InterviewSolution
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InterviewSolution
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A chemist while studying the properties of gaseous `C C l_(2)F_(2)`, a chlorofluorocarbon refrigerant, cooled a 1.25 g sample at constant atmospheric prssure of 1.0 atm from 320 K to 293 K . During cooling the sample volume decreased from 274 to 248 mL. Calculate `Delta H ` and `DeltaU` for the chlorofluorocaron for this process. For `C Cl_(2)F_(2), C_(p) = 80.7 J // ( mol K )`. |
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Answer» `DeltaH = q_(p) ` and `C_(p)` is heat evolved or absorbed per mole for `1^(@) ` fall or rise in temperature . Here, fall in temperature `= 320- 293 = 27 K` Molar mass of `C C l_(2)F_(2) = 12 + 2 xx 35.5 + 2 xx 19 = 121 gmol^(-1)` `:. `Heat evolved from 1.25 g of the sample on being cooled from 320K to 293 K at constant pressure`= ( 80.7) /( 121) xx 1.25 xx 27 J = 22.51J` Further, `Delta H = Delta U + P Delta V = 22.51J` `P Delta V = 1 atm xx (( 248 0 274))/(100) L = - 0.026 L atm = - 0.026 xx 101.325 J = - 2.63 J ` `:. - 22.51 = Delta U - 2.63 J` or ` Delta U = - 22.51 + 2.63 J = - 19.88 J ` |
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