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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

Unknown macro: {dU}
Unknown macro: {T}

</math>

<br>

<math>\Delta S = \frac

Unknown macro: {Delta U}

</math>

<br>

<math>\Delta S = \frac

Unknown macro: {g}
Unknown macro: {298}

</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_

Unknown macro: {univ}

= \Delta S_

Unknown macro: {sys}

+ \Delta S_

Unknown macro: {env}

</math>

<br>

<math>\Delta S_

= \frac

Unknown macro: {298}

</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

Unknown macro: {1}
Unknown macro: {V}

\frac

Unknown macro: {partial M}
Unknown macro: {partial H}

</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

Unknown macro: {partial H}

\right )_

Unknown macro: {T, V}

dH + \left ( \frac

Unknown macro: {partial M}
Unknown macro: {partial V}

\right )_

Unknown macro: {T, H}

dV + \left ( \frac

Unknown macro: {partial T}

\right )_

Unknown macro: {H, V}

dT</math>

<br>

</center>

At constant temperature the above expression can be rearranged to yield the relation below.

<center>

<br>

<math>\frac

Unknown macro: {V}

\left ( \frac

Unknown macro: {partial M}
Unknown macro: {partial H}

\right )_

Unknown macro: {p, T}

= \frac

Unknown macro: {1}

\left [\left ( \frac

Unknown macro: {partial M}
Unknown macro: {partial H}

\right )_

Unknown macro: {T, V}

+ \left ( \frac

Unknown macro: {partial V}

\right )_

Unknown macro: {T, H}

\left ( \frac

Unknown macro: {partial H}

\right )_

Unknown macro: {T, p}

\right]</math>

<br>

</center>

The relation below follows.

<center>

<br>

<math>(\chi_p - \chi_v)_T = \frac

Unknown macro: {1}
Unknown macro: {V}

\left ( \frac

Unknown macro: {partial M}
Unknown macro: {partial V}

\right )_

Unknown macro: {T, H}

\left ( \frac

\right )_

Unknown macro: {T, p}

</math>

<br>

</center>

The quantity below is referred to as the magnetostriction.

<center>

<br>

<math>\lambda_

Unknown macro: {T,p}

= \frac

Unknown macro: {1}
Unknown macro: {V}

\left ( \frac

Unknown macro: {partial V}
Unknown macro: {partial H}

\right )_

</math>

<br>

</center>

The remaining derivative is \left ( \frac

Unknown macro: {partial M}
Unknown macro: {partial V}

\right )_

Unknown macro: {T, H}

. Start with the Maxwell relation below and reduce.

<center>

<br>

<math>\left ( \frac

Unknown macro: {partial V}

\right )_

Unknown macro: {T, H}

= \left ( \frac

Unknown macro: {partial p}
Unknown macro: {partial H}

\right )_

Unknown macro: {V, T}

</math>

<br>

<math> \left ( \frac

Unknown macro: {partial H}

\right )_

Unknown macro: {V, T}

= - \frac{ \left ( \frac

Unknown macro: {partial H}

\right )_{P,T}}{ \left ( \frac

Unknown macro: {partial V}
Unknown macro: {partial P}

\right )_{H, T}}</math>

<br>

<math>\left ( \frac

Unknown macro: {partial p}

\right )_

Unknown macro: {V,T}

= \frac{\lambda_

Unknown macro: {P, T}

}{\beta_{T, H}}</math>

<br>

</center>

The term \beta_

Unknown macro: {T,H}

is the isothermal compressibility under constant applied field and is expressed below.

