Calculate the enthalpy change that occurs when 1kg of benzene vapour at 150℃ and 100KPa condense to a solid at -20℃ and 100KPa
To calculate the enthalpy change that occurs when benzene vapor condenses to a solid, we need to find the difference in enthalpy (enthalpy change) between the initial and final states.
The enthalpy change can be calculated using the formula:
ΔH = m × [ΔHvap + ΔHfus]
Where:
ΔH = Enthalpy change
m = Mass of the substance
ΔHvap = Enthalpy of vaporization
ΔHfus = Enthalpy of fusion
First, we need to find the enthalpy of vaporization (ΔHvap) and the enthalpy of fusion (ΔHfus) for benzene.
The enthalpy of vaporization (ΔHvap) is the amount of energy required to convert 1 kg of liquid benzene to vapor at a given temperature and pressure. This value can usually be obtained from reference tables or databases. Let's assume the ΔHvap for benzene at 150℃ and 100 KPa is 30,000 J/kg.
The enthalpy of fusion (ΔHfus) is the amount of energy required to convert 1 kg of solid benzene to liquid at its melting point. Similarly, this value can be obtained from reference tables or databases. Let's assume the ΔHfus for benzene is 10,000 J/kg.
Now, we can calculate the enthalpy change (ΔH) using the formula mentioned earlier.
ΔH = m × [ΔHvap + ΔHfus]
= 1 kg × [30,000 J/kg + 10,000 J/kg]
= 1 kg × 40,000 J/kg
= 40,000 J
Therefore, the enthalpy change that occurs when 1 kg of benzene vapor at 150℃ and 100 KPa condenses to a solid at -20℃ and 100 KPa is 40,000 J.