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 Hand DeltaU for the chlorofluorocaron for this process. For C Cl_(2)F_(2), C_(p) = 80.7 J // ( mol K ).

A chemist while studying the properties of gaseous C C l_(2)F_(2), a c

1.	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 Hand DeltaU for the chlorofluorocaron for this process. For C Cl_(2)F_(2), C_(p) = 80.7 J // ( mol K ).
Answer» Solution :`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 `