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The most random state of H(2)O system is...

The most random state of `H_(2)O` system is

A

ice

B

Liquid water

C

Steam

D

Randomness is same in all

Text Solution

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The correct Answer is:
To determine the most random state of the water system (H₂O), we will analyze the three states of water: ice, liquid water, and steam. ### Step-by-Step Solution: 1. **Identify the States of Water**: - Water exists in three states: solid (ice), liquid (water), and gas (steam). 2. **Understand the Properties of Each State**: - **Ice (Solid)**: In this state, water molecules are tightly packed in a fixed structure. The molecules can only vibrate around their fixed positions, leading to minimal randomness. - **Liquid Water**: In the liquid state, water molecules are less tightly packed compared to ice. They can move past each other, allowing for more randomness than in ice. However, the movement is still somewhat restricted compared to gases. - **Steam (Gas)**: In the gaseous state, water molecules are far apart from each other and move freely. This allows for the highest degree of randomness and kinetic energy among the molecules. 3. **Compare the Randomness in Each State**: - Randomness in Ice: Very low (molecules are fixed in position). - Randomness in Liquid Water: Moderate (molecules can move but are still somewhat close). - Randomness in Steam: Very high (molecules are far apart and move freely). 4. **Conclusion**: - Since steam has the highest degree of randomness due to the free movement of water molecules, it is the most random state of the water system. 5. **Final Answer**: - The most random state of the H₂O system is **steam (option C)**.
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The least random state of water system is

Entropy is a measure of randomess of system. When a liquid is converted to the vapour state entropy of the system increases. Entropy in the phase transformation is calculated using Delta S = (Delta H)/(T) but in reversible adiabatic process Delta S will be zero. The rise in temperature in isobaric or isochoric process increases the randomness of system, which is given by Delta S = "2.303 n C log"((T_(2))/(T_(1))) C = C_(P) or C_(V) The temperature at whicgh liquid H_(2)O will be in equrilibrium with its vapour is ( Delta H and Delta S for vapourisation are 50 kJ mol^(-1) and 0.15 kJ mol^(-1)K^(-1) )

Knowledge Check

  • The oxidation state of S in H_(2)S_(2)O_(8) is

    A
    `+6`
    B
    `+7`
    C
    `+8`
    D
    0
  • The oxidising state of molybdenum in its oxo complex species [Mo_(2)O_(4)(C_(2)H_(4))_(2)(H_(2)O)]^(2-) is

    A
    `+2`
    B
    `+3`
    C
    `+4`
    D
    `+5`
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