Two non-reactive monoatomic ideal gases have their atomic masses in the ratio `2:3.` The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is `4:3.` The ratio of their densities is

Options

A. `1:4`
B. `1:2`
C. `6:9`
D. `8:9`

Explanation

To find the ratio of the densities of two non-reactive monoatomic ideal gases given their atomic masses and partial pressures, we can follow these steps:

### Step 1: Understand the given information
We have two gases with:
- Atomic masses in the ratio \( m_1 : m_2 = 2 : 3 \)
- Partial pressures in the ratio \( P_1 : P_2 = 4 : 3 \)

### Step 2: Relate atomic mass to molecular weight
Since the atomic mass of a monoatomic gas is numerically equal to its molecular weight, we can write:
\[ \frac{m_1}{m_2} = \frac{2}{3} \]

### Step 3: Use the ideal gas law
From the ideal gas law, we know:
\[ PV = nRT \]
If we express this in terms of density (\( \rho \)), we can rearrange it as:
\[ \rho = \frac{P \cdot m}{R \cdot T} \]
where \( m \) is the molecular weight of the gas.

### Step 4: Establish the relationship for densities
From the equation above, we can express the densities of the two gases as:
\[ \rho_1 = \frac{P_1 \cdot m_1}{R \cdot T} \quad \text{and} \quad \rho_2 = \frac{P_2 \cdot m_2}{R \cdot T} \]

### Step 5: Find the ratio of densities
Now, we can find the ratio of the densities:
\[ \frac{\rho_1}{\rho_2} = \frac{P_1 \cdot m_1}{P_2 \cdot m_2} \]

### Step 6: Substitute the known ratios
Substituting the known ratios:
- \( \frac{P_1}{P_2} = \frac{4}{3} \)
- \( \frac{m_1}{m_2} = \frac{2}{3} \)

We can substitute these into our density ratio:
\[ \frac{\rho_1}{\rho_2} = \frac{4/3 \cdot 2/3}{1} = \frac{8}{9} \]

### Step 7: Conclusion
Thus, the ratio of the densities of the two gases is:
\[ \frac{\rho_1}{\rho_2} = \frac{8}{9} \]

### Final Answer
The ratio of their densities is \( 8 : 9 \).

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