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

Answer» <html><body><p></p>Solution :`DeltaH = q_(p) ` and `C_(p)` is heat evolved or absorbed per mole for `1^(@) ` <a href="https://interviewquestions.tuteehub.com/tag/fall-983217" style="font-weight:bold;" target="_blank" title="Click to know more about FALL">FALL</a> or rise in <a href="https://interviewquestions.tuteehub.com/tag/temperature-11887" style="font-weight:bold;" target="_blank" title="Click to know more about TEMPERATURE">TEMPERATURE</a> . Here, fall in temperature`= 320- 293 = 27 K` <br/> Molar mass of `<a href="https://interviewquestions.tuteehub.com/tag/c-7168" style="font-weight:bold;" target="_blank" title="Click to know more about C">C</a> C l_(2)F_(2) = <a href="https://interviewquestions.tuteehub.com/tag/12-269062" style="font-weight:bold;" target="_blank" title="Click to know more about 12">12</a> + 2 xx 35.5 + 2 xx 19 = 121 gmol^(-1)` <br/> `:. `Heat evolved from 1.25 g of the sample on being cooled from 320K to 293 K at constant <a href="https://interviewquestions.tuteehub.com/tag/pressure-1164240" style="font-weight:bold;" target="_blank" title="Click to know more about PRESSURE">PRESSURE</a>`= ( 80.7) /( 121) xx 1.25 xx 27 J = 22.51J`<br/> Further, `Delta H = Delta U + P Delta V = 22.51J` <br/> `P Delta V = 1 atm xx (( 248 0 274))/(100) L = - 0.026 L atm = - 0.026 xx 101.325 J = -2.63 J ` <br/> `:.- 22.51 = Delta U - 2.63 J` <br/> or ` Delta U = - 22.51 + 2.63 J = - 19.88 J `</body></html>


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