Question 1
A 100 ohm resistor is held at a constant temperature of 300 A current of 10 Amperes is passed through the resistor for 300 seconds. The electrical current is supplied by a reversible electrochemical cell (battery).
What is the change in entropy of the resistor assuming it remains at 300 K?
<p>
</p>
Resistor remains unchanged. There is no change in state function.
<center>
<br>
<math>\Delta S_
= 0</math>
<br>
</center>
What is the change in entropy of the universe (universe = system + surroundings, including the battery)?
<p>
</p>
The net effect of the process is that electrical work is converted to heat in an environment at 300 K
<center>
<br>
<math>Q=W</math>
<br>
<math>Q = I \cdot R \cdot T</math>
<br>
<math>Q = 10 \cdot 100 \cdot 300</math>
<br>
<math>Q = 300 kJ</math>
<br>
<math>\Delta S_
= \frac
</math>
<br>
<math>\Delta S_
= \frac
</math>
<br>
</center>
Note that since the battery is operating reversibly the net contribution of its process to the change in the entropy of the universe is zero.
<p>
</p>
What is the internal energy change of the universe?
<center>
<br>
<math>\Delta E_
= 0</math>
<br>
</center>
This is true since the energy is conserved.
Question 2
Describe accurately the concept of adiabatic demagnetization. Use equations or pictures if needed.
A key property is listed below.
<center>
<br>
<math>\left ( \frac
\right )_T < 0</math>
<br>
</center>
This relation is proven from a Maxwell relation, which follows from a differential.
<center>
<br>
<math>\left ( \frac
\right )_T = \left ( \frac
\right )_H</math>
<br>
<math>\phi = U - TS + PV - MH</math>
<br>
<math>d \phi = -SdT + VdP - MdH </math>
<br>
</center>
Since it is true that <math>\left (\partial M / \partial T \right )_H</math> is less than zero, <math>\left (\partial S / \partial H \right )_T</math> negative. Consider curves of <math>S(T)</math> at constant <math>H</math>.
<center>
Unable to render embedded object: File (Adiabatic_demagnetization.PNG) not found.
</center>
Adiabatic demagnetization consists of two steps.
- Isothermal application of field
- Insulate material and turn off field. There is isentropic demagnetization. The system moves back to a state with <math>H=0</math>, and therefore the temperature drops.
<p>
</p>
For a paramagnetic material that obeys the Curie Law, <math>M = \frac
H</math>, express <math>\left ( \frac
\right )_S</math> in terms of the Curie constant, the heat capacity under constant field and other fundamental properties of the material.
<center>
<br>
<math>\left ( \frac
\right )_S = - \frac{ \left ( \frac
\right )_T }{ \left ( \frac
{ \partial T \right )_H</math>
<br>
<math>\left ( \frac
\right )_T = \left ( \frac
\right )_H </math>
<br>
<math>\left ( \frac
\right )_H = \frac
</math>
<br>
<math>\left ( \frac
\right )_S = \frac{ -T \left ( \frac
\right )_H}
</math>
<br>
<math>M = \frac
H</math>
<br>
<math>\left ( \frac
\right )_H = - \frac
H</math>
<br>
<math>\left ( \frac
\right )_S = \frac
</math>
</center>
Question 3
A box at constant volume and in thermal equilibrium with the environment contains an internal partition which separates two A-B liquid solutions with different concentrations of A. The internal partition has the remarkable property that its displacement (<math>\delta V</math>) and the flow of A species through it are coupled by the relation <math>\delta V = K \delta n_A</math>. The partition is diathermal (lets heat through) and is impermeable to B. Define the equilibrium criterium for A in terms of its chemical potential and the pressure on each side. Explain reasoning and show derivation.
<p>
</p>
The total system is at constant volume and temperature hence its Hemholtz free energy is minimal with resepct to internal degrees of freedom.
<center>
<br>
<math>dF^
= dF^
+ dF^
</math>
<br>
<math>dF^
= -S^
dT^
- p^
dV^
+ \mu_A^
dn_A^
-S^
dT^
- p^
dV^
+ \mu_A^
dn_A^
</math>
<br>
<math> dV^
= -dV^
</math>
<br>
<math>dn_A^
= -dn_A^
</math>
<math>dF^
= - \left ( p^
- p^
\right ) dV^
+ \left ( \mu_A^
- \mu_A^
dn_A^
\right )</math>
<br>
</center>
If <math>dV^
</math> and <math>dn_A^
</math> were independent, equilibrium would require that <math>p^
= p^
</math> and <math>\mu_A^
= \mu_A^
</math>. However <math>dV^
</math> and <math>dn_A^
</math> are not independent.
<center>
<br>
<math>dV^
= k dn_A^
</math>
<br>
<math>dF^
= \left [\left ( \frac{ \mu_A^
- \mu_A^
}
\right )- \left (p^
- p^
\right ) \right] dV^
</math>
<br>
</center>
The expression above is to equal to zero at equilibrium.
<center>
<br>
<math>\mu_A^
- k p^
= \mu_A^
- k p^
</math>
<br>
</center>
Note that if <math>k= \overline
_A</math> is a condition of an osmotic system.
Question 4
At 1 atm pressure, iron undergoes an allotropic (structural) transformation from -iron (fcc) to -iron (bcc) at 912 C (with <math>\gamma</math> stable above 912 C and stable below 912 C). The enthalpy difference (per mole of Fe) between <math>\alpha</math> and <math>\gamma</math> at 912 C is Othe available data is:
<center>
<br>
<math>\Delta \underline H^
= \underline H^
- \underline H^
</math>
<br>
<math>\Delta \underline H^
= -940 \frac
</math>
<br>
</center>
Other available data is below:
<center>
<br>
<math>c_P^
= 40 \frac
</math>
<br>
<math>c_P^
= 33 \frac
</math>
<br>
<math> \underline
^
= 7.346 \times 10^{-6} \frac
</math>
<br>
<math> \underline
^
= 7.284 \times 10^{-6} \frac
</math>
<br>
</center>
Assume that heat capacities are temperature independent in the temperature range of interest.
