Solution to Set 2: Difference between revisions

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==Problem 1==
=Problem 1=
First one is to find the isothermal compressibility of a Van der Waals gas for <math>T > T_c\;</math>.


The Van der Waals equation of state is: <math> \left(P + {N^2a \over V^2}\right) \left(V - Nb\right) = Nk_BT</math>
==Part a==


Solving this for P gives: <math> P = \left({Nk_BT \over V-Nb}\right) - {N^2a \over V^2}</math>
<math>(P+{{aN^2}\over v^2})(v-Nb)=NkT</math>


Then taking the partial derivative with respect to V at constant T: <math> \left( {\partial P \over \partial V} \right)_T =  -{ Nk_BT \over \left( {V-Nb} \right)^2 } + { 2N^2a \over V^3}</math>


Bringing the terms over a common denominator looks like: <math> \left( {\partial P \over \partial V} \right)_T = {2N^2a \left( {V-Nb} \right)^2 - Nk_BTV^3 \over V^3 \left( {V-Nb} \right)^2} </math>
say <math>V={v\over N}</math>


Then finding the negative reciprocal of this function gives the isothermal compressibility: <math> \kappa_T = -\left( {\partial V \over \partial P} \right)_T = - {1 \over {\left( {\partial P \over \partial V} \right)_T}} = {V^3 \left( {V-Nb} \right)^2 \over Nk_BTV^3 - 2N^2a\left( {V-Nb} \right)^2 } </math>


For those of you wondering why the 1/V is missing in the isothermal compressibility equation (it was added to the homework around 5 PM the day the homework was due and is there now), the answer is because it depends on the result desired. The formula used here to solve for the isothermal compressibility gives the total volume change per change in pressure. However, should that 1/V be kept it would give the fractional change in volume per change in pressure. For completeness, the result for the fractional isothermal compressibility is:
<math>Pv+{aN^2v\over v^2}-PNb-{aN^2Nb\over v^2}-NkT=0</math>


<math> \kappa_T = -{1 \over V} \left( {\partial V \over \partial P} \right)_T = - {1 \over V} {1 \over {\left( {\partial P \over \partial V} \right)_T}} = {V^2 \left( {V-Nb} \right)^2 \over Nk_BTV^3 - 2N^2a\left( {V-Nb} \right)^2 } </math>


by multiplying both sides by <math>v^2</math> we get


You can see from this graph how as the temperature approaches the critical temperature for whatever material is being examined, <math> \kappa_T </math> begins to spike at <math> V = V_c </math>. For this material, air, <math> V_c = 3Nb = </math> 1.092*10^-4.


==Problem 2==
<math>{Pv^3}+aN^2V-PNbv^2-aN^3b-NkTv^2=0</math>


===Part a===


We find the critical points for Volume and Temperature when
by dividing both sides by <math>PN^2</math> we get


<math>\frac {\partial P} {\partial V} = 0</math> and <math> \frac {\partial^2 P} {\partial V^2} =0 </math>


<math> {\partial P \over \partial V} =  -{ Nk_BT \over \left( {V-Nb} \right)^2 } + { 2N^2a \over V^3}=0</math>
<math>{v^3\over N^3}+{av\over PN}-{bv^2\over N^2}-{ab\over P}-{kTv^2\over PN^2}=0</math>


<math>{\partial^2 P \over \partial V^2}  =  { Nk_BT \over \left( {V-Nb} \right)^3 } - { 2N^2a \over V^4}=0</math>


Using the two equations to solve for <math>V_c</math> we find that <math>V_c=3 N b</math>
so


Plugging <math>V_c</math> into the first equation it is found that <math>k_B T_c={8 a \over 27 b}</math>


These two critical points are all that is necessary to solve for in order to find the isothermal compressibility.
<math>V^3+V{a\over P}-V^2b-{ab\over P}-V^2{kT\over P}=0</math>


===Part b===
In this part of the homework one is to find that <math>\kappa \left(t\right)</math> is proportional to <math>\left(T - T_c\right)^{-\gamma}</math>.
First remember the identities for the critical volume and temperature:


<math>V_c = 3Nb\!</math>
and combining terms we get


<math>T_c = {8a \over 27 k_B b}</math>


Recall from part 1 that <math> P = \left({Nk_BT \over V-Nb}\right) - {N^2a \over V^2}</math>
<math>V^3-V^2(b+{kT\over P})+V{a\over P}-{ab\over P}=0</math>
Then taking the partial derivative with respect to V at constant T:
<math> \left( {\partial P \over \partial V} \right)_T =  -{ Nk_BT \over \left( {V-Nb} \right)^2 } + { 2N^2a \over V^3}</math>


Here is the best point to evaluate this function at the critical volume and pressure. This changes the function to:
<math> \left( {\partial P \over \partial V} \right)_T =  -{ Nk_BT \over \left( {V_c-Nb} \right)^2 } + { 2N^2a \over V_c^3}</math>


Then one can use the identity for <math>V_c</math> to get:


