Problem 92 - A Complete Resource Book in Chemistry for JEE Main

Question

The change in entropy, in the conversion of one mole of water at \(373 \mathrm{~K}\) to vapour at the same temperature is (Latent heat of vaporization of water = \(\left.2.257 \mathrm{~kJ} \mathrm{~g}^{-1}\right)\) (a) \(99 \mathrm{JK}^{-1}\) (b) \(129 \mathrm{JK}^{-1}\) (c) \(89 \mathrm{JK}^{-1}\) (d) \(109 \mathrm{JK}^{-1}\)

Step-by-Step Solution

Answer

The change in entropy is approximately \(109 \mathrm{JK}^{-1}\), matching option (d).

Step 1: Understand the Formula for Change in Entropy

The change in entropy, denoted as \( \Delta S \), for a phase change can be calculated using the formula: \( \Delta S = \frac{q}{T} \), where \( q \) is the heat absorbed or released in the process, and \( T \) is the temperature in Kelvin at which the process occurs.

Step 2: Convert Latent Heat to Moles

The latent heat of vaporization is given as \(2.257 \mathrm{~kJ} \mathrm{~g}^{-1}\). Since 1 mole of water is 18 grams, we first convert the latent heat into a per mole basis: \( q_{mole} = 2.257 \frac{\mathrm{kJ}}{\mathrm{g}} \times 18 \mathrm{g} = 40.626 \mathrm{kJ/mole} \).

Step 3: Calculate the Change in Entropy

Using the formula from Step 1, substitute \( q = 40.626 \mathrm{kJ/mole} \) and convert to Joules: \( q = 40.626 \times 1000 \mathrm{J/mole} \). Then substitute \( T = 373 \mathrm{K} \): \[ \Delta S = \frac{40,626 \mathrm{J/mole}}{373 \mathrm{K}} \approx 108.96 \mathrm{J/K} \].

Step 4: Determine the Closest Answer Choice

Comparing the calculated \( \Delta S \approx 108.96 \mathrm{J/K} \) to the provided options, it is closest to option (d) \(109 \mathrm{JK}^{-1}\).

Key Concepts

Latent Heat of Vaporization

To understand the phase change of water from liquid to vapor, we first need to grasp the concept of the latent heat of vaporization. This is the amount of heat required to transform a unit mass of a substance from liquid to vapor without changing its temperature. In our example, we're dealing with water where the latent heat of vaporization is given as \(2.257 \text{ kJ g}^{-1}\).

This means that for every gram of water to convert into vapor at its boiling point, specifically at \(373\, \text{K}\) or \(100 \degree \text{C}\), 2.257 kJ of energy is needed. When dealing with larger quantities, such as moles, this value needs to be converted. Since one mole of water weighs 18 grams, we multiply the latent heat by the molar mass to get \(40.626 \text{ kJ/mole}\).

This adjustment is crucial for subsequent calculations because it allows us to equate the energy transformation on a molecular scale.

Entropy Calculation Formula

In any thermodynamic process, changes in entropy give insights into the disorder within a system. During a phase change, like the vaporization of water, entropy increases as liquid molecules move apart into vapor.

To calculate the change in entropy, \( \Delta S \), during this transition, we utilize the formula:
\[ \Delta S = \frac{q}{T} \]
where \( q \) represents the energy absorbed or released, and \( T \) is the absolute temperature in Kelvin during the process. This formula characterizes the relationship between absorbed heat and temperature, providing a way to determine the quantitative increase or decrease in entropy.

In our exercise, with water vaporizing at \(373 \text{ K}\), and using the mole-scaled latent heat of 40.626 kJ (converted to 40626 J), we place these values into the formula to find the change in entropy. By dividing the energy by the temperature, \( \Delta S = \frac{40,626 \text{ J/mole}}{373 \text{ K}} \), we estimate an entropy change of approximately \(108.96 \text{ J/K}\).

Water Vaporization Process

The transition of water from liquid to vapor is a fundamental example that illustrates the concept of phase changes. At its boiling point, water molecules gain enough energy to break free from the liquid's surface and become gas, which is water vapor.

The energy needed for this transformation during boiling is not used to increase temperature but to change the state, showcasing water's latent heat of vaporization. The process is critical in understanding thermodynamic principles since it encompasses energy transfer and entropy change, depicting how heat can instigate phase changes without a shift in temperature.

In practical terms, the understanding of water vaporization informs processes like cooking, meteorology, and even industrial applications where steam generation is paramount. Here, it's crucial to recognize that the energy absorbed in vaporizing water is expressed as a change in the water's entropy, reiterating how energy distribution morphs both state and entropy characteristics.