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SOLUTIONSMANUAL TO ACCOMPANY ATKINS' PHYSICAL CHEMISTRY 125
80.34 80.35 80.36 80.37 80.38 80.39 80.40
0.0
0.5
1.0
VA/(cm3mol
−1)
x A
/(
1
−
x A
)
Figure 5.3
�e data fall on a good straight line and therefore the required integral is simply
the area of the triangle (shown shaded) bounded by xA/(1−xA) = 0 and xA/(1−
xA) = 1; the corresponding values ofVA/(cm3mol
−1) are 80.3406 and 80.3911.
�e area is 12 × base × height
area = 1
2 (80.3911 − 80.3406) × (1 − 0) = 0.02525 cm3mol−1
�e partial molar volume of B is therefore
VB = V∗B − area = 73.99 − 0.02525 = 73.96 cm3mol
−1
P5A.4 �e partial pressure of component A in the vapour is given by pA = yAp,
where p is the total pressure and yA the mole fraction in the vapour.�e mole
fractions of the two components are related by xB = 1 − xA. Using these rela-
tionships the following table is drawn up, and the data are plotted in Fig. 5.4.
xA yA p/(kPa) pA/(kPa) xB yB pB/(kPa)
0 0 36.066 0 1 1 36.066
0.089 8 0.041 0 34.121 1.399 0.910 2 0.959 0 32.722
0.247 6 0.115 4 30.900 3.566 0.752 4 0.884 6 27.334
0.357 7 0.176 2 28.626 5.044 0.642 3 0.823 8 23.582
0.519 4 0.277 2 25.239 6.996 0.480 6 0.722 8 18.243
0.603 6 0.339 3 23.402 7.940 0.396 4 0.660 7 15.462
0.718 8 0.445 0 20.698 9.211 0.281 2 0.555 0 11.488
0.801 9 0.543 5 18.592 10.105 0.198 1 0.456 5 8.487
0.910 5 0.728 4 15.496 11.287 0.089 5 0.271 6 4.209
1 1 12.295 12.295 0 0 0