For the process $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ (1 bar, $\left.273 \mathrm{~K}\right) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$ (1...

2 months ago 1

For the process

$\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ (1 bar,

$\left.273 \mathrm{~K}\right) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$ (1 bar,

$\left.373 \mathrm{~K}\right)$ , the correct set of thermodynamic parameters is

$\Delta G=0, \Delta S=+$ ve

$\Delta G=0, \Delta S=-v e$

$\Delta G=+v e, \Delta S=0$

$\Delta G=-v e, \Delta S=+v e$

Solution:

$\mathrm{H}_2 \mathrm{O}(l)$ and

$\mathrm{H}_2 \mathrm{O}(g)$ both exist together at same temperature and pressure, suggests the state of equilibrium
Hence,

$\quad \Delta G=0$ and conversion of liquid into gas increases disorderness, entropy hence

$\Delta S=+v e$ .

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For the process $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ (1 bar, $\left.273 \mathrm{~K}\right) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$ (1...

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For the process H2O(l)\mathrm{H}_2 \mathrm{O}(\mathrm{l}) (1 bar, 273 K)⟶H2O(g)\left.273 \mathrm{~K}\right) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g}) (1...

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For the process $\mathrm{H}_2 \mathrm{O}(\mathrm{l})$ (1 bar, $\left.273 \mathrm{~K}\right) \longrightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})$ (1...