At `27^@` C, a closed vessel contains equal weights of helium (mol. wt. = 4 ), methane (mol. wt. =16) and sulphur dioxide (mol. wt = 64). The pressure exerted by the mixture is 210 mm. If the partial pressures of helium, methane and sulphur dioxide are `P_(1)`, `P_(2)` and `P_(3)` respectively, which one of the following is correct ?

Options

A. `P_(3) gt P_(2) gt P_(1)`
B. `P_(1) gt P_(2) gt P_(3)`
C. `P_(1) gt P_(3) gt P_(2)`
D. `P_(2) gt P_(3) gt P_(1)`

Answer

To solve the problem, we need to determine the partial pressures of helium, methane, and sulfur dioxide in a closed vessel containing equal weights of each gas at a given temperature. We will use the relationship between partial pressure, mole fraction, and molecular weight.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Temperature = 27°C
- Total pressure (P_total) = 210 mmHg
- Molecular weights:
- Helium (He) = 4 g/mol
- Methane (CH₄) = 16 g/mol
- Sulfur Dioxide (SO₂) = 64 g/mol
- Equal weights of each gas.

2. **Calculate the Number of Moles of Each Gas:**
- Let the weight of each gas be \( W \) grams.
- The number of moles \( n \) of each gas can be calculated using the formula:
\[ n = \frac{W}{\text{Molecular Weight}} \]
- For Helium:
\[ n_1 = \frac{W}{4} \]
- For Methane:
\[ n_2 = \frac{W}{16} \]
- For Sulfur Dioxide:
\[ n_3 = \frac{W}{64} \]

3. **Calculate the Total Number of Moles:**
- The total number of moles \( n_{total} \) is:
\[ n_{total} = n_1 + n_2 + n_3 = \frac{W}{4} + \frac{W}{16} + \frac{W}{64} \]
- To add these fractions, find a common denominator (64):
\[ n_{total} = \frac{16W}{64} + \frac{4W}{64} + \frac{W}{64} = \frac{21W}{64} \]

4. **Calculate the Mole Fractions:**
- Mole fraction \( X_i \) of each gas is given by:
\[ X_i = \frac{n_i}{n_{total}} \]
- For Helium:
\[ X_1 = \frac{n_1}{n_{total}} = \frac{\frac{W}{4}}{\frac{21W}{64}} = \frac{64}{84} = \frac{16}{21} \]
- For Methane:
\[ X_2 = \frac{n_2}{n_{total}} = \frac{\frac{W}{16}}{\frac{21W}{64}} = \frac{64}{336} = \frac{4}{21} \]
- For Sulfur Dioxide:
\[ X_3 = \frac{n_3}{n_{total}} = \frac{\frac{W}{64}}{\frac{21W}{64}} = \frac{1}{21} \]

5. **Calculate the Partial Pressures:**
- The partial pressure \( P_i \) is given by:
\[ P_i = X_i \times P_{total} \]
- For Helium:
\[ P_1 = X_1 \times 210 = \frac{16}{21} \times 210 = 160 \text{ mmHg} \]
- For Methane:
\[ P_2 = X_2 \times 210 = \frac{4}{21} \times 210 = 40 \text{ mmHg} \]
- For Sulfur Dioxide:
\[ P_3 = X_3 \times 210 = \frac{1}{21} \times 210 = 10 \text{ mmHg} \]

6. **Determine the Order of Partial Pressures:**
- From the calculated partial pressures:
- \( P_1 = 160 \text{ mmHg} \) (Helium)
- \( P_2 = 40 \text{ mmHg} \) (Methane)
- \( P_3 = 10 \text{ mmHg} \) (Sulfur Dioxide)
- The order of partial pressures is:
\[ P_1 > P_2 > P_3 \]

### Final Answer:
The correct order of partial pressures is \( P_1 > P_2 > P_3 \) (Helium > Methane > Sulfur Dioxide).

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