Pressure of `1g` of an ideal gas `A` at `27^(@)C` is found to be 2 bar when `2g` of another ideal gas `B` is introduced in the same flask at same temperature the pressure becomes 3 bar. Find a relationship thieir molecular masses .

Options

A. `4:1`
B. `1:4`
C. `1:8`
D. `2:8`

Explanation

To solve the problem, we need to find the relationship between the molecular masses of two ideal gases, A and B, based on the given pressures and masses. Let's break it down step by step.

### Step 1: Gather the Given Information
- Mass of gas A (m_A) = 1 g
- Pressure of gas A (P_A) = 2 bar
- Mass of gas B (m_B) = 2 g
- Total pressure after introducing gas B (P_total) = 3 bar

### Step 2: Determine the Pressure of Gas B
Since the total pressure after introducing gas B is 3 bar, we can find the pressure exerted by gas B (P_B) using the equation:
\[ P_A + P_B = P_{total} \]
Substituting the known values:
\[ 2 \text{ bar} + P_B = 3 \text{ bar} \]
Thus,
\[ P_B = 3 \text{ bar} - 2 \text{ bar} = 1 \text{ bar} \]

### Step 3: Use the Ideal Gas Law
The ideal gas law is given by:
\[ PV = nRT \]
Where:
- P = pressure
- V = volume
- n = number of moles
- R = ideal gas constant
- T = temperature in Kelvin

Since the volume and temperature are constant for both gases, we can write the equations for both gases A and B as follows:

For gas A:
\[ P_A V = n_A R T \]

For gas B:
\[ P_B V = n_B R T \]

### Step 4: Relate the Number of Moles to Mass and Molar Mass
The number of moles (n) can be expressed in terms of mass (m) and molar mass (M):
\[ n = \frac{m}{M} \]

Thus, we can rewrite the equations for gases A and B:
For gas A:
\[ P_A V = \frac{m_A}{M_A} R T \]

For gas B:
\[ P_B V = \frac{m_B}{M_B} R T \]

### Step 5: Divide the Two Equations
Dividing the equation for gas A by the equation for gas B:
\[ \frac{P_A V}{P_B V} = \frac{\frac{m_A}{M_A} R T}{\frac{m_B}{M_B} R T} \]

Since \( V \), \( R \), and \( T \) are constant and cancel out, we have:
\[ \frac{P_A}{P_B} = \frac{m_A / M_A}{m_B / M_B} \]

### Step 6: Substitute the Known Values
Substituting the known values:
\[ \frac{2 \text{ bar}}{1 \text{ bar}} = \frac{1 \text{ g} / M_A}{2 \text{ g} / M_B} \]

This simplifies to:
\[ 2 = \frac{1}{M_A} \cdot \frac{M_B}{2} \]

### Step 7: Rearranging the Equation
Rearranging gives us:
\[ 2 = \frac{M_B}{2M_A} \]
Multiplying both sides by \( 2M_A \):
\[ 4M_A = M_B \]

### Step 8: Relationship Between Molecular Masses
Thus, we find the relationship between the molecular masses:
\[ M_B = 4M_A \]

### Conclusion
The relationship between the molecular masses of gases A and B is:
\[ M_A : M_B = 1 : 4 \]