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Answer :
Final answer:
The boiling point of the substance is equal to the initial temperature (T1).
Explanation:
To find the boiling point of the substance, we can use the Clausius-Clapeyron equation:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Where:
- P1 and P2 are the initial and final pressures, respectively.
- ΔHvap is the enthalpy of vaporization.
- R is the ideal gas constant (8.314 J/mol K).
- T1 and T2 are the initial and final temperatures, respectively.
In this case, we are given the enthalpy of vaporization (ΔHvap = 38.6 kJ/mol) and the entropy of vaporization (ΔSvap = 109 J/mol K). We need to find the boiling point in °C, which corresponds to the final temperature (T2).
Since we are given the values in kJ/mol and J/mol K, we need to convert them to J/mol and J/mol °C, respectively, to match the units of the ideal gas constant (R).
ΔHvap = 38.6 kJ/mol = 38,600 J/mol
ΔSvap = 109 J/mol K
Now, we can rearrange the Clausius-Clapeyron equation to solve for T2:
ln(P2/P1) = (ΔHvap/R) * (1/T1 - 1/T2)
Since we are given the initial pressure (P1) and the atmospheric pressure is usually 1 atm, we can assume P2 = 1 atm.
ln(1/1) = (38,600 J/mol / (8.314 J/mol K)) * (1/T1 - 1/T2)
0 = (38,600 / 8.314) * (1/T1 - 1/T2)
0 = 4649.94 * (1/T1 - 1/T2)
1/T1 - 1/T2 = 0
1/T1 = 1/T2
T1 = T2
Therefore, the boiling point of the substance is equal to the initial temperature (T1).
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