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h1. 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).*
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*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.*
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<math>\Delta H_{bucket} > 0</math> *______* <math>\Delta H_{bucket} < 0</math> *______* <math>\Delta H_{bucket} = 0</math> *______*
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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>.
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*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.*
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<math>\Delta S_{bucket} > 0</math> *______* <math>\Delta S_{bucket} < 0</math> *______* <math>\Delta S_{bucket} = 0</math> *______*
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The stirring is irreversible and the stirring work will be dissipated as heat. Hence the entropy increases.
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*c) Which of the following statements regarding the enthalpy of the surroundings is correct ?*
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<math>\Delta H_{surr} > 0</math> *______* <math>\Delta H_{surr} < 0</math> *______* <math>\Delta H_{surr} = 0</math> *______*
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Same analysis as in part b, but now for the surroundings. It has to perform work, hence the enthalpy of the surroundings decreases.
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*d) What is the minimal entropy change that needs to place in the surroundings. ?*
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<math>\Delta S_{surr} > 0</math> *______* <math>\Delta S_{surr} < 0</math> *______* <math>\Delta S_{surr} = 0</math> *______*
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The minimal entropy change in the surroundings will take place when the work needed for the stirring is produced reversibly.
[Category:Entropy Change]
h1. 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.*
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*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
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<math>G = U - TS - Fl</math>
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<math>dU = TdS + Fdl</math>
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<math>dG = -SdT - ldF</math>
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There is now a potential <math>G(T, F)</math>
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*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 !*
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Write the Clapeyron type equation
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<math>\frac{dF}{dT} = \frac{-\Delta S}{\Delta l}</math>
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<math>\frac{d \sigma}{dT} = \frac{-\Delta H}{T \Delta \epsilon \underline V}</math>
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Integrate to derive the following result.
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<math>\Delta \sigma = - \frac{ \Delta H }{ \Delta \epsilon \underline V } \ln \left ( \frac {T_2}{T_1} \right )</math>
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Use data from the problem.
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<math>\frac{\Delta H}{\Delta \epsilon \underline V } = \frac{300 J/mole}{-0.07 \cdot 8 \cdot 10^-6 \frac{m^3}{mole} }</math>
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<math>\frac{\Delta H}{\Delta \epsilon \underline V } = 530 M Pa</math>
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After having evaluated a constant term, find the upper and lower limits.
* Upper limit:
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<math>70 MPa = 530 MPa \cdot \ln \left ( \frac{T_{upper}}{298 K} \right )</math>
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<math>T_{upper} = 340 K</math>
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* Lower limit:
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<math>-20 MPa = 530 MPa \cdot \ln \left ( \frac{T_{lower}}{298 K} \right )</math>
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<math>T_{upper} = 287 K</math>
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h1. 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.*
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*Derive the equilibrium conditions imposed on the chemical potentials for this system.*
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First find the potential that would be minimal under the conditions described above. There is flowing matter and constant temperature and pressure.
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<math>dG = -SdT + VdP + \sum \mu_i dn_i</math>
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Temperature and pressure is constant, so related terms are equal to zero. Expand <math>dG</math>
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<math>dG^{\alpha} = dG^{\beta}</math>
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<math>dG = \mu_A^{\alpha} dn_A^{\alpha} + \mu_B^{\alpha} dn_B^{\alpha} + \mu_A^{\beta} dn_A^{\beta} + \mu_B^{\beta} dn_B^{\beta}</math>
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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^{\alpha}</math>.
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<math>dn_A^{\alpha} = -dn_A^{\beta}</math>
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<math>dn_A^{\alpha} = -dn_B^{\beta}</math>
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<math>dn_A^{\alpha} = dn_B^{\alpha}</math>
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<math>dG = \mu_A^{\alpha} dn_A^{\alpha} + \mu_B^{\alpha} dn_A^{\alpha} - \mu_A^{\beta} dn_A^{\alpha} - \mu_B^{\beta} dn_A^{\alpha}</math>
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<math>dG = \left [\mu_A^{\alpha} + \mu_B^{\alpha} - \left ( \mu_A^{\beta} + \mu_B^{\beta} \right ) \right]</math>
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<math> \mu_A^{\alpha} + \mu_B^{\alpha} - \left ( \mu_A^{\beta} + \mu_B^{\beta} \right ) = 0</math>
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<math>\mu_A^{\alpha} + \mu_B^{\alpha} = \mu_A^{\beta} + \mu_B^{\beta}</math>
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h1. 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.).*
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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>
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<math>dS = \left ( \frac{\partial S}{\partial V} \right )_T dV + \left ( \frac{\partial S}{\partial T} \right )_V dT</math>
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Manipulate the two partial derivatives to express them in terms of known quantities.
