For the process H2O (ℓ) → H2O (g) at T = 100°C and 1 atmosphere pressure, the correct choice is

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First Law and Various Process

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Question

For the process

H₂O (ℓ) → H₂O (g)

at T = 100°C and 1 atmosphere pressure, the correct choice is

$∆$ S_system < 0 and $∆$ S_surroundings > 0

$∆$ S_system > 0 and $∆$ S_surroundings < 0

$∆$ S_system < 0 and $∆$ S_surroundings < 0

$∆$ S_system > 0 and $∆$ S_surroundings > 0

H₂O(ℓ) → H₂O(g)

For system

Liquid is getting converted into gas $\begin{matrix} {ΔS}_{system} > 0 \end{matrix}$

Process is endothermic ⇒ Heat is given by system

$∆$ S_surrounding $= \frac{q_{surr}}{T}$

where q_surrounding= – $∆$ H

$∵$ q_surr < 0

$∆$ S_surr < 0

This question involves analyzing entropy changes during the phase transition of water from liquid to gas at its boiling point (100°C) and standard pressure (1 atm).

Step 1: Understanding Entropy

Entropy (S) is a measure of disorder or randomness in a system. The change in entropy (ΔS) indicates how the disorder changes during a process.

ΔS > 0 means an increase in disorder (positive entropy change).
ΔS < 0 means a decrease in disorder (negative entropy change).

Step 2: Analyzing ΔS_system

The system is the water itself. The process is:

$H_{2} O (ℓ) \to H_{2} O (g)$

When liquid water turns into water vapor, the molecules become far more disordered and have greater freedom of movement. Therefore, the entropy of the system increases.

$Δ S_{s y s t e m} > 0$

Step 3: Analyzing ΔS_surroundings

The surroundings are everything outside the system. For vaporization to occur, heat must be absorbed from the surroundings. This heat is the enthalpy of vaporization (ΔH_vap), which is positive (endothermic process).

The entropy change of the surroundings is related to the heat transferred (q) and the temperature (T) by the formula:

$Δ S_{s u r r o u n d i n g s} = - \frac{Δ H_{v a p}}{T}$

Since ΔH_vap is positive, the expression -ΔH_vap/T is negative.

$Δ S_{s u r r o u n d i n g s} < 0$

Step 4: Final Answer

Combining both analyses:

ΔS_system > 0 (entropy of water increases)
ΔS_surroundings < 0 (entropy of surroundings decreases)

Therefore, the correct choice is: $Δ S_{s y s t e m} > 0$ and $Δ S_{s u r r o u n d i n g s} < 0$

Detailed Solution

Step 1: Understanding Entropy

Step 2: Analyzing ΔS_system

Step 3: Analyzing ΔS_surroundings

Step 4: Final Answer

Related Topics & Formulae

Detailed Solution

Step 1: Understanding Entropy

Step 2: Analyzing ΔSsystem

Step 3: Analyzing ΔSsurroundings

Step 4: Final Answer

Related Topics & Formulae

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Step 2: Analyzing ΔS_system

Step 3: Analyzing ΔS_surroundings