For the process- H 2 O l - H 2 O g at T -100- C and 1 atmosphere pressure- the correct choiceA- - S system -0 and - S surrounding -0B- - S system 0C- - S system 0 and - S surrounding -0

This problem is based on assumption that total entropy change of universe is zero. At 100^∘C and 1 atmosphere pressure, H_2O ⇌ H_2O (g) is at equilibrium. For e ...

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Byju's Answer

Standard XII

Chemistry

Entropy

For the proce...

Question

For the process, $H_{2} O (l) \to H_{2} O (g) a t T = 100^{\circ} C$ and 1 atmosphere pressure, the correct choice

Δ S_{s y s t e m} > 0 a n d Δ S_{s u r r o u n d i n g} > 0

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Δ S_{s y s t e m} > 0 a n d Δ S_{s u r r o u n d i n g} < 0

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Δ S_{s y s t e m} < 0 a n d Δ S_{s u r r o u n d i n g} > 0

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Δ S_{s y s t e m} < 0 a n d Δ S_{s u r r o u n d i n g} < 0

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Solution

The correct option is B $Δ S_{s y s t e m} > 0 a n d Δ S_{s u r r o u n d i n g} < 0$
This problem is based on assumption that total entropy change of universe is zero.
At $100^{\circ} C$ and 1 atmosphere pressure,
$H_{2} O ⇌ H_{2} O (g)$ is at equilibrium.
For equilibrium, $Δ S_{t o t a l} = 0$
$a n d Δ S_{s y s t e m} + Δ S_{s u r r o u n d i n g} = 0$
As we know during conversion of liquid to gas entropy of system increases, in a similar manner entropy of surrounding decreases.
$∴ Δ S_{s y s t e m} > 0 a n d Δ S_{s u r r o u n d i n g} < 0$

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For the process, H2O(l)→H2O(g) at T=100∘C and 1 atmosphere pressure, the correct choice

For the process, $H_{2} O (l) \to H_{2} O (g) a t T = 100^{\circ} C$ and 1 atmosphere pressure, the correct choice