We have that the vapor pressure at 139.0 ∘C. is mathematically given as
p=0.264atm
From the question we are told
Benzaldehyde, C6H5CHO, has a normal boiling point of 179.0 ∘C and a critical point at 422 ∘C and 45.9 atm.
Estimate its vapor pressure at 139.0 ∘C.
Generally the equation for the is mathematically given as
[tex]ln(\frac{p2}{p1})=\frac{dHvap}{r}(\frac{t_2-t-1}{t1t2})\\\\Therefore\\\\ln(\frac{45.9}{1})=\frac{dHvap}{8.314}(\frac{695-452}{695*452})\\\\dHvap=41057.96J/mol\\\\Hence\\\\ln(\frac{1}{p})=\frac{41057.96}{8.314}\\\\[/tex]
p=0.264atm
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