If 10^-4 L of water is introduced into a 1L flask at 300K, how many moles of water are in the vapour phase when equilibrium is established ?

Vapour pressure of H₂O = 3170 Pa

R = 8.314 JK^-1mol^-1

To determine how many moles of water are in the vapor phase when equilibrium is established in a 1L flask at 300K, we can use the concept of vapor pressure and the ideal gas law. Let's break this down step by step.

Understanding Vapor Pressure

The vapor pressure of a substance is the pressure exerted by its vapor when it is in equilibrium with its liquid phase at a given temperature. For water at 300K, the vapor pressure is given as 3170 Pa. This means that at equilibrium, the water vapor will exert a pressure of 3170 Pa in the flask.

Using the Ideal Gas Law

We can apply the ideal gas law, which is expressed as:

PV = nRT

Where:

• P = pressure (in Pascals)
• V = volume (in cubic meters)
• n = number of moles
• R = ideal gas constant (8.314 J/(K·mol))
• T = temperature (in Kelvin)

Calculating the Number of Moles

First, we need to convert the volume of the flask from liters to cubic meters:

1 L = 0.001 m³

Now, we can rearrange the ideal gas law to solve for the number of moles (n):

n = PV / RT

Substituting the Values

Now, we can substitute the values into the equation:

• P = 3170 Pa
• V = 0.001 m³
• R = 8.314 J/(K·mol)
• T = 300 K

Plugging these into the equation gives:

n = (3170 Pa) * (0.001 m³) / (8.314 J/(K·mol) * 300 K)

Performing the Calculation

Calculating the numerator:

3170 Pa * 0.001 m³ = 3.17 Pa·m³

Now calculating the denominator:

8.314 J/(K·mol) * 300 K = 2494.2 J/mol

Now, we can find n:

n = 3.17 / 2494.2 ≈ 0.00127 moles

Final Result

At equilibrium, the number of moles of water in the vapor phase in the 1L flask at 300K is approximately 0.00127 moles. This calculation illustrates how vapor pressure and the ideal gas law work together to determine the amount of substance in the vapor phase at equilibrium.