Answer the following questions (25 points) (a) Discuss the physical reason(s) as
ID: 593247 • Letter: A
Question
Answer the following questions (25 points) (a) Discuss the physical reason(s) as to why the chemical potential of a pure substance decreases (b) 1. with increasing temperature. (10) The enthalpy of vaporization of a certain liquid was determined at its normal boiling point of 182 K as 14.37 k]/mol. The molar volumes of the liquid and vapor at the boiling points were determined to be 115 cm/mol and 14.6 dm2/mol respectively. i. Use the Clapeyron equation to estimate dp/dT. (5) i. The percentage error in dp/dT if the Clausius-Clapeyron equation were used instead. (5) (c) Suggest likely physical explanation(s) for the error in b(ii). (5)Explanation / Answer
1.a) For any system where the thermodynamic limit exists, we know that the internal energy U, the entropy , the total particle number N and the total volume V are all extensive. Because of this, we know that the Euler relation holds true
U = -PV + + N
and that the chemical potential is just the Gibbs free energy per particle.
G = N
We can use the differential relation of the Gibbs free energy
dG = V dP - d + dN to find expressions for the derivatives of the chemical potential.
d/d = 1/N * dG/d =-/N d/dP = 1/N * dG/dP = V/N
Since N, , and V are all positive, it appears that the chemical potential must increase with pressure at constant temperature, and decrease with temperature at constant pressure. My question is this: the 1D fermi gas is a case where the chemical potential actually increases with temperature for small temperatures, though we would think in all cases the chemical potential should be decreasing with temperature.
b) 1. clapeyron equation dp/dT = (DH)/(T*(Vvapor-Vliquid))
dp/dT = (14.37)/((14600-115)*182)=0.00545 J/cm3 = 5.45 KJ/m3
2. Clausius clapeyron dp/dT = (DH*p)/(R*T^2) = (143700*101325)/(8.314*(182^2))= 5.287KJ/m3
percentage error = (1-(5.287/5.45))*100=2.9 %
c) There are two approximations gives physical explanation to the error
1.The first approximation comes from noting that the volume of a given amount of gas is much larger that the volume of an equivalent amount of solid or liquid.
2. Second approximation we replace Vgas by the volume the gas would have if it were ideal.