Calculate entropy change for vaporization of 1 mole of liquid water to steam at 100 ∘ C if Δ H v =40.8 kJ m o l − 1 . Options: A . &nbspΔSv=107.38 JK−1 mol−1 B . &nbspΔSv=109.48 JK−1 mol−1 C . &nbspΔSv=109.38 JK−1 mol−1 D . &nbspΔSv=107.38 JK−1 mol−1

Answer: Option C : C For entropy change of vaporization can be given as, Δ S v = Δ H v T Given values are, Δ H v = 40.8 × 10 3 J m o l − 1 ; T=373K Therefore Δ S v = ( 40.8 × 10 3 373 ) Δ S v = 109.38 J K − 1 m o l − 1

Calculate entropy change for vaporization of 1 mole of liquid water to steam at

Question

Calculate entropy change for vaporization of 1 mole of liquid water to steam at

100^{\circ} C

Δ H_{v}

=40.8 kJ

m o l^{- 1}

Options:

A . ΔSv=107.38 JK−1 mol−1

B . ΔSv=109.48 JK−1 mol−1

C . ΔSv=109.38 JK−1 mol−1

D . ΔSv=107.38 JK−1 mol−1

Answer: Option C
:
C
For entropy change of vaporization can be given as,

Δ S_{v} = Δ \frac{H_{v}}{T}

Given values are,

Δ H_{v} = 40.8 \times 10^{3} J m o l^{- 1}

; T=373K
Therefore

Δ S_{v} = (40.8 \times \frac{10^{3}}{373})

Δ S_{v} = 109.38 J K^{- 1} m o l^{- 1}

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