As derived in class, the Clausius-Clapeyron equation for the liquid-gas phase bo
ID: 874184 • Letter: A
Question
As derived in class, the Clausius-Clapeyron equation for the liquid-gas phase boundary states that A, HT 1 1 POTL P(T expo with AvaH the enthalpy of vaporization (assumeit to be constant For sublimation, the same equation holds except that AraH must be replaced by AJLH, the sublimation enthalpy. The figure below shows schematically the phase diagram of benzene. Liquid Solid 0.006 H Vapor 273 373 Temperature (K) The following information is provi (i) The boiling temperature at one atmosphere is 353 K. (ii) The vapor pressure of liquid benzene at 20 0C is 1.2x 104 Pa. (iii) The vapor pressure of solid benzene at 44 0C is 137 Pa Civ) The enthalpy of fusion A s 9.95 oleExplanation / Answer
The calculations are as follows,
a) Given are,
P1 = 137 Pa
P2 = 1.2 x 10^4 Pa
T1 = -44 oC = -44 + 273 = 233 K
T2 = 20 oC = 20 + 273 = 293 K
R = 8.314 J/mol.K
Substituting all the values in the Clausius-Clapeyron equation we get,
ln(1.2 x 10^4/137) = (delta Hv/8.314) [(1/233) - (1/293)]
Solving for delta Hv,
delta Hv = 42315 J/mol
= 42.3 kJ/mol
Now let us find the delta Hvap at 353 K,
given values,
P2 = 1 atm
P1 = 1.2 x 10^4 Pa = 0.118 atm
T1 = 293 K
T2 = 353 K
R = 8.314 J/mol.K
Substitute the values we get,
ln(1/0.118) = delta Hvap/8.314 [(1.293) - (1/353)]
Solving for delta Hvap at 353 K we get,
delta Hvap = 30617 J/mol
= 30.62 kJ/mol
therefore, delta Svap = delta Hvap/T
= 30.62 / 353
= 0.09 kJ/mol
(b) Triple point is the temperature and pressure at which all three phases solid, liquid and vapor of a substance coexist in a thermodynamic equilibrium. Thus, at the triple point, delta Hfus + delta Hvap is almost equals to delta Hsublimation.
(c) The pressure at the triple point for the above system is 0.006 atm (value drawn from the graph). It is the reading o the presure scale were all three solid, liquid andvapor phases meet.
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