Unformatted text preview: nknowns will give the point of intersection of the sublimation and vaporization coexistence curve. This leads to T = 175.43 K and p = 0.87 kPa Question 16: (5 points) Which among the following expressions is a correct representation of dS? a) dS = (Cp /T) dT + αV dP b) dS = (CV/T) dT + (κT/α) dV c) dS = (Cp /T) dT  (α/kT) dP d) dS = (CV/T) dT  (κT/α) dV e) dS = (CV/T) dT + (α /kT) dV f) dS = (CV/T) dT + αV dP g) dS = (Cp /T) dT + (α/kT) dV h) dS = (CV/T) dT  αV dP Show your derivations for any credit. dS can be obtained either from the two equations for dU or from the two equations for dH. In the present case, the dU equations given the correct expression for dS. dU = CV dT + (T(∂P/∂T)V – P) dV and (∂P/∂T)V = α/kT dU =  p dV + T dS lead to TdS = CV dT + T (α/kT) dV dS = (CV/T) dT + (α/kT) dV Question 17: (5 points) The change in free energy of a certain process is equal to 4.00 kJ at 400.K. Given that the process is at equilibrium at 200.K and that the entropy change and enthalpy change associated with this process are independent of temperature, calculate the free energy change for this process at 337 K. a) 950. J b) 1080 J c) 1260 J d) 1930 J e) 2740 J f) 4500 J g) 5420 J h) 6280 J Show your calculations. All we need to write is ΔG = ΔH – TΔS at the two temperatures. 4000 J = ΔH – 400.K x ΔS 0 J = ΔH – 200.K x ΔS (since there is equilibrium at 200 K) We solve for ΔH and ΔS and obtain: ΔS =  20 J/K ΔH =  4000 J So at 337 K, we get: ΔG = ΔH – TΔS =  4000 J – ( 20 J/K) x 337 K = 2740 J Question 18: (5 points) Which equation correctly describes the change in the vapor pressure above a pure liquid with temperature? Show your derivation for any credit! !P $ Δ H ! 1 1$ # P1 & T1 &
€ vap
a) l n# P2 & = R # T  T &
#&
#
&
" 1%
" 2 1 % " %
"
% ensity Solid Density Liquid 3
3
(g/cm ) (g/cm ) 1.522 1.458...
View
Full Document
 Spring '07
 AREsker
 Physical chemistry, pH, Trigraph

Click to edit the document details