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 `H_(2)` at `320 K`.

Options

A. `4.0`
B. `4.05`
C. `5.05`
D. `1.05`

Explanation

To solve the problem step by step, we will follow the information provided and apply the gas laws accordingly.

### Step 1: Understand the initial conditions in vessel X
- Total pressure in vessel X = 2 atm at 300 K.
- The vessel contains water (H₂O) and nitrogen gas (N₂).

### Step 2: Determine the initial pressure of nitrogen gas (N₂)
- The problem states that after transferring the contents to vessel Y (which has half the capacity), the pressure of nitrogen gas at 300 K is 3.8 atm.
- Since the total pressure in vessel X was 2 atm, we can infer that the pressure of water vapor in vessel X must be:

\[ P_{H_2O} = P_{total} - P_{N_2} = 2 \, \text{atm} - P_{N_2} \]

### Step 3: Calculate the pressure of nitrogen gas (N₂) in vessel Y at 300 K
- Given that the pressure of N₂ in vessel Y at 300 K is 3.8 atm, we can use this value directly.

### Step 4: Calculate the pressure of nitrogen gas (N₂) at 320 K
- Using the ideal gas law, we know that pressure is directly proportional to temperature (when volume is constant). Therefore, we can calculate the new pressure of N₂ at 320 K using the formula:

\[ P_{N_2, 320} = P_{N_2, 300} \times \frac{T_{320}}{T_{300}} = 3.8 \, \text{atm} \times \frac{320 \, \text{K}}{300 \, \text{K}} \]

- Performing the calculation:

\[ P_{N_2, 320} = 3.8 \times \frac{320}{300} = 3.8 \times 1.0667 \approx 4.05 \, \text{atm} \]

### Step 5: Determine the total pressure in vessel Y at 320 K
- The total pressure observed in vessel Y at 320 K is given as 4.32 atm.

### Step 6: Calculate the pressure of water vapor (H₂O) at 320 K
- The total pressure in vessel Y is the sum of the partial pressures of nitrogen and water vapor:

\[ P_{total} = P_{N_2} + P_{H_2O} \]

- Rearranging gives us:

\[ P_{H_2O} = P_{total} - P_{N_2} \]

- 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 (H₂O) at 320 K is approximately **0.27 atm**.