What is the work done against the atmosphere when `25` grams of water vaprorizes at `373 K` against a constant external pressure of `1 atm` ? Assume t
What is the work done against the atmosphere when `25` grams of water vaprorizes at `373 K` against a constant external pressure of `1 atm` ? Assume that steam obeys perfect gas law. Given that the molar enthalpy of vaporization is `9.72 kcal//"mole"`, what is the change of internal energy in the above process?
A. `1294.0 cal, 11247 cal`
B. `921.4 cal , 11074 cal`
C. `1025.6 cal , 12474.3 cal`
D. `1129.3 cal, 10207 cal`
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Correct Answer - C
Mole of `H_(2)O=1.39`
`Pv=nRT`
`1xxv=1.39xx0.082xx373`
`v=42.51 "lit"`
`w=Pext.dv= 1xx[42.80]atmxxlit. =42.80xx101.325J=(42.80xx101.325)/(4.2)=1025.6 cal`
`Delta H=Delta U+[PDeltav].=12470.6 cal`
`Delta U =Delta H-PDeltav=13500-1025.6=1247.3 cal`
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