Question 1

The Helmholtz free energy (<math>F=U-TS</math>) of a system of atoms (each of which carries a magnetic moment) is found to depend on the magnetic moment of the system as below:

<math>F(M) = A \left ( \frac

Unknown macro: {M}

Unknown macro: {mu}

- \frac

Unknown macro: {1}

Unknown macro: {2}

\right )^2</math>

</center>

The term <math>\sigma</math> is a constant (the Bohr magneton), and <math>F</math> is the free energy per atom, and <math>A</math> is another constant with appropriate units. This relation holds at constant temperature. Compute the relation between applied field <math>H</math> and <math>M</math>.

Need to decide whether to compute assuming constant volume or constant pressure. Since <math>F</math> is provided, a natural choice is constant volume

Option 1

At constant <math>T</math> and <math>V</math>, the following relation is true.

</center>

The term <math>H</math> is a derived quantity in the Hemholtz free energy representation of ths magnetic system. The following expression is true.

<math>H = \left ( \frac

Unknown macro: {partial F }

Unknown macro: { partial M}

\right )_

Unknown macro: {T, V}

</math>

</center>

The expression below is provided in the question.

<math>F(M) = A \left ( \frac

Unknown macro: {mu}

\frac
Unknown macro: {1}

Unknown macro: {2}
\right )^2</math>

</center>

An expression of <math>H</math> is below.

<math>H = \frac

Unknown macro: {2A}

\left ( \frac

Unknown macro: {M}

Unknown macro: {mu}

- \frac

Unknown macro: {1}

Unknown macro: {2}

\right )</math>

</center>

Option 2

If choosing to calculate assuming constant pressure, start with the Gibbs free energy. Below is an expression considering constant temperature and pressure.

</center>

The term <math>H</math> is a derived quantity in the Gibbs free energy representation for this magnetic system.

<math>H = \left ( \frac

Unknown macro: { partial G}

Unknown macro: { partial M}

\right )_

Unknown macro: {T, P}

</math>

</center>

Use the definition of the Gibbs free energy.

<math>G =A \left ( \frac

Unknown macro: {mu}

\frac
Unknown macro: {1}

Unknown macro: {2}
\right )^2 + PV</math>

<math>H = \frac

Unknown macro: {2A}

\left ( \frac

Unknown macro: {M}

Unknown macro: {mu}

- \frac

Unknown macro: {1}

Unknown macro: {2}

\right ) + P \left ( \frac

Unknown macro: { partial V}

Unknown macro: {partial M}

\right )_

Unknown macro: {T, P}

</math>

</center>

Note that <math>H(M)</math> obtained from <math>F</math> agrees with <math>H(M)</math> obtained from <math>G</math> when <math>V</math> is constant, since the term <math>\left ( \frac

Unknown macro: {partial M}

\right )</math> disappears.

Question 2

Some researchers have considered to use phase transitions in a material as a way of rapidly absorbing mechanical energy (work). The idea is that under high enough pressure a substantial amount of work is done on a system when it transforms to a phase with smaller volume.

Some recent results on Absorbium, a previously unknown element, indicate that it might be desirable for such applications. At atmospheric pressure Absorbium undergoes an allotropic reversible phase transition from the <math>\alpha</math> to the <math>\beta</math> form at <math>T_0 = 350K</math>. (The low temperature phase is <math>\alpha</math>). The <math>\beta</math> form has a lower molar volume.

Assume that only reversible processes take place in the material (e.g. no plastic deformation or defect creation). Both phases may be considered incompressible (though there is a volume change when the system transforms from <math>\alpha</math> to <math>\beta</math>.

Clearly state any assumptions you make

DATA:

For the transition from <math>\alpha</math> to <math>\beta</math> at <math>350K</math>: <math>\Delta H=1 \frac
Unknown macro: {kJ}
Unknown macro: {mol}
</math> ; �� = 1 cc/mol




a) Argue that for temperatures below <math>T_0 = 350K</math> the <math>\alpha</math> phase can be transformed to <math>\beta</math> by the application of pressure. How do you know?




Under constant <math>P</math> and <math>T</math> conditions, the relevant potential for this system is the Gibbs free energy.