<center>

<br>

<math>\beta_

Unknown macro: {T, H}

= - \frac

Unknown macro: {1}
Unknown macro: {V}

\left ( \frac

Unknown macro: {partial V}
Unknown macro: {partial p}

\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

Unknown macro: {T,P}

^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_

Unknown macro: {MH_2}

(T_o) + C_p^

Unknown macro: {(MH_2)}

(T_f - T_o )</math>

<br>

<math>H_i = H_M (T_o) + H_

Unknown macro: {H_2}

(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

Unknown macro: {60000}
Unknown macro: {40}

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_

Unknown macro: {Li}

dN_

+ \mu_

Unknown macro: {H}

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_

Unknown macro: {Li}

, \phi, N_H) = U -TS + Vp - \mu_

N_

Unknown macro: {Li}
  • Q \phi</math>

<br>

<math>d \wedge = -SdT + Vdp - N_

d \mu_

Unknown macro: {Li}
  • 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

Unknown macro: {partial wedge}
Unknown macro: {partial T}

\right ){p, \mu

Unknown macro: {Li}

, \phi, N_H}</math>

<br>

<math>V(T, p, \mu_

, \phi, N_H) = \left ( \frac

Unknown macro: {partial p}

\right ){T, \mu

Unknown macro: {Li}

, \phi, N_H}</math>

<br>

<math>-N_

(T, p, \mu_

Unknown macro: {Li}

, \phi, N_H) = \left ( \frac

Unknown macro: {partial wedge}

{\partial \mu_{Li}} \right )_

Unknown macro: {T, p, phi, N_H}

</math>

<br>

<math>Q_

(T, p, \mu_

Unknown macro: {Li}

, \phi, N_H) = \left ( \frac

Unknown macro: {partial wedge}
Unknown macro: {partial phi}

\right ){T, p, \mu

, N_H}</math>

<br>

<math>\mu_

Unknown macro: {H}

_

Unknown macro: {Li}

(T, p, \mu_

, \phi, N_H) = \left ( \frac

Unknown macro: {partial wedge}
Unknown macro: {partial N_H}

\right ){T, p, \mu

Unknown macro: {Li}

, \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

Unknown macro: {partial^2 wedge}
Unknown macro: {partial X_i partial X_j}

= \frac

Unknown macro: {partial X_j partial X_i}

</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

Unknown macro: {partial^2 wedge}
Unknown macro: {partial mu_i partial phi}

\right ) = - \left ( \frac

Unknown macro: {partial Q}
Unknown macro: {partial mu_i}

\right ) = - \left ( \frac{\partial N_{Li}}

Unknown macro: {partial phi}

\right )</math>

<br>

<math>\left ( \frac

Unknown macro: {partial phi^2}

\right ) = - \left ( \frac

Unknown macro: {partial Q}
Unknown macro: {partial phi}

\right )</math>

<br>

<math>\left ( \frac

Unknown macro: {partial^2 wedge}
Unknown macro: {partial p partial phi}

\right ) = \left ( \frac

Unknown macro: {partial V}

\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^

Unknown macro: {circ}

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>\Delta H</math> for evaporation <math>\frac
      Unknown macro: {41 kJ}
    </math>

<p>
</p>

To solve this problem find the temperature at which <math>\Delta G_

Unknown macro: {s right v}

= 0</math>.

<center>

<br>

<math>\Delta G_

= G_v - G_s</math>

<br>

<math>\Delta G_

Unknown macro: {s right v}

= G_v - G_l + G_l - G_s</math>

<br>

</center>

Use the approximations below.

<center>

<br>

<math>\Delta G_

Unknown macro: {s right l}

= \Delta H_

(1 - T/T_

Unknown macro: {s right l}

)</math>

<br>

<math>\Delta G_

Unknown macro: {l right v}

= \Delta H_

(1 - T/T_

Unknown macro: {l right v}

)</math>

<br>

</center>

Solve to find <math>T_

</math>.

<center>

<br>

<math>T_

Unknown macro: {s right v}

= \frac{ \Delta H_

Unknown macro: {s right l}

+ \Delta H_

Unknown macro: {l right v}

}{ \Delta H_

/ T_

Unknown macro: {s right l}

+ \Delta H_

Unknown macro: {l right v}

/ 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^

Unknown macro: {circ}

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

Unknown macro: {partial T}
Unknown macro: {partial l}

\right )_s > 0</math>

<br>

<math>\left ( \frac

Unknown macro: {partial l}

\right )_s = - \frac{ \left ( \frac

Unknown macro: {partial S}

\right )_T }{ \left ( \frac

Unknown macro: {partial S}
Unknown macro: {partial T}

\right )_l</math>

<br>

<math>\left ( \frac

Unknown macro: {partial l}

\right )_s = - \frac{ T \left ( \frac

Unknown macro: {partial l}

\right )_T }

Unknown macro: { C_l }

</math>

<br>

<math>\left ( \frac

Unknown macro: {partial S}

\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

Unknown macro: {partial U}
Unknown macro: {partial l}

\right )_s = f</math>

<br>

</center>

The relation above holds for any mechanically stable thermodynamic system

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