<math> \left( {\partial P \over \partial V} \right)_T =-{ Nk_BT \over \left( {V_c-Nb} \right)^2 } + { 2N^2a \over V_c^3} = -{ Nk_BT \over \left( {3Nb-Nb} \right)^2 }  + { 2N^2a \over \left(3Nb\right)^3}</math>
==part b==


Then reducing:
<math>P={nRT\over (v-nb)^2}-{an^2\over v^2}</math>
<math>-{ Nk_BT \over \left( {3Nb-Nb} \right)^2 } + { 2N^2a \over \left(3Nb\right)^3} = -{ k_BT \over 4Nb^2 } + { 2a \over 27Nb^3} </math>


At this point we can take <math> {k_B \over 4Nb^2} </math> out of both fractions:


<math> {k_B \over 4Nb^2} \left( T - {8a \over 27k_Bb} \right) </math>
Taking the derivative we get


If one looks at the equation for the critical temperature one can see that one has:


<math> {k_B \over 4Nb^2} \left( T - {8a \over 27k_Bb} \right) = {k_B \over 4Nb^2} \left( T - T_c \right)</math>
<math>{dP\over dv}={-nRT\over v-nb}+{2an^2\over v^3}=0</math>


Now one just needs to solve for the compressibility which is the negative reciprocal of our result (as seen in part 1).


<math> \kappa_T = -\left( {\partial V \over \partial P} \right)_T = {1 \over {k_B \over 4Nb^2} \left( T - T_c \right)} = {4Nb^2 \over k_B} \left( T - T_c \right)^{-1}</math>
Multiply the derivative by <math>{-2\over v-nb}</math> to get


So this means that the critical exponent gamma is 1.
 
<math>{2nRT\over (v-nb)^3}-{4an^2\over v^3(v-nb)}=0</math>
 
 
Taking the derivative again we get
 
 
<math>{d^2P\over d^2v}={2nRT\over (v-nb)^3}-{6an^2\over v^4}=0</math>
 
 
By setting the derivatives equal to each other we get
 
 
<math>{2nRT\over (v-nb)^3}-{4an^2\over v^3(v-nb)}={2nRT\over (v-nb)^3}-{6an^2\over v^4}</math>
 
 
Which reduces to
 
 
<math>6(v-nb)=4v</math>
 
 
<math>6v-6nb=4v</math>
 
 
<math>6nb=2v</math>
 
 
<math>3nb=v_c</math>
 
 
 
Now we can say
 
 
<math>{-nRT\over v_c-nb}+{2an^2\over v_c^3}=0</math>
 
 
Plugging <math>v_c</math> in we get
 
 
<math>{-nRT\over (3nb-bn)^2}+{2an^2\over (3nb)^3}=0</math>
 
 
Now we solve for T
 
 
<math>{-nRT\over (2bn)^2}+{2an^2\over 27n^3b^3}=0</math>
 
 
<math>{-RT\over 4b^2n}+{2a\over 27nb^3}=0</math>
 
 
<math>{2a\over 27nb^3}={RT\over 4b^2n}</math>
 
 
<math>{8a\over 27b}=RT</math>
 
 
<math>{8a\over 27bR}=T_c</math>
 
 
 
 
Now <math>T_c</math> and <math>v_c</math> can be plugged in to find <math>P_c</math>
 
 
say
 
 
<math>P={nR({8a\over 27bR})\over 3nb-nb}-{an^2\over (3nb)^2}</math>
 
 
<math>P={8na\over 27b(2nb}-{a\over 9b^2}</math>
 
 
<math>P={4a\over 27b^2}-{a\over 9b^2}</math>
 
 
<math>P={4a\over 27b^2}-{3a\over 27b^2}</math>
 
 
<math>P={a\over 27b^2}</math>
 
 
<math>P_c={a\over 27b^2}</math>
 
 
 
==Part C==
 
Use <math>T_c={8a\over 27bR}, v_c=3nb, P_c={a\over 27b^2}, T_r={T\over T_c}, n=1</math>
 
 
and
 
 
<math>P={RT\over v-b}-{a\over v^2}</math>
 
 
so
 
 
<math>P_rP_c={RT_rT_c\over v_rv_c-b}-{a\over (v_rv_c)^2}</math>
 
 
<math>P_r({a\over 27b^2})={RT_r({8a\over 27bR})\over v_r(3nb)-b}-{a\over v_r^2(3nb)^2}</math>
 
 
<math>P_r({a\over 27b^2})={8aT_r\over 27b^2(3v_r-1)}-{a\over v_r^29b^2}</math>
 
 
<math>P_r({a\over 27})={8aT_r\over 27(3v_r-1)}-{a\over v_r^29}</math>
 
 
<math>P_r({a\over 27})+{a\over v_r^29}={8aT_r\over 27(3v_r-1)}</math>
 
 
<math>{P_r\over 27}+{1\over v_r^29}={8T_r\over 27(3v_r-1)}</math>
 
 
multiply both sides by <math>27(3v_r-1)</math>
 
 
<math>(3v_r-1)({P_r}+{3\over v_r^2})={8T_r}</math>
 
 
<math>(3({v\over v_c})-1)(({P\over P_c})+{3\over ({v_c\over v})^2})={8({T\over T_c})}</math>
 
 
also
 
 
<math>{P_cv_c\over T_c}={({a\over 27b^2})(3nb)\over {8a\over 27b}}</math>
 
 
<math>{P_cv_c\over T_c}={3n\over 8}</math>
 
 
 
