Ncert Exercises & Solutions

The entropy change can be calculated by using the expression $∆ S = \frac{q_{rev}}{T}$ . When water freezes in a glass beaker, the correct statement among the following is:

1.	∆ S (system) decreases but ∆ S (surroundings) remains the same.
2.	∆ S (system) increases but ∆ S (surroundings) decreases.
3.	∆ S (system) decreases but ∆ S (surroundings) increases.
4.	∆ S (system) decreases but ∆ S (surroundings) also decreases.

Hint: Entropy is the degree of randomness.

The entropy change can be calculated by using the expression

∆ S = \frac{q_{rev}}{T}

When water freezes in a glass beaker,

∆ S

(system) decreases because molecules in solid ice are less random than

in liquid water. However, when water freezes to ice, heat is released which is absorbed by the surroundings.

Hence, the entropy of the surroundings increases.

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