pressure (in mmHg) is needed to distill the pure acid at 65°C is 30.4mm of Hg.
Now,
The pressure at 25°C is 4.1 molar, So in order to covert this pressure to atm as follow.
4.1 mbar ×[tex]\frac{1 bar}{1000 mbar}[/tex]×[tex]\frac{1 atm}{1.01325 bar}[/tex] = 0.004025 atm
now ,
according to Clausius- Clapeyraon equation we have,
ln[tex]\frac{P_{2} }{P_{1} }[/tex] = -Δ([tex]\frac{H^{0} }{8.314}[/tex][tex]\frac{1}{298}[/tex] -[tex]\frac{1}{414.5}[/tex])
Where the values in bracket are the values of 1st and 2nd temperature in kelvin.
P₁ and P₂ are the initial and final pressure and R is the gas constant which has value of 8.314.
25°C = 298K
141.5°C =414.5 K
P₁ = 1 atm
P₂ = 0.00405 atm
Hence
-5.51 = Δ H° vap (1.13 × 10⁻⁴)
Δ H° vap = 4.88× 10⁴ J/mol
Hence,
enthalpy of vapor is 4.88× 10⁴ J/mol (equation 1 value)
Again,
In order to find pressure we have to substitute 1 atm for P₁
Here,
T₁ = 65°C = 338K
T₂ = 141.5°C = 414.5 K
R = 8.314 J/mol.K
Now,
Substitute the enthalpy value in equation 1 and calculate pressure at 65°C (338K) as follow :
ln[tex]\frac{P_{2} }{P_{1} }[/tex] = -Δ[tex]\frac{H^{0} }{R}[/tex]([tex]\frac{1}{T_{1} }[/tex] - [tex]\frac{1}{T_{2} }[/tex])
hence,
ln[tex]\frac{P_{2} }{P_{1} }[/tex]= -3.21
[tex]\frac{P_{2} }{1 atm}[/tex] = [tex]e^{-3.21}[/tex]
P₂ = 0.040 atm ×[tex]\frac{760 mm Hg}{1 atm}[/tex]
= 30.4 mm of Hg
From the above conclusion we can say that, pressure (in mmHg) is needed to distill the pure acid at 65°C is 30.4mm of Hg.
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