How many moles of water are in vapour phase present inside the vessel containing `1L` water after sufficient time? (Vapour pressure of water at `27^(@)C=3000Pa,R=(25)/(3)J//mol-K`)

Options

A. `5xx10^(-4)`
B. `120`
C. `1.2xx10^(-3)`
D. None of these

Explanation

To find the number of moles of water in the vapor phase present inside a vessel containing 1 L of water after sufficient time, we can use the ideal gas equation:

\[ PV = nRT \]

Where:
- \( P \) = vapor pressure of water (in Pascals)
- \( V \) = volume of the gas (in cubic meters)
- \( n \) = number of moles of gas
- \( R \) = ideal gas constant (in Joules per mole Kelvin)
- \( T \) = temperature (in Kelvin)

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Vapor pressure of water, \( P = 3000 \, \text{Pa} \)
- Volume of water, \( V = 1 \, \text{L} = 1 \times 10^{-3} \, \text{m}^3 \)
- Ideal gas constant, \( R = \frac{25}{3} \, \text{J/mol·K} \)
- Temperature, \( T = 27^\circ C = 27 + 273 = 300 \, \text{K} \)

2. **Substitute the Values into the Ideal Gas Equation**:
We need to rearrange the ideal gas equation to solve for \( n \):
\[ n = \frac{PV}{RT} \]

3. **Plug in the Values**:
\[ n = \frac{(3000 \, \text{Pa}) \times (1 \times 10^{-3} \, \text{m}^3)}{\left(\frac{25}{3} \, \text{J/mol·K}\right) \times (300 \, \text{K})} \]

4. **Calculate the Denominator**:
First, calculate \( R \times T \):
\[ R \times T = \frac{25}{3} \times 300 = 2500 \, \text{J/mol} \]

5. **Calculate \( n \)**:
Now substitute back into the equation:
\[ n = \frac{3000 \times 1 \times 10^{-3}}{2500} \]
\[ n = \frac{3}{2.5} = 1.2 \times 10^{-3} \, \text{mol} \]

6. **Final Result**:
The number of moles of water in the vapor phase is:
\[ n = 1.2 \times 10^{-3} \, \text{mol} \]