Question 1
A spring with spring constant <math>k</math> is attached on one end to the top of a large insulated container filled with a gas at <math>298K</math>. A basket is connected to the other end of the spring. At rest, the basket hangs <math>1 m</math> above the bottom of the container. A ball of mass <math>m</math> is put in the basket while holding the basket in position. When the basket is let go, the weight of the ball initially pulls the spring down under the influence of gravity, but then the ball rolls out of the basket and falls to the bottom of the box. As a result, the spring wildly oscillates, but ultimately comes to rest. When the box has reached internal equilibrium, calculate the following changes in thermodynamic properties: ( with the initial state being the state where the basket with the ball is held at 1m above the bottom of the box)
<p>
</p>
DATA:
- For the spring the force <math>F = kx</math>, where <math>x</math> is the displacement from the elongation at which the spring has zero force, and <math>F = 50N/m</math>
- <math>m = 1 kg</math>
- The box is large so that its temperature can be considered to be approximately constant.
- No plastic deformation takes place in the ball when it falls
a) the change in the internal energy for the system consisting of the box, spring and mass.
<p>
</p>
Below is the first law.
<center>
<br>
<math>dU = \delta Q + \delta W</math>
<br>
<math> \delta Q = 0 </math>
<br>
<math>\Delta U = W</math>
<br>
<math>\Delta U = -mg \Delta h</math>
<br>
<math>\Delta U = g</math>
<br>
</center>
b) the change in the entropy for the above system
<center>
<br>
<math>\Delta S = \int \frac
</math>
<br>
<math>\Delta S = \frac
</math>
<br>
<math>\Delta S = \frac
</math>
<br>
</center>
c) the change in the entropy for the mass and spring
<p>
</p>
The thermodynamic state of the mass and spring do not change.
<center>
<br>
<math>\Delta S = 0</math>
<br>
</center>
d) the change in the entropy for the universe
<center>
<br>
<math>\Delta S_
= \Delta S_
+ \Delta S_
</math>
<br>
<math>\Delta S_
= \frac
</math>
<br>
</center>
Question 2
Calculate the difference between the constant pressure and constant volume magnetic susceptibilities of an isotropic material. Reduce the answer as much as possible into properties of the system. You may properties defined under constant applied field in the answer (rather than transforming them to properties under constant magnetization)
<p>
</p>
Note: for convenience you may want to absorb <math>\mu_o</math> in the units of <math>H</math> and define a magnetization <math>M</math> that is extensive in nature so that the contribution of the magnetization to the change in internal energy can be written as <math>dU = HdM</math>. In that case, the susceptibility is defined as:
<center>
<br>
<math>\chi = \frac
\frac
</math>
<br>
</center>
Start by considering the magnetization, <math>M</math>, a function of temperature, volume, and applied magnetic field. The problem statement does not indicate whether the constant temperature or constant entropy conditions are appropriate, so either choice is valid. Below is the total differential in this set of coordinates.
<center>
<br>
<math>dM(T, V, H) = \left ( \frac
\right )_
dH + \left ( \frac
\right )_
dV + \left ( \frac
\right )_
dT</math>
<br>
</center>
At constant temperature the above expression can be rearranged to yield the relation below.
<center>
<br>
<math>\frac
\left ( \frac
\right )_
= \frac
\left [\left ( \frac
\right )_
+ \left ( \frac
\right )_
\left ( \frac
\right )_
\right]</math>
<br>
</center>
The relation below follows.
<center>
<br>
<math>(\chi_p - \chi_v)_T = \frac
\left ( \frac
\right )_
\left ( \frac
\right )_
</math>
<br>
</center>
The quantity below is referred to as the magnetostriction.
<center>
<br>
<math>\lambda_
= \frac
\left ( \frac
\right )_
</math>
<br>
</center>
The remaining derivative is \left ( \frac
\right )_
. Start with the Maxwell relation below and reduce.
<center>
<br>
<math>\left ( \frac
\right )_
= \left ( \frac
\right )_
</math>
<br>
<math> \left ( \frac
\right )_
= - \frac{ \left ( \frac
\right )_{P,T}}{ \left ( \frac
\right )_{H, T}}</math>
<br>
<math>\left ( \frac
\right )_
= \frac{\lambda_
}{\beta_{T, H}}</math>
<br>
</center>
The term \beta_
is the isothermal compressibility under constant applied field and is expressed below.