<center>

 

<math>dG = -SdT + VdP</math>

 

</center>

Since <math>\left ( \frac

Unknown macro: {partial G}
Unknown macro: {partial P}

\right )T = V</math> and <math>V

Unknown macro: {alpha}

< V_

Unknown macro: {beta}

</math>, the free energy of the <math>\beta</math> phase can be lowered relative to that of <math>\alpha</math> by applying pressure.




b) Find the amount of work absorbed by the <math>\alpha</math> to <math>\beta</math> transition when it is induced by pressure at <math>T = 300K</math>.




An expression of the change in the Gibbs free energy of transformation <math>\alpha \right \beta</math> is below.

<center>

 

<math>\Delta G^

Unknown macro: {alpha right beta}

(T) \approx \Delta H \left ( 1 - \frac

Unknown macro: {T}

Unknown macro: {T_o}

\right )</math>

 

</center>

The change in the Gibbs free energy of transformation with pressure at constant temperature is found.

<center>

 

<math>\left ( \frac

Unknown macro: {partial P}
\right )_T = V</math>

 

<math>\frac{ \partial \Delta G^

Unknown macro: {alpha right beta}

}

= \Delta V^

Unknown macro: {alpha right beta}
</math>

 

</center>

To find the work absorbed from the environment, it is necessary first to find the transition pressure, <math>P^*</math>, at <math>300 K</math>.

<center>

 

<math>\Delta G(P, 300K) = \Delta G (1, 300K) + P \Delta V^

</math>

 

</center>

The following is true at the transition pressure, <math>P^*</math>, and temperature <math>(300 K)</math>.

<center>

 

<math>\Delta G^

Unknown macro: {alpha right beta}
= 0</math>

 

<math>P^* = - \frac

Unknown macro: {Delta G(1,300K)}

{\Delta V^{\alpha \right \beta}}</math>

 

</center>

An expression of the amount of work absorbed by the environment is below.

<center>

 

<math>W = - \int P^* dV</math>

 

<math>W = - P^* \Delta V^

</math>

 

<math>W = \Delta G(1,300K)</math>

 

</center>

Calculate at <math>300K</math>.

<center>

 

<math>\Delta G (1,300K) = \Delta H \left ( 1 - \frac

Unknown macro: {T}

Unknown macro: {T_o}

\right )</math>

 

<math>\Delta G (1,300K) = \left ( 1000 \frac

Unknown macro: {J}

\right ) \left (1 -\frac

Unknown macro: {300}

Unknown macro: {350}

\right )</math>

 

<math>W = 143 \frac

Unknown macro: {J}

Unknown macro: {mole}

</math>

 

</center>

c) In some cases (e.g. when the transition is completed very quickly) it is more realistic to think of the transition as occurring adiabatically (isentropically). What is the work absorbed from the environment when the transition is induced by pressure at <math>300 K</math>, but occurs adiabatically.




The system undergoes a phase transition at constant pressure and entropy. The relevant potential is enthalpy, <math>H</math>.




It is necessary, as before, to find the pressure at which the transition occurs, under this adiabatic condition. An expression of an equilibrium condition in this case is below.

<center>

 

<math>H^

Unknown macro: {alpha}

(S,P^*) = H^

Unknown macro: {beta}

(S,P^*)</math>

 

</center>

Find <math>P^</math> for which <math>\Delta H (P^) = 0</math>.

<center>

 

<math>\left ( \frac

Unknown macro: {partial Delta H}

Unknown macro: {partial P}

\right )_S = \Delta V</math>

 

<math>P^* = - \frac

Unknown macro: {Delta H(P=1)}

Unknown macro: {Delta V}

</math>

 

</center>

An expression of work absorbed by the environment is below.

<center>

 

<math>W = - \int P^* dV</math>

 

<math>W = - P^* \Delta V^

Unknown macro: {alpha right beta}

</math>

 

<math>W = \Delta H(1)</math>

 

<math>W = 1 \frac

Unknown macro: {mol}
</math>

</center>

Question 3

a) Find the three equations of state for a system with the fundamental equation:
<center>

<math>U = \frac

Unknown macro: {A S^3}

Unknown macro: {NV}

</math>

</center>

In this equation, <math>S</math> is the entropy, <math>N</math> is the number of moles in the system, and <math>V</math> is the volume. <math>A</math> is a constant.

