Question 1
I have a machine stirring in a bucket with liquid. The bucket is under constant pressure and is insulated from the environment (an adiabatic bucket).
<p>
</p>
a) Which of the following statements regarding the enthalpy of the bucket during this process is correct? The bucket is defined as the physical bucket + the liquid in it.
<center>
<br>
<math>\Delta H_
> 0</math> _____ <math>\Delta H
< 0</math> _____ <math>\Delta H
= 0</math> ______
<br>
</center>
In general it is easy to derive that <math>(dH)_P = (\partial Q)_P + (\partial W')_P</math> where <math>\partial W'</math> are all that work terms that are not <math>-pdV</math>. Stirring is such a work term (a form of mechanical work). Hence since work is performed on the bucket <math>W' > 0</math>, hence <math>dH > 0</math>.
<p>
</p>
b) Which of the following statements regarding the entropy of the bucket during this process is correct ? The bucket is defined as the physical bucket + the liquid in it.
<center>
<br>
<math>\Delta S_
> 0</math> _____ <math>\Delta S
< 0</math> _____ <math>\Delta S
= 0</math> ______
<br>
</center>
The stirring is irreversible and the stirring work will be dissipated as heat. Hence the entropy increases.
<p>
</p>
c) Which of the following statements regarding the enthalpy of the surroundings is correct ?
<center>
<br>
<math>\Delta H_
> 0</math> _____ <math>\Delta H
< 0</math> _____ <math>\Delta H
= 0</math> ______
<br>
</center>
Same analysis as in part b, but now for the surroundings. It has to perform work, hence the enthalpy of the surroundings decreases.
<p>
</p>
d) What is the minimal entropy change that needs to place in the surroundings. ?
<center>
<br>
<math>\Delta S_
> 0</math> _____ <math>\Delta S
< 0</math> _____ <math>\Delta S
= 0</math> ______
<br>
</center>
The minimal entropy change in the surroundings will take place when the work needed for the stirring is produced reversibly.
[Category:Entropy Change]
Question 2
A super-elastic single crystal can transform between two phases (�� and ��) which have different unit cells, and hence different shape. At room temperature (298K) a superelastic strain of 7% can be achieved at a uniaxial stress of 30MPa.
<p>
</p>
a) Define the relevant thermodynamic potential which is minimal under conditions of constant applied force and constant temperature. Write the differential of this potential. You can neglect the work performed by/on the atmospheric pressure.
Define the free energy, G
<center>
<br>
<math>G = U - TS - Fl</math>
<br>
<math>dU = TdS + Fdl</math>
<br>
<math>dG = -SdT - ldF</math>
<br>
</center>
There is now a potential <math>G(T, F)</math>
<p>
</p>
b) For its application, the stress needed to achieve the super-elastic strain can not exceed 100MPa or be below 10MPa. Calculate the temperature range in which the material can operate. Clearly state the assumptions made as you derive your result !
<p>
</p>
Write the Clapeyron type equation
<center>
<br>
<math>\frac
= \frac{-\Delta S}
</math>
<br>
<math>\frac
= \frac{-\Delta H}
</math>
<br>
</center>
Integrate to derive the following result.
<center>
<br>
<math>\Delta \sigma = - \frac
\ln \left ( \frac
\right )</math>
<br>
</center>
Use data from the problem.
<center>
<br>
<math>\frac
= \frac
{-0.07 \cdot 8 \cdot 10^-6 \frac
}</math>
<br>
<math>\frac
= 530 M Pa</math>
<br>
</center>
After having evaluated a constant term, find the upper and lower limits.
- Upper limit:
<center>
<br>
<math>70 MPa = 530 MPa \cdot \ln \left ( \frac{T_{upper}}
\right )</math>
<br>
<math>T_
= 340 K</math>
<br>
</center>
- Lower limit:
<center>
<br>
<math>-20 MPa = 530 MPa \cdot \ln \left ( \frac{T_
}
\right )</math>
<br>
<math>T_
= 287 K</math>
<br>
</center>
Question 3
Two system, each containing chemical species A, B and C in different concentrations, are in contact through a semi-permeable wall. The semi-permeable wall does not allow for transport of A, B or C individually, but only allows a pair of molecules A-B to pass through together. The systems can be considered to be at constant temperature and pressure.
<p>
</p>
Derive the equilibrium conditions imposed on the chemical potentials for this system.
<p>
</p>
First find the potential that would be minimal under the conditions described above. There is flowing matter and constant temperature and pressure.
<center>
<br>
<math>dG = -SdT + VdP + \sum \mu_i dn_i</math>
<br>
</center>
Temperature and pressure is constant, so related terms are equal to zero. Expand <math>dG</math>
<center>
<br>
<math>dG^
= dG^
</math>
<br>
<math>dG = \mu_A^
dn_A^
+ \mu_B^
dn_B^
+ \mu_A^
dn_A^
+ \mu_B^
dn_B^
</math>
<br>
</center>
Remember that component C does not enter into the equilibrium because it cannot be moved. Use information in the problem to write down everything in terms of <math>dn_A^
</math>.