<center>
<br>
<math>\beta_
= - \frac
\left ( \frac
\right )_
</math>
<br>
</center>
A result is below. A derivation of the same quantity under conditions of constant entropy would lead to an analogous answer replacing <math>T</math> with <math>S</math> in the relationship below.
<center>
<br>
<math>\left ( \chi_p - \chi_v \right )T = \frac{\lambda
^2}{\beta_{T, H}}</math>
<br>
</center>
Question 3
One mole of metal <math>M</math> (solid) and one mole of <math>H_2</math> gas are in an adiabatic enclosure under constant pressure of <math>1 atm</math>. They are initially at <math>298K</math>. What is the temperature of the system after the metal and hydrogen have reacted to form <math>MH_2</math>?
<p>
</p>
Data:
- <math>\Delta H_f</math> of <math>MH_2</math>: <math>-60 kJ/mole</math> of <math>MH_2</math>
- <math>C_p</math> of <math>H_2</math>: <math>\frac
Unknown macro: {7}Unknown macro: {2}R</math>
- <math>C_p</math> of <math>M</math>: <math>25 \frac
Unknown macro: {J}Unknown macro: {mol cdot K}
</math>
- <math>C_p</math> of <math>MH_2</math>: <math>40 \frac
Unknown macro: {mol cdot K}</math>
The system is an adiabatic enclosure under constant pressure. The enthalpy of the system is therefore constant.
<center>
<br>
<math>dH = \partial Q + V dp</math>
<br>
<math>dH = 0</math>
</center>
Consider a change in enthalpy
<center>
<br>
<math>\Delta H = 0</math>
<br>
<math>H_f - H_i = 0</math>
<br>
</center>
The final state of the system is one mole of <math>MH_2</math> at some final temperature.
<center>
<br>
<math>H_f = H_
(T_o) + C_p^
(T_f - T_o )</math>
<br>
<math>H_i = H_M (T_o) + H_
(T_o)</math>
<br>
</center>
Both <math>H_M(T_o)</math> and <math>H_
(T_o)</math> are the enthalpies of the metal and hydrogen at <math>T_o = 298 K</math>. Solve for <math>T_f</math>.
<center>
<br>
<math>T_f = T_o + \frac{-\Delta H_f}{C_p^
</math>
<br>
<math>T_f = 298 + \frac
K</math>
<br>
<math>T_f = 1798 K</math>
<br>
</center>
Question 4
<math>HLi_x</math> is an electrode material in a <math>Li</math> battery. It consists of a series of inactive elements (represented by <math>H</math>) and <math>Li</math>. The electrode material can exchange <math>Li^</math> ions with a solution that may be considered at constant chemical potential of <math>Li^</math>. In a potentiostatic experiment the electrode is also held under constant electrical potential difference with the electrolyte. The electrode is under constant temperature and pressure.
<p>
</p>
1) write the first law of thermodynamics for the electrode material as <math>dU = ...</math>
<center>
<br>
<math>dU = TdS - pdV + \mu_
dN_
+ \mu_
dN_
+ \phi dQ</math>
<br>
</center>
2) define the relevant thermodynamic potential for the electrode when it is in contact with a large bath of electrolyte containing <math>Li^+</math> and held under constant electrical potential.
<center>
<br>
<math>\wedge (T, p, \mu_
, \phi, N_H) = U -TS + Vp - \mu_
N_
- Q \phi</math>
<br>
<math>d \wedge = -SdT + Vdp - N_
d \mu_
- Q d \phi + \mu_
Unknown macro: {H}dN_H</math>
<br>
</center>
3) define the equations of state for this system
<center>
<br>
<math>-S(T, p, \mu_
, \phi, N_H) = \left ( \frac
\right ){p, \mu
, \phi, N_H}</math>
<br>
<math>V(T, p, \mu_
, \phi, N_H) = \left ( \frac
\right ){T, \mu
, \phi, N_H}</math>
<br>
<math>-N_
(T, p, \mu_
, \phi, N_H) = \left ( \frac
{\partial \mu_{Li}} \right )_
</math>
<br>
<math>Q_
(T, p, \mu_
, \phi, N_H) = \left ( \frac
\right ){T, p, \mu
, N_H}</math>
<br>
<math>\mu_
_
(T, p, \mu_
, \phi, N_H) = \left ( \frac
\right ){T, p, \mu
, \phi}</math>
<br>
</center>
4) how many independent properties are there for this system? Define at least three properties for this system which are not found in simple systems (simple systems are closed systems with only <math>pdV</math> work)
<p>
</p>
Properties of the system are the second partial derivatives of <math>\wedge</math>. There are a total of <math>5^2 = 25</math> partial derivatives, of which <math>10</math> can be eliminated using symmetry (Maxwell relations).
