Question

5. vo Search Screenshots Use the P and P, values from question 4 and Raoult's Law to estimate values of P and P and compare with the CRC Handbook values in question 3. (Questions 4 and 5 show that the phase diagram contains vapor pressure data for each pure component at each Photos - 2018-11-03 (7).png temperature. These data are not always easy to find elsewhere.) See all photos P.-PN Add to Q ♡ ♡ ♡ P. - PN Value for P: Value of Show your work: Os - 2018-11-03 (8).png 4. P. and P, can be approximately determined graphically from the phase diagram in problem 1 above using Dalton's law of partial pressures. 100 90 80 70 60 50 40 30 20 10 O mol % cyclopentane 0 10 20 30 40 50 60 70 80 90 100 mol % glyme P.-P. and P.-P.. Dalton's Law: P.-P. + P.-P.N + P...NY where N and Ny are the vapor phase mole fractions of cyclopentane and glyme, respectively. Assume this time that P -760 torr (sea level). Determine sea level values of P, and P, for the boiling 20 moles cyclopentane / 80 mole % glyme (liquid) mixture.

Answer 1

Gole Triven, A 150 100 90 80 90 60 50 40 30 &010 O mol 0 10 20 30 40 50 60 70 80 90 100 mol of Cyclopentane glyme Dalton's law of partial poessures. Pe = P to Me and Pag Plot Ng Dalton's Law: Prot=Pc + Pg = Plot Ne + Plot Ng where ne and Ng are the vapor phas mole fractions of Cyclopentane and glyme, Plotal = 760 ton (sea level) tot

Determine Sea level values of Pe and Pg for the boiling 20 mole of Cyclopentane 80 mole of glyme Ciquid) mixtyse. -0.2 - Ptotal = 760 toor → Mole fraction of Cyclo Everyonce Pentane. - 20/100 = 0.2 → Mole fraction or glyme & so/100 = 0.8 So, Partial pressure Of Cyclobecome (PC) P 760 tooo x 0.2 20.8 Pentane 152 toro Partial pressure of glyme (Pg) Pq: 760 tour X0.8 O.G08 toso

5 Question 4. Pc and Pg values from - Pc = 152 tor = 608 toro Pe = Pienc So, Pe=Pc/ne - les 153/0.2 Peo s 160 toor Pg - Poganing So, Pojo - Byling -= 608/0.8 Bg" = 760toos Por 760 toor Pg. 760 toor