Problem 201 - A Complete Resource Book in Chemistry for JEE Main

Question

If \(10^{4} \mathrm{dm}^{3}\) of water is introduced into \(1.0 \mathrm{dm}^{3}\) flask at \(300 \mathrm{~K}\), how many moles of water are in the vapour phase when equilibrium is established? (Given: Vapour pressure of \(\mathrm{H}_{2} \mathrm{O}\) at \(300 \mathrm{~K}\) is \(3170 \mathrm{~Pa} ; \mathrm{R}=8.314 \mathrm{JK}^{-1}\) \(\left.\mathrm{mol}^{-1}\right)\) (a) \(5.56 \times 10^{-3} \mathrm{~mol}\) (b) \(1.53 \times 10^{-2} \mathrm{~mol}\) (c) \(4.46 \times 10^{-2} \mathrm{~mol}\) (d) \(1.27 \times 10^{-3} \mathrm{~mol}\)

Step-by-Step Solution

Answer

(d) 1.27 × 10^{-3} mol

Step 1: Determine the Volume Available for Vapour

Given the total capacity of the flask is 1.0 dm³, most of this volume remains available for the vapour, as the remaining liquid water volume is negligible compared to this capacity.

Step 2: Utilize Ideal Gas Law for Vapour

To find the number of moles of water in the vapour phase, apply the ideal gas law formula: \(PV = nRT\). Here, \(P = 3170 \, \text{Pa}\), \(V = 1.0 \, \text{dm}^3 = 1.0 \times 10^{-3} \, \text{m}^3\), \(R = 8.314 \, \text{JK}^{-1} \text{mol}^{-1}\), and \(T = 300 \, \text{K}\). Rearrange to solve for \(n\), the number of moles: \(n = \frac{PV}{RT}\).

Step 3: Substitute the Known Values

Substitute the known values into the equation: \(n = \frac{3170 \, \text{Pa} \times 1.0 \times 10^{-3} \, \text{m}^3}{8.314 \, \text{JK}^{-1}\text{mol}^{-1} \times 300 \, \text{K}}\).

Step 4: Calculate the Moles of Vapour

Perform the calculation: \(n \approx \frac{3170 \times 1.0 \times 10^{-3}}{8.314 \times 300}\). This simplifies to \(n \approx \frac{3.170}{2494.2}\).

Step 5: Determine the Final Solution

By calculating the above expression, we find \(n \approx 1.27 \times 10^{-3} \, \text{mol}\). The correct choice from the given options is (d) 1.27 \times 10^{-3} \, \text{mol}.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental relation in chemistry and physics, representing the behavior of an ideal gas in terms of its pressure, volume, temperature, and amount of substance in moles. This law is given by the equation:

• \[ PV = nRT \]

Where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume that the gas occupies.
• \( n \) is the number of moles of the gas.
• \( R \) is the universal gas constant ( 8.314 \( \text{JK}^{-1}\text{mol}^{-1}\)).
• \( T \) is the temperature in Kelvin.

This equation helps us comprehend how the four variables relate to one another in a medium where gases behave ideally.
In our exercise, this law facilitates the calculation of the moles of water in the vapour phase within a given flask. By rearranging the equation to solve for \( n \), students can compute the moles of vapour phase water accurately under the specified conditions.

Equilibrium

Equilibrium in a chemical context is when the rate of evaporation of the liquid equals the rate of condensation from the vapour phase. When such balance is achieved in a closed system, no net change in the amount of liquid or vapour is observed.
This dynamic state does not mean the molecules stop moving. Rather, the numbers of molecules converting from liquid to vapour and back are equal, keeping the system stable over time.

• In the given exercise, determining when equilibrium is reached helps establish when to measure the number of moles in the vapour phase.
• Water reaching equilibrium in the flask implies the steam generated equals the condensation, maintaining the consistent vapor pressure around 3170 Pa.

This understanding is crucial to accurately applying the Ideal Gas Law and ensuring accurate measurements.

Vapour Phase

The vapour phase refers to the gaseous state that a material may take when it's in equilibrium with its liquid form. In simpler terms, it's when a substance in the liquid state evaporates into gas.
Key attributes of the vapour phase include:

• Formation due to increased temperature or pressure reduction of the liquid.
• Presence when the substance boils or reaches saturation vapour pressure.

For example, in our problem, the vapour phase of water is in equilibrium with liquid water at a vapour pressure of 3170 Pa within a 1 liter flask.
This understanding allows us to utilize the Ideal Gas Law, as we are effectively treating the gaseous molecules of water as an ideal gas within the constraints of the flask's volume and temperature.

Moles of Water

Moles of water indicate the amount of water substance present, providing a link to chemical calculations. A mole is a unit representing \( 6.022 \times 10^{23} \) particles, such as atoms or molecules.

• In gas phase calculations, it quantifies the vapour amount.
• Using this unit, comparisons across chemical reactions and physical states become standardized.

In our scenario, computing the moles present in the vapour phase helps us understand how much water is present as gas within the flask.
To find the moles, once equilibrium is reached, we use the Ideal Gas Law by rearranging it as:

• \[ n = \frac{PV}{RT} \]

Substituting the known values, we derive that \( n \approx 1.27 \times 10^{-3} \text{ mol} \), meaning there is approximately 1.27 millimoles of water in the gaseous state once equilibrium is established.