Answer :
The boiling point of methanol when the external pressure is 0.730 atm is 378.78 K
T₁=338K
ΔH=38.8kJ/mol
P₁=1atm
P₂=0.73atm
we know that,
ln (P₂/P₁)=ΔH/R(1/T₁-1/T₂)
where,
T₁ is initial temperature
T₂ is final temperature
P₁ is initial pressure
P₂ is final pressure
on substituting value we get
T₂ = 378.78 K
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