Question 1
a) Every state function is a conserved quantity
<p>
</p>
<center>
correct __________ not correct __________
</center>
<p>
</p>
Not correct
<p>
</p>
b) The second Law of Thermodynamics states that in any change of state of a system the following holds for the entropy of the system: <math>dS \ge \frac
</math>. For a quasi-static change (but not necessarily reversible), is the temperature T in this expression the temperature of the system or of the environment the system is exchanging heat with? Mark the correct answer below.
<p>
</p>
<center>
T is system temperature __________ T is environment temperature __________
</center>
<p>
</p>
T is system temperature
<p>
</p>
A system is undergoing a reversible expansion from a state <math>I</math> to a state <math>II</math> while remaining in thermal equilibrium with a reservoir at <math>T = 300 K</math>. In the expansion process, the system absorbs <math>600 J</math> of heat from the reservoir.
<p>
</p>
c) Calculate the change in entropy of the system.
<center>
<br>
<math>\Delta S = \frac
</math>
<br>
<math>\Delta S = \frac
</math>
<br>
<math>2 \frac
</math>
<br>
</center>
d) The experiment is now repeated by this time, while the system expands, it is being beaten repeatedly with a hammer (five blows with a <math>3kg</math> hammer). The system arrives at the same state <math>II</math>, now only absorbing <math>450J</math> of heat. Calculate the entropy change of the system in this case.
<p>
</p>
Since entropy is a state function and both processes go from state <math>I</math> to state <math>II</math>, <math>\Delta S</math> is still <math>2 J/K</math>.
<p>
</p>
e) Calculate the entropy change of the universe for the experiment in d).
<p>
</p>
<center>
<br>
<math>\Delta S_
= \Delta S_
+ \Delta S_
</math>
<br>
</center>
The change in the entropy of the system was found in part c). A calculation of the change in the entropy of the environment is below.
<center>
<br>
<math>\Delta S_
= \frac
</math>
<br>
<math>\Delta S_
= \frac{-450J}
</math>
<br>
<math>\Delta S_
= -1.5 \frac
</math>
<br>
</center>
Below is a calculation of the change of entropy in the universe.
<center>
<br>
<math>\Delta S_
= 2-1.5</math>
<br>
<math>\Delta S_
= 0.5 \frac
</math>
<br>
</center>
Question 2
a) Write out the differential of the internal energy for a system that can be magnetized in an applied field. The system can also exchange heat with the environment and change its volume. Write you expression in terms of state variables.
<center>
<br>
<math>dU = TdS - pdV + H dM</math>
<br>
</center>
The term <math>\mu_o</math> is absorbed in the definition of <math>H</math>. The term <math>M</math> is defined as the total extensive magnetization.
<p>
</p>
b) When a magnetic field is applied to a material, it can change its volume. This is referred to as magnetostriction. Write the isentropic magnetostriction in terms of the isothermal magnetostriction and properties of the material (such as compressibility, thermal expansion, <math>M(T)</math>, heat capacity, etc.).
<p>
</p>
Consider magnetostriction under constant <math>x</math>.
<center>
<br>
<math>\lambda_x = \frac
\left ( \frac
\right )_
</math>
<br>
</center>
Use a rule.
<center>
<br>
<math>\lambda_s = \frac
\left ( \frac
\right )_
</math>
<br>
<math>\lambda_s = \frac
\left ( \frac
\right )_T + \frac
\left ( \frac
\right )_H \left ( \frac
\right )_s</math>
<br>
<math>\lambda_T = \frac
\left ( \frac
\right )_T</math>
<br>
<math>\alpha_H = \frac
\left ( \frac
\right )_H</math>
<br>
<math>\left ( \frac
\right )_s = \frac{- \left ( \frac
\right )_T}{\left ( \frac
\right )_H}</math>
<br>
<math>dG = -SdT + VdP - MdH</math>
<br>
<math>\left ( \frac
\right )_T = \left ( \frac
\right )_H</math>
<br>
<math>\left ( \frac
\right )_H = \frac
</math>
<br>
<math>\lambda_s = \lambda_T - \frac
\left ( \frac
\right )</math>
</center>
This solution is reasonable though not perfect. To get real data one would have to make sure that <math>C_H \approx C_
</math> and <math>\alpha_H \approx \alpha_
</math>
<p>
</p>
c) Calculate the change of internal energy with magnetization at constant T and P (i.e., <math>\left ( \frac
\right )_
</math> in terms of measurable quantities). If you use symbols for properties such as magnetostriction, susceptibility, etc., please define them.
