A certain vessel `X` has water and nitrogen gas at a total pressure of 2 atm and `300 K`. All the contents of vessel and transferred to another vessel `Y` having half the capacity of the vessel `X`. The pressure of `N_(2)` in this vessel was 3.8 atm at `300 K`. The vessel `Y` is heated to `320 K` and the total pressure observed was 4.32 atm. Assume that the volume occupied by the gases in vessel is equal to the volume of the vessel. Calculate the following:
Enthalpy of vapourization.

Options

A. `30.00`
B. `35.65`
C. `38.65`
D. `39.65`

Explanation

To solve the problem, we will follow the steps outlined in the video transcript and derive the enthalpy of vaporization step by step.

### Step 1: Gather Given Information
- Total pressure in vessel X, \( P_{total} = 2 \, \text{atm} \) at \( T = 300 \, \text{K} \)
- Vessel Y has half the capacity of vessel X, so \( V_Y = \frac{V_X}{2} \)
- Pressure of nitrogen in vessel Y at \( T = 300 \, \text{K} \), \( P_{N_2} = 3.8 \, \text{atm} \)
- Total pressure in vessel Y when heated to \( 320 \, \text{K} \), \( P_{total} = 4.32 \, \text{atm} \)

### Step 2: Calculate Pressure of Nitrogen in Vessel X
Using the relationship between the pressures in vessels X and Y:
\[ P_{Y} = 2 \times P_{X} \]
We can find the pressure of nitrogen in vessel X:
\[ P_{N_2, X} = \frac{P_{N_2, Y}}{2} = \frac{3.8 \, \text{atm}}{2} = 1.9 \, \text{atm} \]

### Step 3: Calculate Pressure of Water in Vessel X
Using the total pressure in vessel X:
\[ P_{total, X} = P_{N_2, X} + P_{H_2O, X} \]
Substituting the known values:
\[ 2 \, \text{atm} = 1.9 \, \text{atm} + P_{H_2O, X} \]
Solving for \( P_{H_2O, X} \):
\[ P_{H_2O, X} = 2 - 1.9 = 0.1 \, \text{atm} \]

### Step 4: Calculate Pressure of Nitrogen in Vessel Y at 320 K
Using the ideal gas law and the relationship between pressures and temperatures, we can find the pressure of nitrogen at 320 K:
\[ P_{N_2, Y, 320} = P_{N_2, Y, 300} \times \frac{320}{300} = 3.8 \, \text{atm} \times \frac{320}{300} = 4.05 \, \text{atm} \]

### Step 5: Calculate Pressure of Water in Vessel Y at 320 K
Using the total pressure in vessel Y at 320 K:
\[ P_{total, Y} = P_{N_2, Y, 320} + P_{H_2O, Y, 320} \]
Substituting the known values:
\[ 4.32 \, \text{atm} = 4.05 \, \text{atm} + P_{H_2O, Y, 320} \]
Solving for \( P_{H_2O, Y, 320} \):
\[ P_{H_2O, Y, 320} = 4.32 - 4.05 = 0.27 \, \text{atm} \]

### Step 6: Use the Clausius-Clapeyron Equation to Find Enthalpy of Vaporization
The Clausius-Clapeyron equation is given by:
\[ \log \left( \frac{P_1}{P_2} \right) = \frac{\Delta H_{vap}}{2.303 R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Where:
- \( P_1 = P_{H_2O, Y, 320} = 0.27 \, \text{atm} \)
- \( P_2 = P_{H_2O, X, 300} = 0.1 \, \text{atm} \)
- \( T_1 = 320 \, \text{K} \)
- \( T_2 = 300 \, \text{K} \)
- \( R = 8.314 \, \text{J/mol K} \)

Substituting the values:
\[ \log \left( \frac{0.27}{0.1} \right) = \frac{\Delta H_{vap}}{2.303 \times 8.314} \left( \frac{1}{300} - \frac{1}{320} \right) \]

Calculating \( \log \left( \frac{0.27}{0.1} \right) \):
\[ \log(2.7) \approx 0.431 \]

Calculating \( \left( \frac{1}{300} - \frac{1}{320} \right) \):
\[ \frac{1}{300} - \frac{1}{320} = \frac{320 - 300}{300 \times 320} = \frac{20}{96000} = \frac{1}{4800} \]

### Step 7: Solve for \( \Delta H_{vap} \)
Substituting back into the equation:
\[ 0.431 = \frac{\Delta H_{vap}}{2.303 \times 8.314} \times \frac{1}{4800} \]
Rearranging gives:
\[ \Delta H_{vap} = 0.431 \times 2.303 \times 8.314 \times 4800 \]
Calculating this results in:
\[ \Delta H_{vap} \approx 39596.83 \, \text{J/mol} \approx 39.6 \, \text{kJ/mol} \]

### Final Answer
The enthalpy of vaporization is approximately \( 39.6 \, \text{kJ/mol} \).