<center>
<br>
<math>dn_A^
= -dn_A^
</math>
<br>
<math>dn_A^
= -dn_B^
</math>
<br>
<math>dn_A^
= dn_B^
</math>
<br>
<math>dG = \mu_A^
dn_A^
+ \mu_B^
dn_A^
- \mu_A^
dn_A^
- \mu_B^
Unknown macro: {beta}dn_A^
</math>
<br>
<math>dG = \left [\mu_A^
+ \mu_B^
- \left ( \mu_A^
+ \mu_B^
\right ) \right]</math>
<br>
<math> \mu_A^
+ \mu_B^
- \left ( \mu_A^
+ \mu_B^
\right ) = 0</math>
<br>
<math>\mu_A^
+ \mu_B^
= \mu_A^
+ \mu_B^
</math>
<br>
</center>
Question 4
In class we derived an expression for dS in terms of dP and dT. Derive two other expressions for dS: one in terms of dT, dV, another in terms of dV, dP. Write the expressions in terms of heat capacities, compressibilities and coefficients of thermal expansion. Use the symbol �� for thermal expansion, �� for compressibility, and C for heat capacity. If further specification of the property is necessary, please use indices (e.g Cp is constant pressure heat capacity etc.).
<p>
</p>
For <math>S(T, V)</math> start with the differential for <math>S</math> as a function of <math>T</math> and <math>V</math>
<center>
<br>
<math>dS = \left ( \frac
\right )_T dV + \left ( \frac
\right )_V dT</math>
<br>
</center>
Manipulate the two partial derivatives to express them in terms of known quantities.
<center>
<br>
<math>dS = \left ( \frac
\right )_T = \left ( \frac
\right )_V</math>
<br>
<math>\left ( \frac
\right )_V = \frac
{ \left ( \frac
\right )_p \left ( \frac
\right )_T</math>
<br>
<math>\left ( \frac
\right )_V = - \frac{ \left ( \frac
\right )_p }{ \left ( \frac
\right )_T</math>
<br>
<math> \left ( \frac
\right )_V = - \frac
</math>
<br>
<math> \left ( \frac
\right )_V = - \frac
</math>
<br>
</center>
and
<center>
<br>
<math>\left ( \frac
\right )_V = \frac
</math>
<br>
</center>
Combining expression results in the equation below.
<center>
<br>
<math>dS = \left ( \frac
\right ) dV + \left ( \frac
\right ) dT</math>
<br>
</center>
Approach S(V, p) in a similar manner
<center>
<br>
<math>dS = \left ( \frac
\right )_p dV + \left ( \frac
\right )_V dp</math>
<br>
<math>\left ( \frac
\right )_p = \left ( \frac
\right )_p \left ( \frac
\right )_p</math>
<br>
<math>\left ( \frac
\right )_p = \frac
\cdot \frac
{V \alpha_v</math>
<br>
<math>\left ( \frac
\right )_p = \frac
</math>
<br>
</center>
and
<center>
<br>
<math>\left ( \frac
\right )_V = \left ( \frac
\right )_V \left ( \frac
\right )_V</math>
<br>
<math>\left ( \frac
\right )_V = \frac
\cdot \frac
</math>
<br>
</center>
Combine expressions to derive the equation below.
<center>
<br>
<math>dS = \left ( \frac
\right ) dV + \left ( \frac
\right ) dp</math>
<br>
</center>
Question 5
A system has the equation of state H = A T, where A = 100J/K. Assume this equation of state is valid in the temperature range from 1K to 500K. The system is cooled from 298K to 1K by operating a refrigerator with the high temperature heat release at 298K. The low temperature heat absorption cools the system. What is the minimum work required to cool this system from 298 K to 1K ?
Use an ideal refrigerator. Start with the first law and the second law.
<center>
<br>
<math>\delta Q_L + \delta Q_H + \delta W = 0</math>
<br>
<math>\frac
+ \frac
= 0</math>
<br>
<math>\delta Q_H = - \frac
\delta Q_L</math>
<br>
</center>
Combine expression derived from the first and second law.
<center>
<br>
<math>\delta Q_L \left (1 - \frac
\right ) + \delta W = 0</math>
<br>
<math>\delta W = \left ( \frac
- 1 \right ) \delta Q_L</math>
<br>
</center>
Write <math>\delta Q_L</math> in terms of enthalpy.
<center>
<br>
<math>\delta Q_L = dH</math>
<br>
<math>dH = -AdT</math>
<br>
</center>
The term on the right is negative because heat is given off to the heat reservoir at <math>T_L</math>, and <math>dT</math> is negative.
<center>
<br>
<math>\delta W = - A \left ( \frac
- 1 \right ) dT</math>
<br>
<math>W = \int_1^
- A \left ( \frac
- 1 \right ) dT</math>
<br>
<math>W = [AT_H \ln T]_1^
- [AT]_1^
</math>
<br>
<math>W = AT_H \ln 298 - A(297)</math>
<br>
<math>W=140 kJ</math>