If the ration of the masses of `SO_(3)` and `O_(2)` gases confined in a vessel is `1 : 1` , then the ratio of their partial pressure would be

Options

A. `5 : 2`
B. `2 : 5`
C. `2 : 1`
D. `1 : 2`

Explanation

To solve the problem of finding the ratio of the partial pressures of \( SO_3 \) and \( O_2 \) gases when their masses are in the ratio of \( 1:1 \), we can follow these steps:

### Step 1: Define the Masses
Let the mass of \( SO_3 \) be \( m \) and the mass of \( O_2 \) also be \( m \) (since the ratio is \( 1:1 \)).

### Step 2: Calculate the Number of Moles
To find the number of moles of each gas, we use the formula:
\[ \text{Number of moles} = \frac{\text{Given mass}}{\text{Molar mass}} \]

- For \( SO_3 \):
\[ \text{Molar mass of } SO_3 = 32 + (3 \times 16) = 80 \, \text{g/mol} \]
\[ \text{Number of moles of } SO_3 = \frac{m}{80} \]

- For \( O_2 \):
\[ \text{Molar mass of } O_2 = 2 \times 16 = 32 \, \text{g/mol} \]
\[ \text{Number of moles of } O_2 = \frac{m}{32} \]

### Step 3: Calculate the Total Number of Moles
The total number of moles in the vessel is:
\[ \text{Total moles} = \text{Moles of } SO_3 + \text{Moles of } O_2 = \frac{m}{80} + \frac{m}{32} \]

To add these fractions, we need a common denominator:
\[ \text{Common denominator} = 80 \times 32 = 2560 \]
Thus,
\[ \frac{m}{80} = \frac{32m}{2560}, \quad \frac{m}{32} = \frac{80m}{2560} \]
So,
\[ \text{Total moles} = \frac{32m + 80m}{2560} = \frac{112m}{2560} \]

### Step 4: Calculate the Mole Fractions
The mole fraction of each gas is given by:
\[ \text{Mole fraction of } SO_3 = \frac{\text{Moles of } SO_3}{\text{Total moles}} = \frac{\frac{m}{80}}{\frac{112m}{2560}} = \frac{2560}{80 \times 112} = \frac{32}{112} = \frac{2}{7} \]

\[ \text{Mole fraction of } O_2 = \frac{\text{Moles of } O_2}{\text{Total moles}} = \frac{\frac{m}{32}}{\frac{112m}{2560}} = \frac{2560}{32 \times 112} = \frac{80}{112} = \frac{5}{7} \]

### Step 5: Calculate the Partial Pressures
According to Dalton's Law of Partial Pressures, the partial pressure is directly proportional to the mole fraction:
\[ P_{SO_3} \propto \text{Mole fraction of } SO_3, \quad P_{O_2} \propto \text{Mole fraction of } O_2 \]

### Step 6: Find the Ratio of Partial Pressures
The ratio of the partial pressures is:
\[ \frac{P_{SO_3}}{P_{O_2}} = \frac{\text{Mole fraction of } SO_3}{\text{Mole fraction of } O_2} = \frac{\frac{2}{7}}{\frac{5}{7}} = \frac{2}{5} \]

### Final Answer
Thus, the ratio of the partial pressures of \( SO_3 \) to \( O_2 \) is \( \frac{2}{5} \).

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