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 solve the problem step by step, we will use the ideal gas equation and the relationships between pressure, density, and molar mass.

### Step 1: Understand the Ideal Gas Law
The ideal gas law is given by the equation:
\[ PV = nRT \]
Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = temperature

### Step 2: Relate Density to the Ideal Gas Law
We can express the number of moles \( n \) in terms of mass and molar mass:
\[ n = \frac{m}{M} \]
Where:
- \( m \) = mass of the gas
- \( M \) = molar mass of the gas

Substituting this into the ideal gas law gives:
\[ PV = \frac{m}{M} RT \]

Rearranging this, we can express pressure \( P \) in terms of density \( D \):
\[ P = \frac{m}{V} \cdot \frac{RT}{M} = D \cdot \frac{RT}{M} \]
Where \( D = \frac{m}{V} \) is the density of the gas.

Thus, we have:
\[ PM = DRT \]

### Step 3: Set Up the Equations for Two Gases
Let’s denote the two gases as Gas 1 and Gas 2. For Gas 1:
- Pressure = \( P_1 \)
- Molar mass = \( M_1 \)
- Density = \( D_1 \)

For Gas 2:
- Pressure = \( P_2 \)
- Molar mass = \( M_2 \)
- Density = \( D_2 \)

From the ideal gas law, we can write:
\[ P_1 M_1 = D_1 RT \quad \text{(1)} \]
\[ P_2 M_2 = D_2 RT \quad \text{(2)} \]

### Step 4: Divide the Two Equations
Dividing equation (1) by equation (2):
\[ \frac{P_1 M_1}{P_2 M_2} = \frac{D_1}{D_2} \]

### Step 5: Substitute the Given Ratios
From the problem, we know:
- The ratio of the atomic masses (molar masses) is \( \frac{M_1}{M_2} = \frac{2}{3} \)
- The ratio of the partial pressures is \( \frac{P_1}{P_2} = \frac{4}{3} \)

Substituting these ratios into the equation gives:
\[ \frac{4/3 \cdot 2/3}{1} = \frac{D_1}{D_2} \]

### Step 6: Calculate the Ratio of Densities
Calculating the left side:
\[ \frac{4 \cdot 2}{3 \cdot 3} = \frac{8}{9} \]

Thus, we have:
\[ \frac{D_1}{D_2} = \frac{8}{9} \]

### Conclusion
The ratio of the densities of the two gases is:
\[ D_1 : D_2 = 8 : 9 \]