among the gases \( H_2, O_2, NH_3, \) and \( CO_2 \) is:
\[ CO_2 < O_2 < NH_3 < H_2 \]