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Third Law of Thermodynamics

What is the e...

Question

What is the entropy change for the conversion of one gram of ice to water at $273 K$ and one atmospheric pressure?
$(Δ H_{f u s i o n} = 6.025 k J {m o l}^{- 1}$ )

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Solution

$Δ H_{f u s i o n} = 6.025 \times 1000 J {m o l}^{- 1} = \frac{6025}{18} J g^{- 1} = 334.72 J g^{- 1}$
$Δ S_{f u s i o n} = \frac{Δ H_{f u s i o n}}{T} = \frac{334.72}{273} = 1.226 J K^{- 1} {m o l}^{- 1}$

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Similar questions

Q. The entropy change for the conversion of the one gram of ice to water at 273 K and one atmospheric pressure

[Δ H_{f u s i o n} = 6.025 k J m o l^{- 1}]

Q. One mole of ice is converted into water at

273 K

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H_{2} O (s)

H_{2} O (l)

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Q. The entropy change associated with the conversion of

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[Given that : Specific heat of water liquid and water vapour are

4.2 k J K^{- 1} k g^{- 1}

2.0 k J K^{- 1} k g^{- 1}

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344 k j k g^{- 1}

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[Take :

l o g 273 = 2.436, l o g 373 = 2.572, l o g 383 = 2.583]

0^{o} C

, ice and water are in equilibrium and enthalpy change for the process

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