Find three relevant equations of state provided a fundamental equation of the internal energy of a system as a function of extensive properties. Begin with the expression for the first law.

<math>T = \left ( \frac

Unknown macro: {partial U}

Unknown macro: {partial S}

\right )_

Unknown macro: {V, N}

</math>

<math>T = \frac

Unknown macro: {3AS^2}

</math>

<math>-P = \left ( \frac

Unknown macro: {partial U}

Unknown macro: {partial V}

\right )_

Unknown macro: {S, N}

</math>

<math>-P = -\frac

Unknown macro: {AS^3}

Unknown macro: {NV^2}

</math>

<math>\mu = \left ( \frac

Unknown macro: {partial N}

\right )_

Unknown macro: {S, V}

</math>

<math>\mu = -\frac

Unknown macro: {AS^3}

Unknown macro: {N^2V}

</math>

</center>

b) Show that the equations of state are intensive.

Prove that the relation below is true to show that the equations of state found above are intensive properties.

<math>Y(\lambda S, \lambda V, \lambda N ) = Y (S, V, N)</math>

</center>

The term <math>Y</math> is an intensive property of a thermodynamic system or an equation of state from the internal energy representation of the first law. Below are calculations pertaining to this particular problem.

<math>T = \frac

Unknown macro: {3A (lambda S)^2}

Unknown macro: {(lambda N)(lambda V)}

</math>

<math>T = \frac

Unknown macro: {lambda^2 3AS^2}

Unknown macro: {lambda^2 NV}

</math>

<math>T = \frac

Unknown macro: {3AS^2}

Unknown macro: { NV}

</math>

<math>-P = -\frac

Unknown macro: {A(lambda S)^3}

Unknown macro: {(lambda N)(lambda V)^2}

</math>

<math>-P = -\frac

Unknown macro: {lambda^3 3AS^3}

Unknown macro: {lambda^3 NV^2}

</math>

<math>-P = -\frac

Unknown macro: { NV^2}

</math>

<math>\mu = -\frac

Unknown macro: {A(lambda S)^3}

Unknown macro: {(lambda N)^2(lambda V)}

</math>

<math>\mu = -\frac

Unknown macro: {lambda^3 3AS^3}

Unknown macro: {lambda^3 N^2V}

</math>

<math>\mu = -\frac

Unknown macro: {AS^3}

Unknown macro: { N^2V}

</math>

</center>

Question 4

Air is enclosed in a spherical soap bubble of radius <math>r</math>. The soap film does not let air molecules through. The pressure outside the bubble is <math>p_o</math>. Using the conditions of equilibrium, it is possible to show that the air pressure inside the bubble is <math>p_i = p_o + \frac

Unknown macro: {2 sigma}

Unknown macro: {r}

</math>, where <math>r</math> is the radius of the film, and <math>\sigma</math>* is the interfacial energy between the air and soap film. ( <math>\sigma</math> *is the conjugate variable to the surface area of the bubble (<math>A</math>)).

What is the heat capacity of a collection of bubbles under constant outside pressure (per mole of gas)? Write in terms of properties of the gas and the interface (<math>Cp</math>, *<math>CV</math>,* <math>\sigma</math>, etc.)

Air can be treated as an ideal gas for this problem.

Note that the heat capacity for this system is neither <math>C_p</math> nor <math>C_V</math> since the volume and pressure inside the bubble is not constant upon heating. Heat capacities in general are defined by an expression below.

<math>C_X = T \left ( \frac

Unknown macro: {partial S}

Unknown macro: {partial T}

\right )_X</math>

</center>

In this question, it is necessary to find the condition <math>X</math> that remains constant and which is not <math>V</math> or <math>P</math>.

For a simple system, it is possible to express the change in entropy as a function of <math>(T, P)</math>.

<math>dS = \left ( \frac

Unknown macro: { partial S }

\right )_P dT + \left ( \frac

Unknown macro: { partial S }

Unknown macro: {partial P}

\right )_T dP</math>

</center>

It is possible to take the partial derivative of the above expression, with constant <math>X</math>.

