Calculate the entropy change in JK-1 mol -1 for vaporisation of liquid water to steam at 100- C- Given that heat of vaporisation is 40-8 kJ mol-1-

We know that, Δ S=Δ H_vT Here given that, Δ H=40.8 kJ/mol, T=373K putting the values, Δ S=40.8×10^3373=109.38 JK^ 1mol^ 1 Theory: Entropy Calculation with ...

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Standard XIII

Chemistry

Entropy

Calculate the...

Question

Calculate the entropy change (in $J K^{- 1} m o l^{- 1}$ ) for vaporisation of liquid water to steam at $100^{\circ} C$ . Given that heat of vaporisation is $40.8 k J m o l^{- 1}$ .

109.38

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100.38

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120.38

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129.38

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Solution

The correct option is A 109.38
We know that,
$Δ S = \frac{Δ H_{v}}{T}$
Here given that,
$Δ H = 40.8 k J / m o l, T = 373 K$
putting the values,
$Δ S = \frac{40.8 \times 10^{3}}{373} = 109.38 J K^{- 1} {mol}^{- 1}$

Theory:

Entropy Calculation with phase transformation:
Phase change occurs at constant Pressure & Temperature and is considered reversible if it occurs at its transition temperature.

Entropy change for Vaporisation:
At constant pressure $q = q_{p}$ = $Δ H$
$Δ S_{v a p}$ = $\frac{Δ H_{v a p}}{T_{b}}$ where $T_{b}$ is boiling point of a substance.

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Entropy

Standard XIII Chemistry

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Calculate the entropy change (in JK−1 mol −1) for vaporisation of liquid water to steam at 100∘C. Given that heat of vaporisation is 40.8 kJmol−1.

Calculate the entropy change (in $J K^{- 1} m o l^{- 1}$ ) for vaporisation of liquid water to steam at $100^{\circ} C$ . Given that heat of vaporisation is $40.8 k J m o l^{- 1}$ .