Question:
Identify the correct statements from the following
\[
\text{I. At }0\,K,\text{ the entropy of pure crystalline materials approaches zero.}
\]
\[
\text{II. Entropy for the process }H_2O(l)\longrightarrow H_2O(g)\text{ decreases.}
\]
III. Gibbs energy is a state function.
Identify the correct statements from the following
\[ \text{I. At }0\,K,\text{ the entropy of pure crystalline materials approaches zero.} \] \[ \text{II. Entropy for the process }H_2O(l)\longrightarrow H_2O(g)\text{ decreases.} \] III. Gibbs energy is a state function.
\[ \text{I. At }0\,K,\text{ the entropy of pure crystalline materials approaches zero.} \] \[ \text{II. Entropy for the process }H_2O(l)\longrightarrow H_2O(g)\text{ decreases.} \] III. Gibbs energy is a state function.
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Entropy increases when a substance changes from solid \(\rightarrow\) liquid \(\rightarrow\) gas because molecular randomness increases in that order.
Updated On: Jun 22, 2026
- i, ii, iii
- i, iii Only
- ii, iii Only
- i, ii Only
Show Solution
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The Correct Option is B
Solution and Explanation
Step 1: Analyze statement I.
According to the third law of thermodynamics: \[ \text{The entropy of a perfectly crystalline substance at }0\,K\text{ is zero.} \] Thus, for pure crystalline materials, \[ S\rightarrow 0 \quad \text{as} \quad T\rightarrow 0\,K \] Therefore, statement I is correct.
Step 2: Analyze statement II.
The process given is \[ H_2O(l)\longrightarrow H_2O(g) \] This is the conversion of liquid water into water vapour.
Gas molecules possess much greater randomness and disorder than liquid molecules.
Hence, entropy increases during vaporisation: \[ \Delta S\gt 0 \] Therefore, the statement saying entropy decreases is incorrect.
Thus, statement II is false.
Step 3: Analyze statement III.
Gibbs free energy is defined as \[ G=H-TS \] Since enthalpy \((H)\), temperature \((T)\), and entropy \((S)\) are state functions, Gibbs free energy is also a state function.
Therefore, statement III is correct.
Step 4: Final conclusion.
The correct statements are: \[ \boxed{\text{i and iii only}} \]
According to the third law of thermodynamics: \[ \text{The entropy of a perfectly crystalline substance at }0\,K\text{ is zero.} \] Thus, for pure crystalline materials, \[ S\rightarrow 0 \quad \text{as} \quad T\rightarrow 0\,K \] Therefore, statement I is correct.
Step 2: Analyze statement II.
The process given is \[ H_2O(l)\longrightarrow H_2O(g) \] This is the conversion of liquid water into water vapour.
Gas molecules possess much greater randomness and disorder than liquid molecules.
Hence, entropy increases during vaporisation: \[ \Delta S\gt 0 \] Therefore, the statement saying entropy decreases is incorrect.
Thus, statement II is false.
Step 3: Analyze statement III.
Gibbs free energy is defined as \[ G=H-TS \] Since enthalpy \((H)\), temperature \((T)\), and entropy \((S)\) are state functions, Gibbs free energy is also a state function.
Therefore, statement III is correct.
Step 4: Final conclusion.
The correct statements are: \[ \boxed{\text{i and iii only}} \]
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