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h1. 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?*
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Resistor remains unchanged. There is no change in state function.
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<math>\Delta S_{resistor} = 0</math>
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*What is the change in entropy of the universe (universe = system + surroundings, including the battery)?*
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The net effect of the process is that electrical work is converted to heat in an environment at 300 K
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<math>Q=W</math>
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<math>Q = I \cdot R \cdot T</math>
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<math>Q = 10 \cdot 100 \cdot 300</math>
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<math>Q = 300 kJ</math>
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<math>\Delta S_{env} = \frac{300 kJ}{300 K}</math>
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<math>\Delta S_{env} = \frac{1 kJ}{K}</math>
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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.
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*What is the internal energy change of the universe?*
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<math>\Delta E_{universe} = 0</math>
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This is true since the energy is conserved.
h1. Question 2
*Describe accurately the concept of adiabatic demagnetization. Use equations or pictures if needed.*
A key property is listed below.
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<math>\left ( \frac{ \partial S }{ \partial H } \right )_T < 0</math>
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This relation is proven from a Maxwell relation, which follows from a differential.
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<math>\left ( \frac{\partial S}{\partial H} \right )_T = \left ( \frac{\partial H}{\partial T} \right )_H</math>
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<math>\phi = U - TS + PV - MH</math>
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<math>d \phi = -SdT + VdP - MdH </math>
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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>.
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!Adiabatic_demagnetization.PNG!
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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.
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</p>
*For a paramagnetic material that obeys the Curie Law,* <math>M = \frac{\kappa V}{T} H</math>*, express* <math>\left ( \frac{ \partial T }{ \partial H } \right )_S</math> *in terms of the Curie constant, the heat capacity under constant field and other fundamental properties of the material.*
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<math>\left ( \frac{\partial T}{\partial H} \right )_S = - \frac{ \left ( \frac{ \partial S }{ \partial H } \right )_T }{ \left ( \frac{ \partial S }{ \partial T \right )_H</math>
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<math>\left ( \frac{\partial S}{\partial H} \right )_T = \left ( \frac{\partial M}{\partial T} \right )_H </math>
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<math>\left ( \frac{\partial S}{\partial T} \right )_H = \frac{C_H}{T} </math>
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<math>\left ( \frac{\partial T}{\partial H} \right )_S = \frac{ -T \left ( \frac{\partial M}{\partial T} \right )_H}{C_H} </math>
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<math>M = \frac{\kappa V}{T} H</math>
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<math>\left ( \frac{\partial M}{\partial T} \right )_H = - \frac{\kappa V}{T^2} H</math>
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<math>\left ( \frac{ \partial T }{ \partial H } \right )_S = \frac{ \kappa V H }{ C_H T}</math>
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h1. 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.*
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The total system is at constant volume and temperature hence its Hemholtz free energy is minimal with resepct to internal degrees of freedom.
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<math>dF^{tot} = dF^{\alpha} + dF^{\beta}</math>
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<math>dF^{tot} = -S^{\alpha} dT^{\alpha} - p^{\alpha} dV^{\alpha} + \mu_A^{\alpha} dn_A^{\alpha} -S^{\beta} dT^{\beta} - p^{\beta} dV^{\beta} + \mu_A^{\beta} dn_A^{\beta}</math>
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<math> dV^{\alpha} = -dV^{\beta}</math>
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<math>dn_A^{\alpha} = -dn_A^{\beta}</math>
<math>dF^{tot} = - \left ( p^{\alpha} - p^{\beta} \right ) dV^{\alpha} + \left ( \mu_A^{\alpha} - \mu_A^{\beta} dn_A^{\alpha} \right )</math>
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If <math>dV^{\alpha}</math> and <math>dn_A^{\alpha}</math> were independent, equilibrium would require that <math>p^{\alpha} = p^{\beta}</math> and <math>\mu_A^{\alpha} = \mu_A^{\beta}</math>. However <math>dV^{\alpha}</math> and <math>dn_A^{\alpha}</math> are not independent.
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<math>dV^{\alpha} = k dn_A^{\alpha}</math>
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<math>dF^{tot} = \left [\left ( \frac{ \mu_A^{\alpha} - \mu_A^{\beta} }{k} \right )- \left (p^{\alpha} - p^{\beta} \right ) \right] dV^{\alpha}</math>
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The expression above is to equal to zero at equilibrium.
