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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Thermodynamics

Calculate the entropy change for the conversion of 1 gm of ice to water at 273 K and one atm pressure. The enthalpy of fusion of ice is 6.025 kJ/mol

$\begin {array} {1 1} (A)\;-12.227\: J/K & \quad (B)\;12.227 \: J/K \\ (C)\;1.227 \: J/K & \quad (D)\;22.1\: J/K \end {array}$

1 Answer

$ \Delta S_{fusion} = \large\frac{\Delta H_{fusion}}{T_f}$$ =\large\frac{ 6025\: J/mol}{ 273K}$$= 22.1\: J/K-mol$
Now, entropy change for conversion of 1 mol, i.e, 18g of ice into water at $273\: K = 22.1 J/K$
So, Entropy change for converting 1g of ice into water = $\large\frac{22.1J/K}{ 18}$$ = 1.227 J/K$
Ans : (C)
answered Mar 16, 2014 by thanvigandhi_1
 

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