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 need to analyze the conditions given for the two containers X and Y filled with the same gas. We will use the Ideal Gas Law, which states that:

\[ PV = nRT \]

Where:
- \( P \) = pressure
- \( V \) = volume
- \( n \) = number of moles
- \( R \) = universal gas constant
- \( T \) = absolute temperature

### Step-by-Step Solution:

1. **Identify the conditions for containers X and Y**:
- For container X:
- Pressure \( P_X = 3P_Y \)
- Volume \( V_X = 3V_Y \)
- Temperature \( T_X = 3T_Y \)
- For container Y:
- Pressure \( P_Y = P \)
- Volume \( V_Y = V \)
- Temperature \( T_Y = T \)

2. **Apply the Ideal Gas Law for container X**:
\[ P_X V_X = n_X R T_X \]
Substituting the values:
\[ (3P)(3V) = n_X R (3T) \]
Simplifying:
\[ 9PV = 3n_X RT \]
Dividing both sides by 3:
\[ 3PV = n_X RT \quad \text{(Equation 1)} \]

3. **Apply the Ideal Gas Law for container Y**:
\[ P_Y V_Y = n_Y R T_Y \]
Substituting the values:
\[ P V = n_Y R T \quad \text{(Equation 2)} \]

4. **Relate the number of moles to mass**:
The number of moles \( n \) can be expressed in terms of mass \( m \) and molar mass \( M \):
\[ n = \frac{m}{M} \]
For container X:
\[ n_X = \frac{m}{M} \quad \text{(mass of gas in X is } m \text{ grams)} \]
For container Y:
\[ n_Y = \frac{m_Y}{M} \quad \text{(mass of gas in Y is } m_Y \text{ grams)} \]

5. **Substituting the expressions for \( n_X \) and \( n_Y \) into Equations 1 and 2**:
- From Equation 1:
\[ 3PV = \frac{m}{M} RT \]
- From Equation 2:
\[ PV = \frac{m_Y}{M} RT \]

6. **Equate the two equations**:
Since both equations equal \( PV \):
\[ 3PV = \frac{m}{M} RT \quad \text{and} \quad PV = \frac{m_Y}{M} RT \]
Dividing the first equation by the second:
\[ \frac{3PV}{PV} = \frac{m}{m_Y} \]
This simplifies to:
\[ 3 = \frac{m}{m_Y} \]
Rearranging gives:
\[ m_Y = \frac{m}{3} \]

### Final Answer:
The mass of gas in container Y is \( \frac{m}{3} \) grams.