A 110.1 gram sample of an unknown substance (MM = 189.50 g/mol) is cooled from 233.0 °C to 147.6 °C. (heat capacity of liquid = 1.62 J/g・°C; heat capacity of gas = 1.04 J/g・°C; ∆Hvap = 78.11 kJ/mol; normal boiling point, Tb = 160.3 °C) How much energy (in kJ) is absorbed/released to cool the gas?

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A 110.1 gram sample of an unknown substance (MM = 189.50 g/mol) is cooled from 233.0 °C to 147.6 °C. (heat capacity of liquid = 1.62 J/g・°C; heat capacity of gas = 1.04 J/g・°C; ∆Hvap = 78.11 kJ/mol; normal boiling point, Tb = 160.3 °C) How much energy (in kJ) is absorbed/released to cool the gas?

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Step 1

Mass of the unknown substance= 110.1 g

Heat capacity of liquid = 1.62 J/g・°C

Heat capacity of the gas = 1.04 J/g.ºC

The given substance is cooled from Temperature: 233.0ºC -> 147.6ºC.

The complete process comprises of the following sub-processes:

Cooling the gas from 233.0ºC to 160.3ºC.

Condensation of the gas at its boiling point, Tb = 160.3ºC.

Cooling the liquid from 160.3ºC to 147.6ºC.

 

 

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