<math>\left ( \frac

Unknown macro: {partial T}

\right )_X = \left ( \frac

Unknown macro: { partial S }

\right )_P + \left ( \frac

Unknown macro: { partial S }

Unknown macro: {partial P}

\right )_T \left ( \frac

Unknown macro: { partial P }

Unknown macro: {partial T}

\right )_X</math>

</center>

Note that this looks like the 'Zemansky Rule'

<math>C_X = C_P + T \left ( \frac

Unknown macro: {partial P}

\right )_T \left ( \frac

Unknown macro: { partial P }

Unknown macro: {partial T}

\right )_X</math>

</center>

To find <math>C_x</math> as a function of known properties of the system, it is necessary to manipulate the expression above. Start by using a Maxwell Relation considering an ideal gas.

<math>\left ( \frac

Unknown macro: { partial S }

\right )_T = - \left ( \frac

Unknown macro: { partial V }

Unknown macro: {partial T}

\right )_P</math>

<math>\left ( \frac

Unknown macro: { partial S }

Unknown macro: {partial P}

\right )_T = - \frac

Unknown macro: {nR}

Unknown macro: {P}

</math>

<math>C_X = C_P - V \left ( \frac

Unknown macro: { partial P }

\right )_X</math>

</center>

There is a need to find <math>\left ( \frac

Unknown macro: { partial P }

Unknown macro: {partial T}

\right )_X</math>.

<math>\left ( \frac

Unknown macro: {partial T}

\right )_X = - \frac{ \left ( \frac

Unknown macro: { partial X }

\right )_P }{ \left ( \frac

Unknown macro: { partial X }

Unknown macro: {partial P}

\right )_T</math>

</center>

To find <math>dX</math>, use the condition for mechanical equilibrium between the gas inside the bubble and the environment.

<math> P_i = P_o + \frac

Unknown macro: {r}

</math>

<math>dP_i - \frac

Unknown macro: {2 sigma}

Unknown macro: {r^2}

dr = 0</math>

<math>V = \frac

Unknown macro: {4}

Unknown macro: {3}

\pi r^3</math>

<math>dP - \frac

Unknown macro: {3}

\frac

Unknown macro: {sigma}

Unknown macro: {r}

\frac

Unknown macro: {dV}

Unknown macro: {V}

= 0</math>

<math>\left ( \frac

Unknown macro: {partial P}

Unknown macro: {partial T}

\right )_X = - \frac{ \left ( \frac

Unknown macro: { partial P }

\right )_P - \frac

Unknown macro: {2 sigma}

Unknown macro: {3 r V}

\left ( \frac

Unknown macro: {partial V}

Unknown macro: {partial T}

\right )_P}{ \left ( \frac

Unknown macro: { partial P }

\right )_T - \frac

Unknown macro: {2 sigma}

Unknown macro: {3 r V}

\left ( \frac

Unknown macro: {partial V}

Unknown macro: {partial P}

\right )_T}</math>

<math>\left ( \frac

Unknown macro: {partial T}

\right )_X = - \frac{ 0 - \frac

Unknown macro: {3 r V}

\left ( \frac

Unknown macro: {partial V}

Unknown macro: {partial T}

\right )_P}{1 - \frac

Unknown macro: {2 sigma}

\left ( \frac

Unknown macro: {partial V}

Unknown macro: {partial P}

\right )_T}</math>

</center>

The following is true regarding an ideal gas.

<math>\frac

Unknown macro: {V}

\left ( \frac

Unknown macro: { partial V}

Unknown macro: {partial T}

\right )_p = \frac

Unknown macro: {NR}

Unknown macro: {PV}

</math>

<math>\frac

Unknown macro: {1}

\left ( \frac

Unknown macro: { partial V}

Unknown macro: {partial T}

\right )_p = \frac

Unknown macro: {1}

Unknown macro: {T}

</math>

<math>\frac

Unknown macro: {V}

\left ( \frac

Unknown macro: {partial P}

\right )_T = \frac{-NRT}

Unknown macro: {P^2V}

</math>

<math>\frac

Unknown macro: {1}

Unknown macro: {V}

\left ( \frac

Unknown macro: { partial V}

\right )_T = -\frac

Unknown macro: {1}

Unknown macro: {P}

</math>

</center>

Combine expressions.

