Recent content by swmmr1928

I agree that Chapter 10 does not describe any examples with deviations from Raoult's Law. I was not clear in my first post. Raoult's Law for Binary system at Constant Temperature P_{t}(x)=P_{2}^{sat}+(P_{1}^{sat}-P_{2}^{sat})x_{1} \\ P_{t}(y)=\frac{1}{y_{1}/P_{1}^{sat}+y_{2}/P_{2}^{sat}}...

The book I am reading, Smith Van Ness Abbott has several figures of Pressure vs Composition for Vapor Liquid Equilibrium of a Binary system. It often includes a dashed straight line to represent Raoult's Law. What confuses me is that only the liquid phase ( P-x1 ) is said to exhibit...

I don't understand how some terms are derived. How did the last term of 3-47 originate? How did 3-49 get so many terms from just one term in 3-41? Why integrate from V to infinite? That is not intuitive. Thas a functions are unusual because the absolute values of U,H,S cannot be computed...

##\frac {\partial U} {\partial V}_T=\sum\nolimits_{i=1}^n \frac {\partial E_i*p_i} {\partial V}\overset{Product rule}=\sum\nolimits_{i=1}^n E_i\frac {\partial p_i} {\partial V}+p_i\frac {\partial E_i} {\partial V}## ##E_i*\frac {\partial p_i} {\partial V}\overset{From post 40}=-p_i E_i*[ \frac...

I'll back up: ## E_i*\frac {\partial p_i} {\partial V}=-p_i E_i*[ \frac {\partial ln(Q)} {\partial V} + B*\frac {\partial E} {\partial V}]=-p_i E_i*[ B*\overline{P} - B*\overline{P}] =0 ##Right?

##\sum\nolimits_{i=1}^n -E_i p_i \overline{P} - U*\overline{P}=1/B*[{\overline{P}-\sum\nolimits_{i=1}^n p_i*\overline{P}}]##Wrong: ##-\overline{P}[\sum\nolimits_{i=1}^n E_i p_i + U]=1/B*[{\overline{P}}-\sum\nolimits_{i=1}^n p_i*\overline{P}]## ##-\overline{P}[\sum\nolimits_{i=1}^n E_i p_i +...

## \frac {\partial ln(Q)} {\partial V}=\overline{P}*β ## ## \frac {\partial E_i} {\partial V}=-\overline{P} ## ## \sum\nolimits_{i=1}^n E_i p_i *(-\frac {\partial E_i} {\partial V}) - U*\overline{P}=kT*[{\overline{P}+\sum\nolimits_{i=1}^n p_i*\frac {\partial E_i} {\partial V}}-p_i*E_i*[\frac...

##e^{-B*E_i}*\frac {\partial} {\partial V} \frac {1} {\sum\nolimits_{j=1}^n e^{-B*E_j}}=e^{-B*E_i}*\frac {\partial} {\partial V} \sum\nolimits_{j=1}^n e^{B*E_j}= e^{-B*E_i}*\sum\nolimits_{j=1}^n B*e^{B*E_j}*\frac {\partial E_j} {\partial V}## Edit: Maybe I should avoid expressing Q explicitly...

##\frac {\partial p_i} {\partial V}=B*E_i-(B/Q)*e^{-B*E_i}*\frac {\partial E_i} {\partial V}##

## \frac {\partial p_i} {\partial V}=exp(-B*E_i)*\frac {\partial} {\partial V} \frac{1} {Q} + 1/Q* \frac {\partial exp(-B*E_i)} {\partial V}## ## \frac {\partial exp(-B*E_i)} {\partial V}= exp(-B*E_i)*(-B)* \frac{\partial E_i}{ \partial V}## This B will hopefully cancel the kT Sum of ##\frac...

##p_i \frac {\partial E_i} {\partial V} + E_i \frac {\partial p_i} {\partial V}## ##p_i=exp(-B*E_i)/Q(N,T,V)=exp(-B*E_i)/exp(-B*E_i)## ##\frac {\partial p_i} {\partial V}=0## ##\frac {\partial {\sum\nolimits_{i=1}^n E_i p_i}} {\partial V}=\sum\nolimits_{i=1}^n p_i \frac {\partial E_i} {\partial V}##

If I use the expression for U that expresses the probability in E_i and exp(-BE_i) and Q(N,V,T) then Product and quotient rule?

Instead it is ##\frac {\partial {\sum\nolimits_{i=1}^n E_i p_i}} {\partial V}## And the other ##\sum\nolimits_{i=1}^n E_i p_i*(-\frac {\partial E_i} {\partial V})## Can these be differentialted? If both the probability and energy are functions of Volume, i can use product rule

Compare this ##\sum\nolimits_{i=1}^n E_i *(-\frac {\partial E_i} {\partial V})*p_i## to ##{\frac {\partial U} {\partial V}}_{N,T}=\frac {\partial} {\partial V} { {\sum\nolimits_{i=1}^n E_i}}## ##\sum\nolimits_{i=1}^n E_i *(-\frac {\partial E_i} {\partial V})*p_i## ##\sum\nolimits_{i=1}^n \frac...

##\sum\nolimits_{i=1}^n -\frac {\partial E_i} {\partial V}*E_i*p_i=\sum\nolimits_{i=1}^n -\frac {\partial E_i} {\partial V}*E_i*\frac {e^{-β*E_i}} {Q(N,V,T)}## ##{\frac {\partial U} {\partial V}}_{N,T}## ##U=\frac{\sum\nolimits_{i=1}^n E_i*e^{-E_i/kT}} {Q}## ##Q=\sum\nolimits_{i=1}^n...