The density of gas `A` is twice that of B at the same temperature the molecular weight of gas B is twice that of A. The ratio of pressure of gas A and B will be :

Options

A. `1:6`
B. `1:1`
C. `4:1`
D. `1:4`

Explanation

To solve the problem, we need to find the ratio of the pressures of gas A and gas B given the conditions about their densities and molecular weights. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
- The density of gas A (D_A) is twice that of gas B (D_B):
\[ D_A = 2D_B \]
- The molecular weight of gas B (M_B) is twice that of gas A (M_A):
\[ M_B = 2M_A \]
- Both gases are at the same temperature (T).

### Step 2: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- P is the pressure,
- V is the volume,
- n is the number of moles,
- R is the gas constant,
- T is the temperature.

### Step 3: Relate Density and Moles
We can express the number of moles (n) in terms of mass (m) and molecular weight (M):
\[ n = \frac{m}{M} \]
Also, we know that volume (V) can be expressed in terms of density (D):
\[ V = \frac{m}{D} \]

### Step 4: Substitute in the Ideal Gas Law
Substituting for n and V in the ideal gas law gives:
\[ P \left(\frac{m}{D}\right) = \left(\frac{m}{M}\right)RT \]
This simplifies to:
\[ P = \frac{mRT}{V} = \frac{mRT}{m/D} = \frac{D \cdot RT}{M} \]
Thus, we have:
\[ P = \frac{D \cdot R \cdot T}{M} \]

### Step 5: Set Up the Ratio of Pressures
For gases A and B, we can write:
\[ P_A = \frac{D_A \cdot R \cdot T}{M_A} \]
\[ P_B = \frac{D_B \cdot R \cdot T}{M_B} \]

### Step 6: Find the Ratio \(\frac{P_A}{P_B}\)
Now, substituting the expressions for \(P_A\) and \(P_B\):
\[ \frac{P_A}{P_B} = \frac{\frac{D_A \cdot R \cdot T}{M_A}}{\frac{D_B \cdot R \cdot T}{M_B}} = \frac{D_A \cdot M_B}{D_B \cdot M_A} \]
Since \(R\) and \(T\) are constants, they cancel out.

### Step 7: Substitute the Known Values
Using the relationships established:
- \(D_A = 2D_B\)
- \(M_B = 2M_A\)

Substituting these into the ratio:
\[ \frac{P_A}{P_B} = \frac{(2D_B) \cdot (2M_A)}{D_B \cdot M_A} \]
This simplifies to:
\[ \frac{P_A}{P_B} = \frac{4D_B \cdot M_A}{D_B \cdot M_A} = 4 \]

### Final Answer
Thus, the ratio of the pressures of gas A to gas B is:
\[ \frac{P_A}{P_B} = 4:1 \]