A `0.5 dm^3` flask contains gas A and 2 `dm^3` flask contains gas B at the same temperature. If density of A = `4 gdm^-3`, that of B = `2 gdm^-3` and molar mass of A is half of that of B, then the ratio of pressure exerted by gas is

Options

A. `P_A / P_B = 0.5`
B. `P_A / (P_B) = 2`
C. `P_A / (P_B) = 4`
D. `P_A / (P_B) = 0.25`

Answer

To find the ratio of pressure exerted by gas A and gas B, we can follow these steps:

### Step 1: Write down the given data
- Volume of gas A, \( V_A = 0.5 \, \text{dm}^3 \)
- Volume of gas B, \( V_B = 2 \, \text{dm}^3 \)
- Density of gas A, \( D_A = 4 \, \text{g/dm}^3 \)
- Density of gas B, \( D_B = 2 \, \text{g/dm}^3 \)
- Molar mass of gas A, \( M_A = \frac{1}{2} M_B \)
- Both gases are at the same temperature, \( T_A = T_B = T \)

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

### Step 3: Relate number of moles to density and molar mass
The number of moles \( n \) can be expressed as:
\[ n = \frac{m}{M} \]
Where \( m \) is the mass and \( M \) is the molar mass.

The mass can also be expressed in terms of density:
\[ m = D \times V \]
Thus,
\[ n = \frac{D \times V}{M} \]

### Step 4: Substitute into the ideal gas law
Substituting \( n \) into the ideal gas law gives:
\[ PV = \frac{D \times V}{M} RT \]
Rearranging gives:
\[ P = \frac{D \times R \times T}{M} \]

### Step 5: Calculate pressure for both gases
For gas A:
\[ P_A = \frac{D_A \times R \times T}{M_A} \]
For gas B:
\[ P_B = \frac{D_B \times R \times T}{M_B} \]

### Step 6: Find the ratio of pressures
Now, we can find the ratio of the pressures:
\[ \frac{P_A}{P_B} = \frac{\frac{D_A \times R \times T}{M_A}}{\frac{D_B \times R \times T}{M_B}} \]
This simplifies to:
\[ \frac{P_A}{P_B} = \frac{D_A \times M_B}{D_B \times M_A} \]

### Step 7: Substitute the known values
Substituting the known values:
- \( D_A = 4 \, \text{g/dm}^3 \)
- \( D_B = 2 \, \text{g/dm}^3 \)
- \( M_A = \frac{1}{2} M_B \)

Thus, we can write:
\[ \frac{P_A}{P_B} = \frac{4 \times M_B}{2 \times \frac{1}{2} M_B} \]
This simplifies to:
\[ \frac{P_A}{P_B} = \frac{4 \times M_B}{2 \times \frac{1}{2} M_B} = \frac{4}{1} = 4 \]

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