<p>
</p>
Let <math>T_o</math> be the reversible transformation temperature, <math>911 C</math>, and <math>T</math> be the actual transformation temperature, <math>870 C</math>.
<p>
</p>
(a) What is the heat of transformation when one mole of Fe transforms from <math>\gamma</math> to <math>\alpha</math> at T = 870 C and at 1 atm.
<p>
</p>
In general, the expression below of enthalpy is true.
<center>
<br>
<math>dH = Cp dT</math>
<br>
<math>H(T) - H(T_o) = \int_
^T C_p dT</math>
<br>
<math>H(T) - H(T_o) = C_p (T - T_o)</math>
<br>
<math>\Delta \underline H^
(T) = \underline H^
(T) - \underline H^
(T)</math>
<br>
<math>\Delta \underline H^
(T) = \Delta \underline H^
(T_o) + \Delta C_p^
(T - T_o)</math>
<br>
<math>\Delta H^
(T) = -1234 \frac
</math>
<br>
</center>
(b) What is the entropy of transformation of one mole of Fe at T = 870 C and 1 atm.
<p>
</p>
In general, the equation below relating entropy to heat capacity is true.
<center>
<br>
<math>dS = \frac
dT</math>
<br>
<math>S(T) - S(T_o) = \int_
^T \frac
dT</math>
<br>
<math>S(T) - S(T_o) = C_p \ln \frac
</math>
<br>
<math>\Delta S^
(T) = \Delta S^
(T_o) + \Delta C_p^
\ln \left ( \frac
\right )</math>
<br>
<math>\Delta S^
(T) = -1.046 \frac
</math>
<br>
</center>
(c) What is the change in the entropy of the universe when one mole of iron transformas from <math>\gamma</math> to <math>\alpha</math> at 870 C and at 1 atm.
<center>
<br>
<math>\Delta_
= \Delta S_
+ \Delta S_
</math>
<br>
<math>\Delta S_
= -1.046 \frac
</math>
<br>
<math>\Delta S_
= \frac
</math>
<br>
<math>\Delta S_
= \frac{- \Delta H^
(T)}
</math>
<br>
<math>\Delta S_
= \frac
</math>
<br>
<math>\Delta S_
= 1.08 \frac
</math>
<br>
</center>
The term <math>Q</math> is the heat given off to the environment by the transforming solid at the transformation temperature T = 870 C.
<center>
<br>
<math>\Delta S_
= -1.046 + 1.08</math>
<br>
<math>\Delta S_
= 0.034</math>
<br>
</center>
The change of entropy is and should be greater than zero since the transformation occurs spontaneously and irreversibly.
<p>
</p>
(d) What is the equilibrium transformation temperature at P = 1000 atm.
<center>
<br>
<math>\frac
= \frac
</math>
<br>
<math>P^* - P^*_o = \frac
\ln \left ( \frac
\right )</math>
<br>
<math> \Delta H = \Delta \underline H^
</math>
<br>
<math> \Delta H = -940 </math>
<br>
<math> \Delta V = \Delta \underline V^
</math>
<br>
<math> \Delta V = \Delta \underline V^
- \underline V^
</math>
<br>
<math> \Delta V = 0.062 \cdot 10^{-6} </math>
<br>
<math> P_o^* = 101327 Pa </math>
<br>
<math> P_o^* = 101327000 Pa </math>
<br>
<math> T_o^* = 1185 K </math>
<br>
<math>P^* - P_o^* = \frac
\ln \left ( \frac
\right )</math>
<br>
<math>T = 1177 K</math>
<br>
</center>
Question 5
A small amount of liquid water in equilibrium with its pure vapor is contained in a piston of volume <math>V_o</math> and at a temperature of 25 C. The volume <math>V_o</math> is such that there is one mole of vapor. The system is slowly compressed isothermally down to a volume <math>\frac
</math>.
<p>
</p>
Data:
- At 25 C the vapor pressure of water is 0.03 atm.
- The enthalpy of evaporation for water is about 40 kJ / mole and can be taken as independent of temperature and pressure for this problem.
- You can ignore the contribution of the liquid water to the volume of the container.
<p>
</p>
(a) What is the final pressure of the system?
<p>
</p>
Since temperature remains at 25 C, pressure of <math>H_2 O</math> remains at 0.03 atm. This is the equilibrium pressure at that temperature.
(b) What is the entropy change of the system?
<p>
</p>
The terms <math>P</math>, <math>R</math>, and <math>T</math> are constant and the volume is reduced by one half. Therefore, one-half of the moles have condensed.
<center>
<br>
<math>\Delta \underline S = \frac
\frac
</math>
<br>
<math>\Delta \underline S = \frac
\left ( \frac
\right )</math>
<br>
<math>\Delta \underline S = 67 \frac
</math>
<br>
</center>
Question 6
The partial molar volume of species when present in a liquid solution is smaller thant the volume of in a pure solid state. Will the solubility of in the liquid solution increase or decrease under pressure? Explain briefly.
<center>
<br>
<math>\frac
= \overline V_i</math>
<br>
</center>
In solution, <math>\overline V_i</math> is smaller than that of that of the solid. Therefore, the chemical potential in solid will rise more with pressure than chemical potential in liquid. The maximum possible concentration will increase.