Two separate bulbs contain ideal gas `A` and `B`. The density of a gas `A` is twice that of a gas `B`. The molecular mass of `A` is half that of gas `B`. The two gases are at the same temperature. The ratio of the pressure of `A` to that gas `B` is

Options

A. 2
B. `1//2`
C. 4
D. `1//4`

Explanation

To solve the problem, we need to find the ratio of the pressures of two ideal gases A and B, given their densities and molecular masses.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The density of gas A (DA) is twice that of gas B (DB):
\[ D_A = 2D_B \]
- The molecular mass of gas A (MA) is half that of gas B (MB):
\[ M_A = \frac{1}{2}M_B \]
- Both gases are at the same temperature (T).

2. **Using the Ideal Gas Law**:
The density of an ideal gas can be expressed using the formula:
\[ D = \frac{PM}{RT} \]
where P is the pressure, M is the molar mass, R is the ideal gas constant, and T is the temperature.

3. **Setting Up the Ratios**:
For gases A and B, we can write:
\[ D_A = \frac{P_A M_A}{RT} \quad \text{and} \quad D_B = \frac{P_B M_B}{RT} \]

4. **Taking the Ratio of Densities**:
From the densities, we have:
\[ \frac{D_A}{D_B} = \frac{P_A M_A}{P_B M_B} \]
Substituting the values we know:
\[ \frac{2D_B}{D_B} = \frac{P_A \left(\frac{1}{2}M_B\right)}{P_B M_B} \]
Simplifying gives:
\[ 2 = \frac{P_A \cdot \frac{1}{2} M_B}{P_B M_B} \]

5. **Canceling Molar Mass**:
We can cancel \(M_B\) from both sides:
\[ 2 = \frac{P_A \cdot \frac{1}{2}}{P_B} \]
Rearranging gives:
\[ 2P_B = \frac{1}{2}P_A \]

6. **Finding the Pressure Ratio**:
Multiplying both sides by 2:
\[ 4P_B = P_A \]
Therefore, the ratio of the pressures is:
\[ \frac{P_A}{P_B} = 4 \]

### Final Answer:
The ratio of the pressure of gas A to that of gas B is:
\[ \frac{P_A}{P_B} = 4 \]