A `V dm^(3)` flask contains gas `A` and another flask of `2V dm^(3)` contains gas `B` at the same temperature If density of gas `A` is `3.0gdm^(-3)` and of gas `B` is `1.5g dm^(-3)` and mo1 wt of `A =1//2` wt pf `B` then the ratio of pressure exerted by gases is .

Options

A. `(P_(A))/(P_(B))=2`
B. `(P_(A))/(P_(B))=1`
C. `(P_(A))/(P_(B))=4`
D. `(P_(A))/(P_(B))=3`

Explanation

To solve the problem step by step, we will use the ideal gas law and the relationships between density, molecular weight, and pressure.

### Step 1: Write down the known values
- Volume of gas A, \( V_A = V \, \text{dm}^3 \)
- Volume of gas B, \( V_B = 2V \, \text{dm}^3 \)
- Density of gas A, \( D_A = 3.0 \, \text{g/dm}^3 \)
- Density of gas B, \( D_B = 1.5 \, \text{g/dm}^3 \)
- Molecular weight of gas A, \( M_A = \frac{1}{2} M_B \)

### Step 2: Use the ideal gas equation
The ideal gas equation is given by:
\[ PV = nRT \]
where \( P \) is pressure, \( V \) is volume, \( n \) is the number of moles, \( R \) is the gas constant, and \( T \) is temperature.

### Step 3: Relate density to pressure
The number of moles \( n \) can be expressed in terms of mass and molecular weight:
\[ n = \frac{W}{M} \]
where \( W \) is the mass of the gas.

Using the definition of density \( D \):
\[ D = \frac{W}{V} \implies W = D \cdot V \]
Substituting this into the ideal gas equation gives:
\[ P = \frac{D \cdot V \cdot R \cdot T}{M} \]

### Step 4: Write the expressions for pressures of gases A and B
For gas A:
\[ P_A = \frac{D_A \cdot V_A \cdot R \cdot T}{M_A} \]
For gas B:
\[ P_B = \frac{D_B \cdot V_B \cdot R \cdot T}{M_B} \]

### Step 5: Substitute the known values
Substituting the known values into the equations:
\[ P_A = \frac{3.0 \cdot V \cdot R \cdot T}{M_A} \]
\[ P_B = \frac{1.5 \cdot 2V \cdot R \cdot T}{M_B} \]

### Step 6: Substitute \( M_A \) in terms of \( M_B \)
Since \( M_A = \frac{1}{2} M_B \):
\[ P_A = \frac{3.0 \cdot V \cdot R \cdot T}{\frac{1}{2} M_B} = \frac{6.0 \cdot V \cdot R \cdot T}{M_B} \]

### Step 7: Write the expression for \( P_B \)
Now substituting \( P_B \):
\[ P_B = \frac{1.5 \cdot 2V \cdot R \cdot T}{M_B} = \frac{3.0 \cdot V \cdot R \cdot T}{M_B} \]

### Step 8: Find the ratio of pressures
Now, we can find the ratio of pressures \( \frac{P_A}{P_B} \):
\[ \frac{P_A}{P_B} = \frac{\frac{6.0 \cdot V \cdot R \cdot T}{M_B}}{\frac{3.0 \cdot V \cdot R \cdot T}{M_B}} = \frac{6.0}{3.0} = 2 \]

### Step 9: Conclusion
The ratio of the pressures exerted by gases A and B is:
\[ \frac{P_A}{P_B} = 2 \]