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<math>\mu_A^{\alpha} - k p^{\alpha} = \mu_A^{\beta} - k p^{\beta}</math>
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Note that if <math>k= \overline{V}_A</math> is a condition of an osmotic system.
h1. 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:*
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<math>\Delta \underline H^{\gamma \right \alpha} = \underline H^{\alpha} - \underline H^{\gamma}</math>
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<math>\Delta \underline H^{\gamma \right \alpha} = -940 \frac{J}{mole}</math>
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*Other available data is below:*
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<math>c_P^{\alpha} = 40 \frac{J}{mole \cdot K}</math>
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<math>c_P^{\gamma} = 33 \frac{J}{mole \cdot K}</math>
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<math> \underline{V}^{\alpha} = 7.346 \times 10^{-6} \frac{m^3}{mole}</math>
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<math> \underline{V}^{\alpha} = 7.284 \times 10^{-6} \frac{m^3}{mole}</math>
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</center>
*Assume that heat capacities are temperature independent in the temperature range of interest.*
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</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>*.*
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</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.*
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</p>
In general, the expression below of enthalpy is true.
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<math>dH = Cp dT</math>
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<math>H(T) - H(T_o) = \int_{T_o}^T C_p dT</math>
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<math>H(T) - H(T_o) = C_p (T - T_o)</math>
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<math>\Delta \underline H^{\gamma \right \alpha} (T) = \underline H^{\alpha} (T) - \underline H^{\gamma} (T)</math>
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<math>\Delta \underline H^{\gamma \right \alpha} (T) = \Delta \underline H^{\gamma \right \alpha} (T_o) + \Delta C_p^{\gamma \right \alpha} (T - T_o)</math>
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<math>\Delta H^{\gamma \right \alpha} (T) = -1234 \frac{J}{mole}</math>
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*(b) What is the entropy of transformation of one mole of Fe at T = 870 C and 1 atm.*
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</p>
In general, the equation below relating entropy to heat capacity is true.
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<math>dS = \frac{Cp}{T} dT</math>
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<math>S(T) - S(T_o) = \int_{T_o}^T \frac{Cp}{T} dT</math>
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<math>S(T) - S(T_o) = C_p \ln \frac{T}{T_o}</math>
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<math>\Delta S^{\gamma \right \alpha} (T) = \Delta S^{\gamma \right \alpha} (T_o) + \Delta C_p^{\gamma \right \alpha} \ln \left ( \frac{T}{T_o} \right )</math>
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<math>\Delta S^{\gamma \right \alpha} (T) = -1.046 \frac{J}{mole K}</math>
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*(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.*
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<math>\Delta_{universe} = \Delta S_{solid} + \Delta S_{environment}</math>
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<math>\Delta S_{solid} = -1.046 \frac{J}{mole K}</math>
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<math>\Delta S_{environment} = \frac{Q}{T}</math>
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<math>\Delta S_{environment} = \frac{- \Delta H^{\gamma \right \alpha}(T)}{T}</math>
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<math>\Delta S_{environment} = \frac{1234}{1143}</math>
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<math>\Delta S_{environment} = 1.08 \frac{J}{mole K}</math>
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The term <math>Q</math> is the heat given off to the environment by the transforming solid at the transformation temperature T = 870 C.
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<math>\Delta S_{universe} = -1.046 + 1.08</math>
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<math>\Delta S_{universe} = 0.034</math>
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The change of entropy is and should be greater than zero since the transformation occurs spontaneously and irreversibly.
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</p>
*(d) What is the equilibrium transformation temperature at P = 1000 atm.*
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<math>\frac{\partial P^*}{\partial T^*} = \frac{ \Delta H }{ T^* \Delta V }</math>
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<math>P^* - P^*_o = \frac{\Delta H}{\Delta V} \ln \left ( \frac{ T^* }{ T_o^* } \right )</math>
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<math> \Delta H = \Delta \underline H^{\gamma \right \alpha}</math>
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<math> \Delta H = -940 </math>
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<math> \Delta V = \Delta \underline V^{\gamma \right \alpha} </math>
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<math> \Delta V = \Delta \underline V^{\alpha} - \underline V^{\alpha} </math>
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<math> \Delta V = 0.062 \cdot 10^{-6} </math>
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<math> P_o^* = 101327 Pa </math>
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<math> P_o^* = 101327000 Pa </math>
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<math> T_o^* = 1185 K </math>
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<math>P^* - P_o^* = \frac{\Delta H}{\Delta V} \ln \left ( \frac{T^*}{T_o^*} \right )</math>
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<math>T = 1177 K</math>
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</center>
h1. 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{V_o}{2}</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.*
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</p>
*(a) What is the final pressure of the system?*
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</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?*
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</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.
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<math>\Delta \underline S = \frac{1}{2} \frac{\Delta \underline H}{T} </math>
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<math>\Delta \underline S = \frac{1}{2} \left ( \frac{40000}{298} \right )</math>
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<math>\Delta \underline S = 67 \frac{J}{K}</math>
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</center>
h1. 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.*
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<math>\frac{ \partial \mu_i }{ \partial p } = \overline V_i</math>
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</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.
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