|
|
| Line 1: |
Line 1: |
| ==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>
| | say <math>V={v\over N}</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>
| | <math>Pv+{aN^2v\over v^2}-PNb-{aN^2Nb\over v^2}-NkT=0</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>
| | by multiplying both sides by <math>v^2</math> we get |
|
| |
|
| 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^3}+aN^2V-PNbv^2-aN^3b-NkTv^2=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 dividing both sides by <math>PN^2</math> we get |
|
| |
|
| | <math>{v^3\over N^3}+{av\over PN}-{bv^2\over N^2}-{ab\over P}-{kTv^2\over PN^2}=0</math> |
|
| |
|
| 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.
| | so |
|
| |
|
| ==Problem 2==
| | <math>V^3+V{a\over P}-V^2b-{ab\over P}-V^2{kT\over P}=0</math> |
|
| |
|
| ===Part a===
| | and combining terms we get |
|
| |
|
| We find the critical points for Volume and Temperature when
| | <math>V^3-V^2(b+{kT\over P})+V{a\over P}-{ab\over P}=0</math> |
| | |
| <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>{\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>
| |
| | |
| 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.
| |
| | |
| ===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>
| |
| | |
| <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>
| |
| 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>
| |
| | |
| Then reducing:
| |
| <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>
| |
| | |
| 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>
| |
| | |
| 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>
| |
| | |
| So this means that the critical exponent gamma is 1.
| |
Problem 1
Part a
say 
by multiplying both sides by 
we get
by dividing both sides by 
we get
so
and combining terms we get
