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:
Pressure of water vapour at `320 K`.

Options

A. `0.27`
B. `0.32`
C. `4.0`
D. `1.0`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the initial conditions
We have a vessel \( X \) containing water and nitrogen gas at a total pressure of 2 atm at a temperature of 300 K. The pressure of nitrogen gas in vessel \( Y \) at 300 K is given as 3.8 atm.

### Step 2: Calculate the pressure of nitrogen gas in vessel \( Y \)
Since the volume of vessel \( Y \) is half that of vessel \( X \), when the contents are transferred, the pressure of nitrogen gas will change due to the change in volume.

Using the ideal gas law, we know that pressure is directly proportional to temperature when the volume is constant. Thus, we can use the following relation:

\[ P_{N_2} \propto T \]

At 300 K, the pressure of nitrogen is 3.8 atm. We can find the pressure of nitrogen at 320 K using the formula:

\[ P'_{N_2} = P_{N_2} \times \frac{T'}{T} \]

Substituting the values:

\[ P'_{N_2} = 3.8 \, \text{atm} \times \frac{320 \, \text{K}}{300 \, \text{K}} = 3.8 \, \text{atm} \times \frac{32}{30} = 4.05 \, \text{atm} \]

### Step 3: Calculate the total pressure in vessel \( Y \) at 320 K
The total pressure in vessel \( Y \) at 320 K is given as 4.32 atm.

### Step 4: Calculate the pressure of water vapor
The total pressure in vessel \( Y \) is the sum of the partial pressures of nitrogen gas and water vapor:

\[ P_{total} = P'_{N_2} + P_{H_2O} \]

Rearranging the equation to find the pressure of water vapor:

\[ P_{H_2O} = P_{total} - P'_{N_2} \]

Substituting the known values:

\[ P_{H_2O} = 4.32 \, \text{atm} - 4.05 \, \text{atm} = 0.27 \, \text{atm} \]

### Final Answer
The pressure of water vapor at 320 K is **0.27 atm**.

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