A chemist while studying the properties of gaseous `C_(2)Cl_(2)F_(2)`, a chlorofluoro carbon refrigerant, cooled a `1.25 g` sample at constant atmospheric pressure of `1.0 atm` from `320 K` to `290 K`. During cooling, the sample volume decreased from `274` to `248 mL`. Calculate `Delta H` and `Delta U` for the chlorofluoro carbon for this process. For `C_(2)Cl_(2)F_(2)`, `C_(P) = 80.7 J mol^(-1) K^(-1)`.

Explanation

Fall in temperature = 320 - 293 = 27 K
Molar mass of `C_(2) CI_(2)F_(2) = 2 xx 12 + 2 xx 35.5 + 2 xx 19 = 133`
`C_(p) = 80.7 J mol^(-1)K^(-1)`
or = `(80.7)/(133) = 0.6068 J g^(-1)K^(-1)`
Heat evolved from 1.25 g of sample on being cooled
`q_(P) = m xx C_(p) xx DeltaT`
`= 1.25 g xx 0.6068 J K^(-1) g^(-1) xx 27 K`
`= 20.48 J`
`therefore DeltaH = -20.48 J`
Now `DeltaH = Delta U + pDeltaV`
`pDeltaV = 1atm xx(248 -274)/(1000)L`
` = - 0.026L atm`
` = -0.026 xx 101.325 = - 2.63 J`
` - 20.48 = DeltaU - 2.63`
`DeltaU = -20.48 + 2.63`
`= -17.84 J`.