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Chemistry - Thermodynamics - I Question with Solution | TestHub

ChemistryThermodynamics - IFirst law of ThermodynamicsEasy2 minPYQ_2020

ChemistryEasysingle choice

1 mol of ice is converted to liquid at 273 K; $H_{2} O$ (s) and $H_{2} O$ (l) have entropies 38.20 and 60.03 $J m o l^{- 1} d e g$ . Enthalpy changes in the conversion will be

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Answer:

Solution:

$^{Δ S = S_{f i n a l} - S_{i n i t i a l}}$
$= (60.03 - 38.20) J m o l^{- 1} K^{- 1}$
$= 21.83 J m o l^{- 1} K^{- 1}$
Also $Δ S = \frac{Δ H_{f u s i o n}}{T}$
or $Δ H_{f u s i o n} = T Δ S = 273 K \times 21.83 J m o l^{- 1} K^{- 1}$
$= 5959.59 J m o l^{- 1}$ .

Stream:NTA_ABHYAS_NEETSubject:ChemistryTopic:Thermodynamics - ISubtopic:First law of Thermodynamics

⏱ 2mℹ️ Source: PYQ_2020

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