You are experimenting with an ethanol and water mixture with a mole fraction of

Question

You are experimenting with an ethanol and water mixture with a mole fraction of ethanol equal to 0.39. See graph.

Based on the equation above, what is the value of the correct ?Hvap to use in this experiment?

You measure the starting pressure and temperature of your system as 1.08 atm and 24.0°C, respectively. What is the pressure of the ethanol-water system at a new temperature of 34.7°C?

AHrap (k/mol) Hvap versus Xethanol 42.5 42.0 41.5 41.0 40.5 40.0- 39.5 39.0 y-3.7052x+ 42.516 R" = 0.9663 0.2 0.4 0.6 0.8 1.0 Mole Fraction of Ethanol

Explanation / Answer

when x= 0.3 delH= -3.7052*0.3+42.516 =41.40444 Kj/mol

From Clasius Clayperon equation ln (P2/P1) =(delH/) *(1/T1-1/T2)

P2 to be evaluted at T2= 34.7 deg.c =34.7+273.15=307.85K

P1=1.08 atm and T1- 24+273.15 K= 297.15K R= gas constant =8.314 J/mol.K

ln (P2/1.08)= (41.40444*1000/8.314)*(1/297.15-1/307.85) =0.582513

P2/1.08= exp(0.582513)= 1.791

P2 =1.08*1.791= 1.933 atm

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.