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Find the entropy change when 90 g of H(2...

Find the entropy change when 90 g of `H_(2)O` at `10^(@)`C was converted into steam at `100^(@)`C. [Given `C_(P)(H_(2)O)=75.29 JK^(-1)mol^(-1) and Delta H_("vap")=43.932 JK^(-1)mol^(-1)`]

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For entropy change of vaporisation, `Delta_(V)S^(Theta) = (Delta_(V)H^(Theta))/(T)`
Given, `Delta_(V)H^(Theta) = 40.8 xx 10^(3)J ,T = 373 K`
`Delta_(V)S^(Theta) = (40.8 xx 10^(3))/(373) = 109.38 J K^(-1) mol^(-1)`
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What is the value of change in internal energy at 1 atm in the process? H_(2)O(l,323K)rarrH_(2)O(g,423K) Given : C_(v,m)(H_(2)O,l)=75.0JK^(-1) mol^(-1):" "C_(p,m)(H_(2)O,g)=33.314JK^(-1)mol^(-1) DeltaH_(vap)"ar 373 K"=40.7KJ//mol (a) 42.91kJ//mol (b) 43086kJ//mol (c) 42.6kJ//mol (d) 49.6kJ//mol

The internal energy change (in J) when 90 g of water undergoes complete evaporation at 100^(@)C is ______ . (Given : Delta H_("vap") for water at 373 K = 41 kJ/ mol , R = 8.314 JK^(-1) mol^(-1) )

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Calcualte the enthalpy change on freezing of 1.0 mole of water at 10.0^(@)C to ice at -10^(@) C. Delta_(fs)H=6.03 kJ mol^(-1) at 0^(@)C . C_(p)[H_(2)O(l)] = 75.3 J mol^(-1) K^(-1), C_(P)[H_(2)O(s)] = 36.8 Jmol^(-1)K^(-1)

Calculate the entropy change accompanying the following change of state H_(2)O (s, 10^(@)C, 1 atm) rarr H_(2)O(l, 10^(@)C, 1atm) C_(P) for ice = 9 cla deg^(-1) mol^(-1) C_(P) for H_(2)O = 18 cal deg^(-1) mol^(-1) Latent heat of fustion of ice = 1440 cal mol^(-1) at 0^(@)C .

Calculate the entropy change accompanying the following change of state H_(2)O (s, -10^(@)C, 1 atm) rarr H_(2)O(l, 10^(@)C, 1atm) C_(P) for ice = 9 cal deg^(-1) mol^(-1) C_(P) for H_(2)O = 18 cal deg^(-1) mol^(-1) Latent heat of fustion of ice = 1440 cal mol^(-1) at 0^(@)C .

At 0^(@)C , ice and water are in equilibrium and Delta H^(@) = 6.00 kJ/mol for the process H_(2)O(s) hArr H_(2)O(I) Value of Delta S^(@) for the conversion of ice to liquid water is {:((1),10.15 JK^(-1)mol^(-1),(2),17.25 JK^(-1)mol^(-1)),((3),21.98 JK^(-1)mol^(-1),(4),30.50 JK^(-1)mol^(-1)):}

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