<center>
<br>
<math>\left ( \frac
\right )_
= \left ( \frac
\right )_
</math>
<br>
<math>\left ( \frac
\right )_
= T \left ( \frac
\right )_
- p \left ( \frac
\right )_
+ H</math>
<br>
</center>
Find <math>\left ( \frac
\right )_
</math>.
<center>
<br>
<math>\left ( \frac
\right )_
= \left ( \frac
\right )_
\left ( \frac
\right )_
</math>
<br>
<math>\left ( \frac
\right )_
= \left ( \frac
\right )_
</math>
<br>
<math>\chi = \frac
\left ( \frac
\right )_
</math>
<br>
<math>\left ( \frac
\right )_
= \frac
\left ( \frac
\right )_
</math>
<br>
</center>
Find <math>\left (\frac
\right )_
</math>.
<center>
<br>
<math>\left (\frac
\right )_
= \left ( \frac
\right )_
\left ( \frac
\right )_
</math>
<br>
<math>\left (\frac
\right )_
= \frac
</math>
<br>
<math>\left (\frac
\right )_
= \frac
</math>
<br>
</center>
Combine relations
<center>
<br>
<math>\left ( \frac
\right )_
= \frac
\left ( \frac
\right )_
- \frac
+ H</math>
<br>
</center>
Question 3
<math>1 kg</math> of steam at <math>100 C</math> and a pressure of one atmosphere is supercooled to <math>80 C</math> by being brought into contact with a reservoir at that temperature. While the steam does not condense immediately when it reaches <math>80 C</math>, it does after a couple of hours.
<p>
</p>
(Latent heat of vaporization is <math>2257 kJ/kg</math>, average heat capacity of water is <math>4.2 kJ/kg \cdot K</math> and the heat capacity of steam is approximately <math>2.0 kJ/kg \cdot K</math>)
<p>
</p>
(a) Calculate the total amount of heat exchanged between the water and the environment starting from the point that the steam was supercooled to <math>80 C</math>.
<center>
Unable to render embedded object: File (Supercooled_water_--_enthalpy_vs_temperature.PNG) not found.
<br>
<math>Q = \Delta H (T=80)</math>
<br>
<math>H^
(80) = H^
(100) + \int_
^
C_p^
dT</math>
<br>
<math>H^
(80) = H^
(100) + \int_
^
C_p^
dT</math>
<br>
<math>\Delta H (80) = H^
(80) - H^
(80)</math>
<br>
<math>\Delta H (80) = \Delta H (100) + \int_
^
\left ( C_p^
- C_p^
Unknown macro: {steam}\right )dT</math>
<br>
<math>\Delta H^
= H^
- H^
</math>
<br>
<math>\Delta H = -2257 \frac
</math>
<br>
<math>\Delta H (80) = \Delta H(100) - \left ( C_p^
- C_p^
\right ) \left (100 - 80 \right )</math>
<br>
<math>\Delta H (80) = -2257 - (4.2-2.0)(20)</math>
<br>
<math>\Delta H (80) = -2301 \frac
</math>
<br>
</center>
(b) What is the entropy change of the water during the same period
<p>
</p>
The same figure applies as in part a with the y-axis as entropy and with different slopes.
<center>
<br>
<math>\Delta S_
= \int_
^
\frac{C_p^{steam}}
dT - \frac{\Delta H^
}
+ \int_
^
\frac{C_p^{liq}}
dT</math>
<br>
<math>\Delta S_
= \frac{\Delta H^{vap}}
dT - \left ( C_p^
- C_p^
\right ) \ln \left ( \frac
\right )</math>
<br>
<math>\Delta S_
= \frac{-2257}
- 2.2 \ln \left (\frac
\right )</math>
<br>
<math>\Delta S_
= -6.0509 - 0.121</math>
<br>
<math>\Delta S_
= -6.172 \frac
</math>
<br>
</center>
(c) By how much does the entropy of the universe increase?
