A container X has volume double that of container Y and both are connected by a thin tube. Both contains same ideal gas. The temperature os X is 200K and that of Y is 400K. If mass of gas in X is m then in Y it will be.

Options

A. m/8
B. m/6
C. m/4
D. m/2

Answer

To solve the problem, we will use the ideal gas law and the relationship between pressure, volume, and temperature for the two containers.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Volume of container X (Vx) = 2 * Volume of container Y (Vy)
- Temperature of container X (Tx) = 200 K
- Temperature of container Y (Ty) = 400 K
- Mass of gas in container X (Mx) = m
- We need to find the mass of gas in container Y (My).

2. **Use the Ideal Gas Law:**
The pressure in both containers is the same since they are connected by a thin tube. The ideal gas law can be expressed as:
\[ P = \frac{nRT}{V} \]
Where:
- P = Pressure
- n = Number of moles
- R = Universal gas constant
- T = Temperature
- V = Volume

3. **Set Up the Pressure Equations:**
For container X:
\[ P_x = \frac{n_x R T_x}{V_x} \]
For container Y:
\[ P_y = \frac{n_y R T_y}{V_y} \]
Since \(P_x = P_y\), we can equate these two equations:
\[ \frac{n_x R T_x}{V_x} = \frac{n_y R T_y}{V_y} \]

4. **Cancel R from Both Sides:**
\[ \frac{n_x T_x}{V_x} = \frac{n_y T_y}{V_y} \]

5. **Substitute Known Values:**
Substitute \(V_x = 2V_y\):
\[ \frac{n_x T_x}{2V_y} = \frac{n_y T_y}{V_y} \]
Now, we can cancel \(V_y\) from both sides:
\[ \frac{n_x T_x}{2} = n_y T_y \]

6. **Express Number of Moles in Terms of Mass:**
The number of moles \(n\) can be expressed as:
\[ n_x = \frac{M_x}{M_m} \quad \text{and} \quad n_y = \frac{M_y}{M_m} \]
where \(M_m\) is the molar mass (constant for both gases since they are the same).

7. **Substituting Moles into the Equation:**
Substitute \(n_x\) and \(n_y\) into the equation:
\[ \frac{M_x}{M_m} \cdot T_x / 2 = \frac{M_y}{M_m} \cdot T_y \]
Cancel \(M_m\):
\[ \frac{M_x T_x}{2} = M_y T_y \]

8. **Rearranging to Find Mass in Container Y:**
\[ M_y = \frac{M_x T_x}{2 T_y} \]

9. **Substituting the Known Values:**
Substitute \(M_x = m\), \(T_x = 200 K\), and \(T_y = 400 K\):
\[ M_y = \frac{m \cdot 200}{2 \cdot 400} \]
Simplifying this gives:
\[ M_y = \frac{m \cdot 200}{800} = \frac{m}{4} \]

### Final Answer:
The mass of gas in container Y is \(M_y = \frac{m}{4}\).