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<math>dS = \left ( \frac{\partial S}{\partial V} \right )_T = \left ( \frac{\partial p}{\partial T} \right )_V</math>
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<math>\left ( \frac{ \partial p }{ \partial T } \right )_V = \frac{ -1 }{ \left ( \frac{ \partial T }{ \partial V } \right )_p \left ( \frac{ \partial V }{ \partial p } \right )_T</math>
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<math>\left ( \frac{ \partial p }{ \partial T } \right )_V = - \frac{ \left ( \frac{ \partial V }{ \partial T } \right )_p }{ \left ( \frac{ \partial V }{ \partial p } \right )_T</math>
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<math> \left ( \frac{ \partial p }{ \partial T } \right )_V = - \frac{ -V \alpha_v }{ -V \beta_T } </math>
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<math> \left ( \frac{ \partial p }{ \partial T } \right )_V = - \frac{ \alpha_v }{ \beta_T } </math>
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and
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<math>\left ( \frac{\partial S}{\partial T} \right )_V = \frac{c_v}{T}</math>
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Combining expression results in the equation below.
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<math>dS = \left ( \frac{ \alpha_v }{ \beta_T } \right ) dV + \left ( \frac{ c_v }{ T } \right ) dT</math>
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Approach S(V, p) in a similar manner
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<math>dS = \left ( \frac{\partial S}{\partial V} \right )_p dV + \left ( \frac{\partial S}{\partial p} \right )_V dp</math>
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<math>\left ( \frac{\partial S}{\partial V} \right )_p = \left ( \frac{\partial S}{\partial T} \right )_p \left ( \frac{\partial T}{\partial V} \right )_p</math>
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<math>\left ( \frac{\partial S}{\partial V} \right )_p = \frac{c_p}{T} \cdot \frac{1}{V \alpha_v</math>
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<math>\left ( \frac{\partial S}{\partial V} \right )_p = \frac{c_p}{T V \alpha_v}</math>
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and
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<math>\left ( \frac{\partial S}{\partial p} \right )_V = \left ( \frac{\partial S}{\partial T} \right )_V \left ( \frac{\partial T}{\partial p} \right )_V</math>
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<math>\left ( \frac{\partial S}{\partial p} \right )_V = \frac{c_v}{T} \cdot \frac{\beta_T}{\alpha_v}</math>
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Combine expressions to derive the equation below.
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<math>dS = \left ( \frac{ c_p }{ T V \alpha_v } \right ) dV + \left ( \frac{ c_v \beta_T}{ T \alpha_v } \right ) dp</math>
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h1. 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.
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<math>\delta Q_L + \delta Q_H + \delta W = 0</math>
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<math>\frac{\delta Q_H}{T_H} + \frac{\delta Q_L}{T_L} = 0</math>
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<math>\delta Q_H = - \frac{T_H}{T_L} \delta Q_L</math>
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Combine expression derived from the first and second law.
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<math>\delta Q_L \left (1 - \frac{T_H}{T_L} \right ) + \delta W = 0</math>
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<math>\delta W = \left ( \frac {T_H}{T_L} - 1 \right ) \delta Q_L</math>
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Write <math>\delta Q_L</math> in terms of enthalpy.
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<math>\delta Q_L = dH</math>
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<math>dH = -AdT</math>
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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.
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<math>\delta W = - A \left ( \frac{T_H}{T_L} - 1 \right ) dT</math>
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<math>W = \int_1^{298} - A \left ( \frac{T_H}{T_L} - 1 \right ) dT</math>
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<math>W = [AT_H \ln T]_1^{298} - [AT]_1^{298}</math>
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<math>W = AT_H \ln 298 - A(297)</math>
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<math>W=140 kJ</math>
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