Question 1
A beaker of water (100g) at 20 C (1 atm) is heated to a temperature of 100 C by placing it on a reservoir at 100 C. The reservoir is at constant temperature.
<p>
</p>
DATA:
- Heat capacity for water: 4.184J/g
- Enthalpy of evaporation for water: 41kJ/mole
- Molar mass for water: 18g/mole
<p>
</p>
a) Calculate the entropy change of the beaker of water
<p>
</p>
Solution to this proplem depends on whether it was considered whether the water would boil.
<center>
<br>
<math>C_p = 4.184 \frac
</math>
<br>
<math>\Delta H_
= 41 \frac
</math>
<br>
<math>M_
= 18 \frac
</math>
<br>
<math>dS = \frac
</math>
<br>
<math>dS = 100 \cdot C_p \frac
</math>
<br>
<math>\Delta S_
= 100 \cdot C_p \ln \left ( \frac
\right )</math>
<br>
</center>
Below is the numberical calculation without boiling.
<center>
<br>
<math>\Delta S_
= 100 \cdot 4.184 \cdot \ln \left ( \frac
\right )</math>
<br>
<math>\Delta S_
= 101.004 \frac
</math>
<br>
</center>
There is a calculation below considering the case of the water boiling
<center>
<br>
<math>\Delta S_
= 100 C_p \ln \left ( \frac
\right ) + \frac{\Delta H_
}{T_{evap}}</math>
<br>
<math>\Delta S_
= 100 \cdot 4.184 \cdot \ln \left ( \frac
\right ) + \frac
\cdot 100</math>
<br>
<math>\Delta S_
= 711.668 \frac
</math>
<br>
</center>
b) What is the entropy change of the universe?
<p>
</p>
First, find the entropy change of the reservoir. Because the reservoir is at constant temperature, it is known that <math>\Delta S = \frac{Q_{res}}{T_{res}}</math>. It is known that heat that flowed into the beaker flowed out of the reservoir.
<center>
<br>
<math>Q_
= -100 C_p \left (T_2 - T_1 )</math>
<br>
<math>-100 \cdot 4.184 \cdot (100 - 20)</math>
<br>
<math>Q_
= -3.347 \cdot 10^4 J</math>
<br>
<math>\Delta S = \frac{Q_{res}}
</math>
<br>
<math>\Delta S = -89.737 \frac
</math>
<br>
</center>
Below is a calculation if boiling is included.
<center>
<br>
<math>\Delta S_
= \frac{Q_{res}}{T_{res}} - \frac{\Delta H_{evap}}{T_{evap}}</math>
<br>
<math>\Delta S_
= -700.396 \frac
</math>
<br>
</center>
Calculate the entropy change of the universe first without boiling.
<center>
<br>
<math>\Delta S_
= \Delta S_
+ \Delta S_
</math>
<br>
<math>\Delta S_
= 11.267 \frac
</math>
<br>
</center>
And calculate the change with boiling.
<center>
<br>
<math>\Delta S_
= 11.267 \frac
</math>
<br>
</center>
The total entropy change of the universe is positive, as expected. Note there is no effect of boiling on the total entropy change
<p>
</p>
c) Assume now that an ideal Carnot engine is inserted between the reservoir and the beaker of water. Calculate the entropy change of the universe in this case.
<p>
</p>
A carnot engine gives reversible heat transfer, so the entropy change of the universe will be zero
Question 2
An insulating (non-conducting) cylinder is clamped between two rigid and charged walls. The charge produces a constant electrical field in the material equal to E, which produces a polarization P in the material. The contribution of this polarization to the internal energy is EdP. The whole assembly is at constant temperature.
<p>
</p>
a) Write down the relevant thermodynamic potential that is minimal for these boundary conditions as a Legendre transform of the energy.
<center>
<br>
<math>dU = TdS + V \sigma_
d \epsilon_
+ V \sigma_
d \epsilon_
+ V \sigma_
d \epsilon_
+ E dP</math>
<br>
</center>
Determine a function of <math>T</math>, <math>\epsilon_
</math>, <math>\sigma_
</math>, <math>\sigma_
</math>, and <math>E</math>.
