The vapor pressure, P, of a certain I was measured at two temperatures, T If you

Question

The vapor pressure, P, of a certain I was measured at two temperatures, T If you were going to graphically determine the enthalpy of vaporizaton, AHvp, for this liquid, what points would you plot? The data is shown in this table. Number T(K) P (kPa) 325 3.98 875 5.10 To avoid rounding errors, use three significant figures in the x values and four point 1: Number NOTE: Keep the pressure units in kPa. the y values point 2: Determine the rise, run, and slope of the line formed by these points. ise run slope Number Phat is the enthalpy of vaporization of this liquid? Number J/ mol

Explanation / Answer

The given data

T(K). . . P(kPa )
325 . . . 3.98
875 . . . 5.10

For graphically determination   

From the Clausius-Clapeyron equation

ln(P2/P1) = Hvap/R (1/T1 – 1/T2)

. . . . . . . . . . . . . . X. . . . . . . . . . . . . . . . . . . .Y . . . . . .
Point 1 = 1/325 = 0.00308*** . . . ln(3.98) = 1.381**

Point 2 = 1/875 = 0.00114*** . . . .ln(5.10) = 1.629**

Rise y2 – y1
= 1.629 - 1.381 = 0.248

Run x2 –x1
= 0.00114 – 0.00308 = –0.00194

Slope = rise / run
0.248 / (–0.00194) = –127.835

ln(P2/P1) = Hvap/R (1/T1 – 1/T2)

ln(5.10 / 3.98) = Hvap/8.314 (1/325 – 1/875)

0.2479 = (Hvap/8.314) (0.00193)

Hvap = 1068.15 J/mol

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