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प्रश्न
The entropy change can be calculated by using the expression ∆S = `q_(rev)/T`. When water freezes in a glass beaker, choose the correct statement amongst the following :
विकल्प
∆S (system) decreases but ∆S (surroundings) remains the same.
∆S (system) increases but ∆S (surroundings) decreases.
∆S (system) decreases but ∆S (surroundings) increases.
∆S (system) decreases and ∆S (surroundings) also decreases.
उत्तर
∆S (system) decreases but ∆S (surroundings) increases.
Explanation:
During the process of freezing energy is released, which is absorbed by the surroundings.
∴ `∆S_("system") = - q_(rev)/T`;
`∆S_("surroundings") = q_(rev)/T`
Therefore, the entropy of the system decreases and that of the surroundings increases.
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संबंधित प्रश्न
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(iv) Burning carbon in oxygen to give carbon dioxide.
Enthalpy diagram for a particular reaction is given in figure. Is it possible to decide spontaneity of a reaction from given diagram. Explain.
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| Reaction | Entropy change |
| (i) A liquid vapourises | (a) ∆S = 0 |
| (ii) Reaction is non-spontaneous at all temperatures and ∆H is positive |
(b) ∆S = positive |
| (iii) Reversible expansion of an ideal gas | (c) ∆S = negative |
Assertion (A): A liquid crystallises into a solid and is accompanied by decrease in entropy.
Reason (R): In crystals, molecules organise in an ordered manner.
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