<math>\left ( \frac

Unknown macro: {partial P}

Unknown macro: {partial T}

\right )_X = \frac{ \frac

Unknown macro: {2 sigma}

{3 r T}}{1 - \frac

{3 r P}}</math>

</center>

An expression of the specific heat is below.

<math>C_X = C_P - V \frac{ \frac

Unknown macro: {2 sigma}

{3 r T}}{1 - \frac

{3 r P}}</math>

<math>C_X = C_P - V \frac{ \frac

{T}}{\frac

Unknown macro: {3 r P}

Unknown macro: {2 sigma}

- 1}</math>

<math>C_X = C_P - \frac{ \frac

Unknown macro: {PV}

{T}}{\frac

Unknown macro: {2 sigma}

1}</math>

<math>C_X = C_P - \frac

Unknown macro: {nR}

{\frac

Unknown macro: {3 r P}

- 1}</math>

</center>

Use the expression <math>P = P_o + \frac

Unknown macro: {2 sigma}

Unknown macro: {r}

</math>.

<math>C_X = C_P - \frac

Unknown macro: {nR}

{\frac

Unknown macro: {3 r P_o}

+ 2}</math>

</center>

Note that when <math>\sigma = 0</math>, <math>C_X = C_P</math>.

Question 5

a) The only way to transfer heat from high temperature to low temperature is by an irreversible process.

CORRECT___________ NOT CORRECT_________

Incorrect

</center>

It is possible to transfer heat reversibly from a high temperature thermal reservoir to a low temperature thermal reservoir. An example is the Carnot Cycle. The first three stages of this cycle are below.

Isothermal Expansion: Heat is withdrawn from a hot region to an intermediate system, or working fluid, at the same temperature.
Adiabatic Expansion: The system decreases its temperature at constant entropy.
Isothermal Compression: The system is isothermally compressed while in contact with the cold reservoir. The heat taken from the hot region is transferrred to the cold region in this step.

b) If the thermal expansion is negative, the constant pressure heat capacity is smaller than the constant volume heat capacity.

CORRECT___________ NOT CORRECT_________

Incorrect

</center>

Using mathematical manipulations and Maxwell relations, an expression below can be derived.

<math>C_P - C_V = \frac

Unknown macro: {TValpha^2}

Unknown macro: {beta_T}

</math>

</center>

The difference between <math>C_P</math> and <math>C_V</math> is alwatys positive.

c) Enthalpy is a conserved quantity

CORRECT___________ NOT CORRECT_________

Incorrect

</center>

An example of this is when work is converted to heat. Think of stirring of a liquid. When a liquid is stirred, its temperature raises, increasing its enthalpy. In this case, there are no obvious sources of enthalpy.

d) In an adiabatic expansion of a material the temperature always goes down.

CORRECT___________ NOT CORRECT_________

Incorrect

</center>

Think of the adiabatic free expansion of an ideal gas. In this case, since the energy of the system remains constant, there cannot be a change in its temperature. It is possible to show the expressions below regarding a general case.

<math>\left ( \frac

Unknown macro: {partial T}

Unknown macro: {partial V}

\right )_S = \frac{ - \left ( \frac

Unknown macro: {partial S}

\right )_T }{ \left ( \frac

Unknown macro: {partial S}

\right )_V}</math>

<math>\left ( \frac

Unknown macro: {partial T}

Unknown macro: {partial V}

\right )_S = \frac{ - \left ( \frac

Unknown macro: {partial P}

\right )_V }{ \frac

Unknown macro: {C_V}

{T}}</math>

<math>\left ( \frac

Unknown macro: {partial T}

Unknown macro: {partial V}

\right )_S = \frac

Unknown macro: {T alpha}

Unknown macro: {C_V beta_T}

</math>

<math> dT = \frac

Unknown macro: {C_V beta_T}

dV</math>

</center>

Note that the change in temperature, <math>dT</math>, depends on the sign of <math>\alpha</math>, for any given change in volume, <math>dV</math>.

e) The entropy of a system can decrease

CORRECT___________ NOT CORRECT_________

Correct

</center>

An example of this is solidification. In this case, the system reduces its entropy when it transforms to the solid phase. The environment, however, increases its entropy due to the removal of heat from the solidfying system.

Child pages

3.20 Exam QuestionsExam12003

Question 1

Question 2

Question 3

Question 4

Question 5