<center>
<br>
<math>\Phi = U - TS - V \sigma_
d \epsilon_
- V \sigma_
d \epsilon_
- EP</math>
<br>
<math>d \Phi = -S dT + V \sigma_
d\epsilon_
- V \epsilon_
Unknown macro: {yy}- V \epsilon_
d \sigma_
Unknown macro: {zz}- PdE</math>d \sigma_
<br>
</center>
b) Write down the equations of state derived from this potential. Clearly indicate what the variables are in these equations of state.
<center>
<br>
<math>-S = f_1 (T, \epsilon_
, \sigma_
, \sigma_
, E)</math>
<br>
<math>V \sigma_
= f_2 (T, \epsilon_
, \sigma_
, \sigma_
, E)</math>
<br>
<math>-V \epsilon_
= f_3 (T, \epsilon_
, \sigma_
, \sigma_
, E)</math>
<br>
<math>-V \epsilon_
= f_4 (T, \epsilon_
, \sigma_
, \sigma_
, E)</math>
<br>
<math>-P = f_5 (T, \epsilon_
, \sigma_
, \sigma_
, E)</math>
<br>
</center>
Question 3
The objective of this question is to quantify the difference between the isothermal elastic stretching of a material and the isentropic stretching.
<p>
</p>
Consider a material of length l and surface area A to which a force is being applied as shown in Figure 1. This force causes only elastic deformation. Poison effects can be neglected.
<p>
</p>
Note: The Young���s modulus E of a material can be defined as <math>l \left ( \frac
\right )</math>. The stress is the force per unit area.
<p>
</p>
a) Derive a relation between the isentropic Young���s modulus and the isothermal Young���s modulus in terms of well known properties of the material (Note: ���well known properties��� are second derivatives of G: thermal expansion, compressibilities, heat capacities).
<p>
</p>
Start with the following expression to find <math>\frac
\left ( \frac
L \right )_S</math>.
<center>
<br>
<math>dL = \left ( \frac
L \right )_T d \sigma + \left ( \frac
L \right )_
dT</math>
<br>
<math>\frac
\left ( \frac
L \right )_S = \frac
\left ( \frac
L \right )_T + \frac
\left ( \frac
L \right )_
\left ( \frac
T \right )_
</math>
<br>
<math>\frac
\left ( \frac
L \right )_S = \frac
</math>
<br>
<math>\frac
\left ( \frac
L \right )_T = \frac
</math>
<br>
<math>\frac
\left ( \frac
L \right )_
= \alpha_L</math>
<br>
<math>\left ( \frac
T \right )_S = -\frac{\left ( \frac
S \right )_T} {\left (\frac
S \right )_{\sigma}}</math>
<br>
<math>\frac
= \frac
+ \alpha_L \left [-\frac{\left ( \frac
S \right )_T}{\left ( \frac
S \right )_{\sigma}} \right]</math>
<br>
</center>
Find a relation using the Maxwell Relation of the following potential.
<center>
<br>
<math>d \phi = -S dT - A L d \sigma</math>
<br>
<math>\left ( \frac
S \right )_T = A \left ( \frac
L \right )_
</math>
<br>
<math>\left ( \frac
S \right )_T = A L \alpha_L</math>
<br>
<math>\left ( \frac
S \right )_
= \frac{C_{\sigma}}
</math>
<br>
</center>
Substitute everything in.
<center>
<br>
<math>\frac
= \frac
- \frac
{C_{\sigma}}</math>
<br>
</center>
b) Estimate the relative difference between the isentropic Young���s modulus and the isothermal Young���s modulus for a typical metal (e.g. Al, Fe, Cu etc.). (Note: show clearly what estimates you use for the properties). Answer should be as % difference between isentropic and isothermal modulus.
<p>
</p>
Estimate the relative difference between the two. Assume that <math>T = 300K</math> and <math><E></math> is roughly <math>100 GPa</math>.