<center>
<br>
<math>\Delta S_
= \Delta S_
- \Delta S_
</math>
<br>
<math>\Delta S_
= -6.172 \frac
</math>
<br>
<math>\Delta S_
= \frac
</math>
<br>
<math>\Delta S_
= \frac
</math>
<br>
<math>\Delta S_
= 6.518 - 6.172</math>
<br>
<math>\Delta S_
= 0.346 \frac
</math>
<br>
</center>
Question 4
Consider two identical bodies of heat capacity <math>C_p</math> and with negligible thermal expansion coefficients. The two bodies are placed in thermal contact in an adiabatic enclosure and have initial temperature <math>T_1</math> and <math>T_2</math>, respectively.
<p>
</p>
(a) What is the final temperature?
The term <math>W</math> is zero. Since the system is isolated, the following equation is true:
<center>
<br>
<math>\Delta U = \Delta U_1 + \Delta U_2</math>
<br>
<math>C_p (T_f - T_1) + C_p (T_f - T_2) = 0</math>
<br>
<math>T_f = \frac
</math>
<br>
</center>
Now consider these two bodies being brought into thermal equilibrium by a Carnot engine operating between them (the bodies are still in an adiabatic enclosure). The size of the cycle is small so that the temperatures of the bodies behave as reservoirs during one cycle.
<p>
</p>
(b) What is the entropy change of the universe for this second process?
<p>
</p>
Since the Carnot engine is reversible and the bodies don't exchange heat with the environment, <math>\Delta S_
= 0</math>.
<p>
</p>
(c) What is the final temperature of this process?
<center>
<br>
<math>\Delta S = \Delta S_1 + \Delta S_2</math>
<br>
<math>\Delta S = 0</math>
<br>
<math>\Delta S_1 = \int_
^
\frac
dT</math>
<br>
<math>\Delta S_1 = C_p \ln \left ( \frac
\right )</math>
<br>
<math>\Delta S_2 = \int_
^
\frac
dT</math>
<br>
<math>\Delta S_2 = C_p \ln \left ( \frac
\right )</math>
<br>
</center>
From the first equation, it is seen that <math>\Delta S_1 = - \Delta S_2</math>. This means that <math>T_F = \sqrt
</math>.
Question 5
In bulk form the material nanoexcitium undergoes a phase transition from <math>\alpha</math> to <math>\beta</math> at <math>320 K</math> with an enthalpy of transformation of <math>10 \frac
</math> (~<math>1 \frac
</math>).
<p>
</p>
It is observed that when nanoexcitium is made in small particle form and put in solution that its transformation temperature can change. More importantly, a marked change of transition temperature is noticed when a magnetic field is applied to the solution. One has theorized that this is due to different surface reconstructions, the surface atoms of <math>\alpha</math> and <math>\beta</math> forms have very different magnetic moments and therefore have different magnetizations in a magnetic field. (The atoms below the surface inside the particle have no magnetic moments).
<p>
</p>
a) Write down an approximate Clausius-Clapeyron equation for the shift of the transition temperature with applied magnetic field in terms of the difference in surface magnetization of the particles in the <math>\alpha</math> and <math>\beta</math> form (you can neglect the size dependence of the enthalpy of transformation).
<center>
<br>
<math>dU = TdS - pdV + HdM</math>
<br>
</center>
Neglect <math>-pdV</math>. The term <math>M</math> is the total magnetization. Define <math>G</math>.
<center>
<br>
<math>G = U - TS - HM</math>
<br>
<math>dG = -SdT - MdH</math>
<br>
</center>
Use Clausius Clapeyron along phase boundary <math>dG^
= dG^
</math>.
<center>
<br>
<math>\frac
= - \frac
</math>
<br>
</center>
b) Show that for spherical particles the shift in transition temperature with magnetic field scales with <math>\frac
</math>, where <math>r</math> is the radius of the particles.
<p>
</p>
The term <math>\Delta M</math> is equal to <math>M^
- M^
</math>. It scales with the surface and is proportional to <math>r^2</math>. The term <math>\Delta S</math> is extensive and is proportional to <math>r^3</math>. Therefore, the following is true.
<center>
<br>
<math>\frac
\prop \frac
</math>
<br>
</center>