Question

A chemist while studying the properties of gaseous $C_{2}Cl_{2}F_{2}$, a chlorofluorocarbon refrigerant, cooled 1.25 g sample at constant atmospheric pressure of 1.0 atm from 320 K to 293 K. During cooling, the sample volume decreased from 274 to 248 ml. Calculate ΔH and ΔU for the chlorofluorocarbon for this process. The value of molar heat capacity is 80.7 J $K^{-1}$ $mol^{-1}$.

Answer 1

From the given

Mass of  ${ C }_{ 2 }{ Cl }_{ 2 }{ F }_{ 2 }$ = 1.25 g

Pressure = 1.0 atm

$Change\quad in\quad temperature=\quad \Delta T\quad =\quad 293\quad K\quad -\quad 320\quad K\\ \quad \quad \quad \\ \quad \quad \quad \quad \quad =\quad -27\quad K$

$Change\quad in\quad volume,\quad \Delta V\quad =\quad 248\quad mL\quad -\quad 274\quad mL\\ \\\Delta V=\quad -26\quad mL$

${ C }_{ p,m }\quad =\quad 80.7\quad J/mol\quad K$

$Molar\quad mass\quad of\quad { C }_{ 2 }{ Cl }_{ 2 }{ F }_{ 2 }\quad =\quad 24\quad +\quad 71.0\quad +\quad 38.0\quad \\ \\Molar\quad mass\quad =\quad 133\quad g\quad { mol }^{ -1 }$

Let’s calculate the number of moles

$Number\quad of\quad moles\quad of\quad { C }_{ 2 }{ Cl }_{ 2 }{ F }_{ 2 }\quad =\quad n\quad =\quad \frac { 1.25\quad g }{ 133\quad g\quad mo{ l }^{ -1 } } \quad =\quad \frac { 1.25 }{ 133 } mol$

Since,$\quad \Delta H\quad =\quad { q }_{ p }\quad =\quad n\quad \times \quad { C }_{ p,m }\quad \times \quad \Delta T$

Substituting the all given values,

$\Delta H\quad =\quad \left( \frac { 1.25 }{ 133 } mol \right) \quad \times \quad \left( { 80.7\quad J/mol\quad K}\right) \quad \times \quad (-27\quad K)\quad =\quad -20.5\quad J$

$\Delta U\quad =\quad { q }_{ V }\quad =\quad n\quad \times \quad { C }_{ v }\quad \times \quad \Delta T$

Here,

${ C }_{ v,m }\quad =\quad { C }_{ p,m }\quad -\quad R\\ \\=\quad (80.7\quad -\quad 8.31)\quad J/mol\quad K\\ \\ { C }_{ v,m }\quad =\quad 72.4\quad J/mol\quad K$

Substituting the all given values

$\Delta U\quad =\quad \left( \frac { 1.25 }{ 133 } \quad mol \right) \quad \times \quad \left({ 72.4\quad J/mol\quad K}\right) \quad \times \quad (-27\quad K)\\ \\\quad \quad \quad \quad =\quad -18.4\quad J$