<center>
<br>
<math>\alpha_L \approx 10^{-5} \frac
</math>
<br>
<math>\frac{C_{\sigma}}
\approx \frac
</math>
<br>
<math>\frac
= \frac
\frac
\frac
</math>
<br>
<math>\frac
= \frac
</math>
<br>
<math>\frac
= \frac
</math>
<br>
<math>\frac
{C_{\sigma}} = 7.1 \cdot 10^{-15}</math>
<br>
<math>\frac
= - \frac
{C_{\sigma}}</math>
<br>
<math>E_s \cdot E_T = <E>^2</math>
<br>
<math>\frac
= \frac
{C_{\sigma}} \cdot <E></math>
<br>
<math>\frac
= -7.1 \cdot 10^{-15} <E></math>
<br>
<math>\frac
\approx -1%</math>
<br>
</center>
Question 4
One mole of ideal gas, initially at 10 atm pressure and 298K in a thermally insulated cylinder, is expanded by displacing a piston. The final pressure of the gas is 1atm. The pressure in the environment is 1 atm. The piston is connected to an electrical generator which converts the work performed by the piston into electrical energy (with no losses). This electrical energy is used to heat the gas in the piston through a resistance heater.
<p>
</p>
DATA
- The gas can be treated as ideal with heat capacities Cv = 5/2 R and Cp = 7/2 R
<p>
</p>
a) What is the final temperature in the gas?
<center>
<br>
<math>\mbox
= P_o</math>
<br>
<math>\mbox
= -PdV</math>
<br>
<math>\mbox
= P_o dV</math>
<br>
<math>\mbox
= (P - P_o) dV</math>
<br>
</center>
The work done on the generator is fed into the gas as heat.
<center>
<br>
<math>dU_
= C_V dT</math>
<br>
<math>dU_
= (P - P_o) dV - PdV</math>
<br>
<math>dU_
= - P_o dV </math>
<br>
</center>
Solve the expression for the final temperature. It is known that the final pressure is <math>P_o</math>.
<center>
<br>
<math>C_V (T_2 - T_1) = - -P_o (V_2 - V_1)</math>
<br>
<math>C_V (T_2 - T_1) = - P_o \left ( \frac
- \frac
\right )</math>
<br>
<math>( C_V + R ) T_2 = \left ( \frac
R + C_V \right ) T_1</math>
<br>
<math>T_2 = \left ( \frac{\frac
+ C_V}
\right ) T_1</math>
<br>
<math>T_2 = \frac
T_1</math>
<br>
</center>
b) What is the change in entropy of the surroundings (universe ��� system)?
<p>
</p>
Define the environment as everything except the piston/gas assembly and the generator. There was no heat flow to the environment so the entropy of the environment does not change.
c) Is this a reversible process?
No. The entropy of the system increase, so the total entropy change is greater than zero and therefore irreversible.
Question 5
Titanium forms an hcp (��) phase at low temperature but transforms at 1155K to a bcc (��) phase. At 1943 K this ��-phase melts. Estimate the metastable melting point of the �� phase.
<p>
</p>
DATA:
- For ��: ��Sm = 9.6 J/mol-K for the �� -. �� transition ��S = 3.43J/mol-K
Find where <math>\Delta G_
(T) = 0</math>.
<center>
<br>
<math>\Delta G_
= \Delta G_
+ \Delta G_
</math>
<br>
<math>\Delta G_
= \Delta H_
+ \Delta H_
- T \cdot (\Delta S_
Unknown macro: {epsilon-beta}+ \Delta S_
)</math>
<br>
<math>\Delta H_
= T_
\Delta S_
</math>
<br>
<math>\Delta H_
= T_
\Delta S_
</math>
<br>
<math>T_M = \frac{\Delta H_
+ \Delta H_{\beta-L}}{\Delta S_
+ \Delta S_{\beta-L}}</math>
<br>
<math>T_M = 1735.6 K</math>
<br>
</center>