Question

The normal boiling point of cyclohexane is 80.74 °C.
At 20.00 °C, the vapour pressure of cyclohexane is 0.1026 atm.
What is the vapour pressure of cyclohexane when the temperature is 30.00 °C?

Answer 1

We have to solve this problem by using integrated form of Clausius-Clapeyron equation....

Given, At, T = 20°C . P = 0.1026 atue l = 0.1026 atu 220 +273 l 293k T = 30.00'c ,Py = ? = (30+273)K = 303k And, at normal boiding poing To 1 To=80.74°C The rapour pressure, (80.74 +273k Py = 1 atm 3 353.74 k we know the integrated form of claussion- elaypeyrowaequation, di ( 12 ) = Oh (+ / - / - 0 - Where ana heat of vaposization of e liquid R= 8.314 Julmont Now we have to use the value T = 2934 , P = 0.1026 atur To = 353.74k, P = latu in equation is to evaluate AH=? tanto luffon) = levering ati -to) or, la ce a fost atin) = Che cosaam 553170 or, ln(9.74) = 0 (0.003413 -0.002827) ore 2.276 = 4 x 0.00058667

or 2.276 = 1H X 0.000586 8.314] mol or AH = 2.276x8.314 I mot 0.000586 . : AH = 32291.23 I moll Now how turn value, E T=2934, P = 0.1026 atur Te = 3034, P2 = ? and an=32291223 I mail use this value in equo, we en conta do cs to) ox, heu le = 322.01,23 (1 2032 2033) = 3883.958 (0.003413–0.003300) = (3883.958 x 0.000113) = 0.4389 0.4389 0.1026 atne e = 1.551 :P2 = (4.551 %0.1026) atma 20.1591 afur : vapour pressure of cyclohexane at 30°c is p=0.1591 atar