When 1.8 g of steam at the normal boiling point of water is converted into water at the same temperature, enthalpy and entropy changes respectively will be $(Δ H_{v a p} (f o r w a t e r = 4.08 k J m o l^{- 1})$

$4.08 k J, 10.93 J K^{- 1}$

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$- 4.08 k J, - 40.8 j k^{- 1}$

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$- 4.08 k J, - 10.93 J K^{- 1}$

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$4.08 k J, 40.8 J K^{- 1}$

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Solution

The correct option is C
$- 4.08 k J, - 10.93 J K^{- 1}$

$Δ H = (- 40.8) \times \frac{1.8}{18} = - 4.08 k J;$
$Δ S = \frac{Δ H}{T} = \frac{- 4.08 \times 10^{3}}{373.15} = - 10.93 J K^{- 1}$

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Similar questions

When 1.8 g of steam at the normal boiling point of water is converted into water at the same temperature, enthalpy and entropy changes respectively will be $(Δ H_{v a p} (f o r w a t e r = 4.08 k J m o l^{- 1})$

Q. When 1 g of water at

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gets converted into steam at the same temperature, the change in volume is approximately

Q. When

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of water at

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gets converted into steam at the same temperature, the change in volume is approximately :

Q. At 373 K, steam and water are in equilibrium and

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(100 k P a)

will be [The density of water and steam are

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and

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When 1.8 g of steam at the normal boiling point of water is converted into water at the same temperature, enthalpy and entropy changes respectively will be (ΔHvap(for water=4.08kJ mol−1)

The correct option is C −4.08 kJ,−10.93 JK−1 ΔH=(−40.8)×1.818=−4.08 kJ; ΔS=ΔHT=−4.08×103373.15=−10.93 JK−1

When 1.8 g of steam at the normal boiling point of water is converted into water at the same temperature, enthalpy and entropy changes respectively will be $(Δ H_{v a p} (f o r w a t e r = 4.08 k J m o l^{- 1})$

The correct option is C
$- 4.08 k J, - 10.93 J K^{- 1}$

$Δ H = (- 40.8) \times \frac{1.8}{18} = - 4.08 k J;$
$Δ S = \frac{Δ H}{T} = \frac{- 4.08 \times 10^{3}}{373.15} = - 10.93 J K^{- 1}$