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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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Calculate the entropy change when 36 g of liquid water evaporates at $373\: K( \Delta H_{vap} = 40.63\: kJ/mol)$

$\begin {array} {1 1} (A)\;109 \: J/K-mol & \quad (B)\;-109\: J/K-mol \\ (C)\;-218\: J/K-mol & \quad (D)\;218\: J/K-mol \end {array}$

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$ \Delta S_{vap} = \large\frac{\Delta H_{vap}}{ T_b}$$ = \large\frac{40.63 \times 1000J/mol}{ 373K}$$ = 109 \: J/K-mol$
So, Entropy change for evaporation of 36 g of water = $\large\frac{109 \times 36}{18}$$ = 218\: J/K$
Ans : (D)
answered Mar 16, 2014 by thanvigandhi_1 1 flag
 

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