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Standard XII

Chemistry

Law of Mass Action

H 2 O s → H 2...

Question

$H_{2} O (s) \to H_{2} O (l)$

This phase transition is carried out at constant temperature and pressure, then work done during the process :

W < 0

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W > 0

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W = 0

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can't determined

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Solution

The correct option is B
W > 0

Density of ice is less than water.

So, volume of $H_{2} O (l)$ is less than volume of $H_{2} O (s)$

$\Rightarrow △ V = (V_{2} - V_{1}) = - v e$

W = $- P △ V = (-) (-) = + v e$

$△ W > 0$

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Similar questions

H_{2} O (s) \to H_{2} O (l)

This phase transition is carried out at constant temp and pressure then work done during the process:

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H2O(s)→ H2O(l) This phase transition is carried out at constant temperature and pressure, then work done during the process :

The correct option is B W > 0 Density of ice is less than water. So, volume of H2O(l) is less than volume of H2O(s) ⇒△V=(V2−V1)=−ve W =−P△V=(−)(−)=+ve △W>0

$H_{2} O (s) \to H_{2} O (l)$

This phase transition is carried out at constant temperature and pressure, then work done during the process :

The correct option is B
W > 0

Density of ice is less than water.

So, volume of $H_{2} O (l)$ is less than volume of $H_{2} O (s)$

$\Rightarrow △ V = (V_{2} - V_{1}) = - v e$

W = $- P △ V = (-) (-) = + v e$

$△ W > 0$