<center>
<br>
<math>\frac
= \frac
</math>
<br>
</center>
Properties that are not found in simple systems are any second partial derivative involving a variable not found in a simple system.
<center>
<br>
<math>\left ( \frac
\right ) = - \left ( \frac
\right ) = - \left ( \frac{\partial N_{Li}}
\right )</math>
<br>
<math>\left ( \frac
\right ) = - \left ( \frac
\right )</math>
<br>
<math>\left ( \frac
\right ) = \left ( \frac
\right )</math>
<br>
</center>
Because the variable <math>N_H</math>, the amount of "inactive" elements is fixed, you could treat it either as a parameter or a variable when specifying the first law (i.e., the system is closed to exchange of <math>H</math>). For example, you may only be interested in the potential <math>\wedge</math> "per mole <math>H</math>" treating <math>N_H</math> as a parameter. In this case only <math>4^2 - 6 = 10</math> independent parameters would remain. Answers of both types were considered valid.
Question 5
At <math>1 atm</math> pressure the melting point of water is <math>0^
C</math> and the boiling point is <math>100^
C</math>. Calculate the metastable sublimation point of solid <math>H_2O</math> *(i.e. the temperature at which the solid would evaporate if it could be prevented from melting first).
'''<p>
</p>
Data:
- <math>\Delta H</math> for melting <math>6 \frac
Unknown macro: {kJ}Unknown macro: {mole}</math>
</math>
- <math>\Delta H</math> for evaporation <math>\frac
Unknown macro: {41 kJ}
- <math>\Delta H</math> for evaporation <math>\frac
<p>
</p>
To solve this problem find the temperature at which <math>\Delta G_
= 0</math>.
<center>
<br>
<math>\Delta G_
= G_v - G_s</math>
<br>
<math>\Delta G_
= G_v - G_l + G_l - G_s</math>
<br>
</center>
Use the approximations below.
<center>
<br>
<math>\Delta G_
= \Delta H_
(1 - T/T_
)</math>
<br>
<math>\Delta G_
= \Delta H_
(1 - T/T_
)</math>
<br>
</center>
Solve to find <math>T_
</math>.
<center>
<br>
<math>T_
= \frac{ \Delta H_
+ \Delta H_
}{ \Delta H_
/ T_
+ \Delta H_
/ T_
</math>
<br>
<math>T_
\approx 356 K</math>
<br>
</center>
Note that at constant pressure the transistion temperature must be bracketed between <math>0^
C</math> and <math>100^
C</math>.
Question 6
I am trying to determine the properties of a new solid NewSol. When it is stretched adiabatically, its temperature increases.
<p>
</p>
If NewSol is stretched isothermally does its entropy increase or decrease?
<center>
<br>
______S increases ___X__S decreases
<br>
<math>\left ( \frac
\right )_s > 0</math>
<br>
<math>\left ( \frac
\right )_s = - \frac{ \left ( \frac
\right )_T }{ \left ( \frac
\right )_l</math>
<br>
<math>\left ( \frac
\right )_s = - \frac{ T \left ( \frac
\right )_T }
</math>
<br>
<math>\left ( \frac
\right )_T < 0</math>
<br>
</center>
If NewSol is stretched adiabatically, does its internal energy decrease, increase, or stay the same
<center>
<br>
_X_U increases ______U decreases _____U remains the same
<br>
<math>\left ( \frac
\right )_s = f</math>
<br>
</center>
The relation above holds for any mechanically stable thermodynamic system