In two containers X and Y same gas is filled. If the pressure, volume and absolute temperature of gas in X are three times as compared to that in Y and if the mass of X is `mg`, the mass of Y is

Options

A. `mg`
B. `m//3 g`
C. `m//2g`
D. `2 mg`

Explanation

To solve the problem, we will use the ideal gas equation and the relationships given in the question. Let's break it down step by step.

### Step 1: Write down the ideal gas equation
The ideal gas equation is given by:
\[ PV = nRT \]
where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = absolute temperature

### Step 2: Relate moles to mass
The number of moles \( n \) can also be expressed as:
\[ n = \frac{m}{M} \]
where:
- \( m \) = mass of the gas
- \( M \) = molar mass of the gas

### Step 3: Set up equations for containers X and Y
For container X:
- Pressure: \( P_x = 3P_y \)
- Volume: \( V_x = 3V_y \)
- Temperature: \( T_x = 3T_y \)
- Mass: \( m_x = mg \)

For container Y, we need to find the mass \( m_y \).

### Step 4: Write the ideal gas equation for both containers
For container X:
\[ P_x V_x = n_x R T_x \]
Substituting the values:
\[ (3P_y)(3V_y) = \frac{m_x}{M} R (3T_y) \]
This simplifies to:
\[ 9P_y V_y = \frac{mg}{M} R (3T_y) \]

For container Y:
\[ P_y V_y = n_y R T_y \]
Substituting the values:
\[ P_y V_y = \frac{m_y}{M} R T_y \]

### Step 5: Divide the equations
Now, we will divide the equation for container X by the equation for container Y:
\[ \frac{9P_y V_y}{P_y V_y} = \frac{\frac{mg}{M} R (3T_y)}{\frac{m_y}{M} R T_y} \]
This simplifies to:
\[ 9 = \frac{mg \cdot 3}{m_y} \]

### Step 6: Solve for \( m_y \)
Rearranging the equation gives:
\[ m_y = \frac{mg \cdot 3}{9} \]
This simplifies to:
\[ m_y = \frac{mg}{3} \]

### Final Answer
The mass of gas in container Y is:
\[ m_y = \frac{mg}{3} \]