==Part D==
 
<math>K=({dv\over dP})_T</math>
 
 
<math>K_T={1\over v}({dv\over dP})</math>
 
 
<math>{dP\over dv}</math> at <math>v_c={-nkT\over 2n^2b^2}+{2an^2\over 27n^3b^3}</math>
 
 
<math>={-kT\over 4nb^2}+{2a\over 27nb^3}</math>
 
 
<math>={1\over 4nb^2}(-kT+{8a\over 27b})</math>
 
 
<math>={1\over 4nb^2}({8a\over 27b}-kT)\approx T-T_c</math>

Latest revision as of 14:07, 30 April 2011

Problem 1

Part a

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (P+{{aN^2}\over v^2})(v-Nb)=NkT}


say Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V={v\over N}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Pv+{aN^2v\over v^2}-PNb-{aN^2Nb\over v^2}-NkT=0}


by multiplying both sides by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v^2} we get


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {Pv^3}+aN^2V-PNbv^2-aN^3b-NkTv^2=0}


by dividing both sides by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle PN^2} we get


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {v^3\over N^3}+{av\over PN}-{bv^2\over N^2}-{ab\over P}-{kTv^2\over PN^2}=0}


so


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V^3+V{a\over P}-V^2b-{ab\over P}-V^2{kT\over P}=0}


and combining terms we get


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V^3-V^2(b+{kT\over P})+V{a\over P}-{ab\over P}=0}


part b

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P={nRT\over (v-nb)^2}-{an^2\over v^2}}


Taking the derivative we get


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {dP\over dv}={-nRT\over v-nb}+{2an^2\over v^3}=0}


Multiply the derivative by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {-2\over v-nb}} to get


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {2nRT\over (v-nb)^3}-{4an^2\over v^3(v-nb)}=0}


Taking the derivative again we get


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {d^2P\over d^2v}={2nRT\over (v-nb)^3}-{6an^2\over v^4}=0}


By setting the derivatives equal to each other we get


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {2nRT\over (v-nb)^3}-{4an^2\over v^3(v-nb)}={2nRT\over (v-nb)^3}-{6an^2\over v^4}}


Which reduces to


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 6(v-nb)=4v}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 6v-6nb=4v}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 6nb=2v}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3nb=v_c}


Now we can say


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {-nRT\over v_c-nb}+{2an^2\over v_c^3}=0}


Plugging Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v_c} in we get


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {-nRT\over (3nb-bn)^2}+{2an^2\over (3nb)^3}=0}


Now we solve for T


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {-nRT\over (2bn)^2}+{2an^2\over 27n^3b^3}=0}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {-RT\over 4b^2n}+{2a\over 27nb^3}=0}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {2a\over 27nb^3}={RT\over 4b^2n}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {8a\over 27b}=RT}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {8a\over 27bR}=T_c}



Now Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v_c} can be plugged in to find Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_c}


say


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P={nR({8a\over 27bR})\over 3nb-nb}-{an^2\over (3nb)^2}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P={8na\over 27b(2nb}-{a\over 9b^2}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P={4a\over 27b^2}-{a\over 9b^2}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P={4a\over 27b^2}-{3a\over 27b^2}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P={a\over 27b^2}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_c={a\over 27b^2}}


Part C

Use Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T_c={8a\over 27bR}, v_c=3nb, P_c={a\over 27b^2}, T_r={T\over T_c}, n=1}


and


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P={RT\over v-b}-{a\over v^2}}


so


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_rP_c={RT_rT_c\over v_rv_c-b}-{a\over (v_rv_c)^2}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_r({a\over 27b^2})={RT_r({8a\over 27bR})\over v_r(3nb)-b}-{a\over v_r^2(3nb)^2}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_r({a\over 27b^2})={8aT_r\over 27b^2(3v_r-1)}-{a\over v_r^29b^2}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_r({a\over 27})={8aT_r\over 27(3v_r-1)}-{a\over v_r^29}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P_r({a\over 27})+{a\over v_r^29}={8aT_r\over 27(3v_r-1)}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {P_r\over 27}+{1\over v_r^29}={8T_r\over 27(3v_r-1)}}


multiply both sides by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 27(3v_r-1)}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (3v_r-1)({P_r}+{3\over v_r^2})={8T_r}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (3({v\over v_c})-1)(({P\over P_c})+{3\over ({v_c\over v})^2})={8({T\over T_c})}}


also


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {P_cv_c\over T_c}={({a\over 27b^2})(3nb)\over {8a\over 27b}}}


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {P_cv_c\over T_c}={3n\over 8}}


Part D

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K=({dv\over dP})_T}



at




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