- #1
laughingnahga
- 2
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Using line equation y = -4058.7x + 16.10 with the experiment pressure measured in kPa instead of atm.
I already solved for heat of vaporization thus:
(-4058.7K)(-8.314 J/mol [itex]\ast[/itex]K)=heat of vaporization = 33444 J/mol = 33.44 kJ/mol.
Also, the vapor pressure for normal boiling point:
(1atm)(101325 Pa/1atm)(1 kPa/1000 Pa)=101.32 kPa
I attempted to solve for T by rearranging the linear form of the Clausius-Clapeyron equation, ln Pvap = (-[itex]\Delta[/itex]H/R)(1/T) + ln [itex]\beta[/itex] into something like this (ln101.32 kPa)(-33444 J/mol [itex]\div[/itex] 8.413 J/mol * K)[itex]\div[/itex]1 - ln16.10 = T but I got something insane like 21857.7 K which even with subtracting 273 to get C is no where near close to the answer choices provided.
Any help with the last part would be greatly appreciated.
I already solved for heat of vaporization thus:
(-4058.7K)(-8.314 J/mol [itex]\ast[/itex]K)=heat of vaporization = 33444 J/mol = 33.44 kJ/mol.
Also, the vapor pressure for normal boiling point:
(1atm)(101325 Pa/1atm)(1 kPa/1000 Pa)=101.32 kPa
I attempted to solve for T by rearranging the linear form of the Clausius-Clapeyron equation, ln Pvap = (-[itex]\Delta[/itex]H/R)(1/T) + ln [itex]\beta[/itex] into something like this (ln101.32 kPa)(-33444 J/mol [itex]\div[/itex] 8.413 J/mol * K)[itex]\div[/itex]1 - ln16.10 = T but I got something insane like 21857.7 K which even with subtracting 273 to get C is no where near close to the answer choices provided.
Any help with the last part would be greatly appreciated.