Two vessels connected by a valve of negligible volume. One container (I) has 2.8 g of `N_(2)` at temperature `T_(1) (K)`. The other container (II) is completely evacuated. The container (I) is heated to `T_(2)(K)`while container (II) is maintained at `T_(2)//3(K)`. volume of vessel (I) is half that of vessel (II). If the valve is opened then what is the mass ratio of `N_(2)` is both vessel `(W_(I)//W_(II))`?

Options

A. `1:2`
B. `1:3`
C. `1:6`
D. `3:1`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the number of moles of nitrogen in vessel I.
Given:
- Mass of nitrogen in vessel I, \( W_1 = 2.8 \, \text{g} \)
- Molar mass of nitrogen, \( M = 28 \, \text{g/mol} \)

Using the formula for moles:
\[ n = \frac{W}{M} \]
\[ n_1 = \frac{2.8 \, \text{g}}{28 \, \text{g/mol}} = 0.1 \, \text{mol} \]

### Step 2: Define the volumes of the vessels.
Let:
- Volume of vessel I, \( V_1 = V \)
- Volume of vessel II, \( V_2 = 2V \) (since vessel I is half the volume of vessel II)

### Step 3: Write the ideal gas equations for both vessels.
For vessel I:
\[ P_1 V_1 = n_1 R T_1 \quad \text{(before opening the valve)} \]
After the valve is opened, the number of moles in vessel I will be \( n_1 - x \):
\[ P_1 V_1 = (0.1 - x) R T_2 \]

For vessel II:
\[ P_2 V_2 = n_2 R T_2/3 \quad \text{(where \( n_2 = x \))} \]
\[ P_2 (2V) = x R \frac{T_2}{3} \]

### Step 4: Relate the pressures in both vessels.
From the ideal gas law, we can express the pressures:
\[ P_1 = \frac{(0.1 - x) R T_2}{V} \]
\[ P_2 = \frac{x R \frac{T_2}{3}}{2V} \]

### Step 5: Set the pressures equal after the valve is opened.
Since the pressure will equalize when the valve is opened:
\[ \frac{(0.1 - x) R T_2}{V} = \frac{x R \frac{T_2}{3}}{2V} \]

### Step 6: Cancel out common terms and solve for \( x \).
Cancel \( R \) and \( T_2 \) from both sides:
\[ (0.1 - x) = \frac{x}{6} \]
Multiply through by 6:
\[ 6(0.1 - x) = x \]
\[ 0.6 - 6x = x \]
Combine like terms:
\[ 0.6 = 7x \]
\[ x = \frac{0.6}{7} \approx 0.0857 \, \text{mol} \]

### Step 7: Calculate the mass of nitrogen in vessel II.
Using the number of moles transferred to vessel II:
\[ W_2 = x \times M = \frac{0.6}{7} \times 28 \approx 2.4 \, \text{g} \]

### Step 8: Calculate the mass of nitrogen remaining in vessel I.
The remaining moles in vessel I:
\[ n_1 - x = 0.1 - \frac{0.6}{7} = \frac{0.1 \times 7 - 0.6}{7} = \frac{0.7 - 0.6}{7} = \frac{0.1}{7} \]
Calculating the mass:
\[ W_1 = \left(\frac{0.1}{7}\right) \times 28 \approx 0.4 \, \text{g} \]

### Step 9: Calculate the mass ratio \( \frac{W_1}{W_2} \).
\[ \frac{W_1}{W_2} = \frac{0.4 \, \text{g}}{2.4 \, \text{g}} = \frac{1}{6} \]

### Final Answer:
The mass ratio of \( N_2 \) in both vessels \( \frac{W_1}{W_2} \) is \( \frac{1}{6} \).

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