CHEM-443, Fall 2013, Section 010 Student Name______Midterm 1 October 2, 2013

Directions: Please answer each question to the best of your ability. Make sure your response is legible, precise, includes relevant dimensional units (where appropriate), logically presented (include non-mathematical language if necessary to convey your intent clearly and transparently), and correct/accurate. You are free to use your Equations Handbook, Calculator, blunt/sharp writing instrument, and brain (your choice). If you have any doubt(s) about the meaning/intent of a question(s), please ask immediately so you do not wander off on an incorrect path! Please leave all responses to be graded on the exam sheet. on scratch paper will not be considered.

Problem Points

1 /10

2 /15

3 /30 4 /10 5 /15 6 /20 Total /100

1 (10 Points) Matching / Fill-In. Provide the proper response from the right column in the blanks provided in the left column. Keep in mind that some blanks in the left column may require multiple selections from the choices in the right column.

1. Joule-Thomson expansion for cooling is a reversible A internal energy process. True or False?______H______2. Functional mathematical relationships between state B entropy variables such as pressure, volume, temperature are called ______K______. 3. The Joule Coefficient for an is ____P C state function ______4. Molar properties are ______L______, being D isenthalpic independent of system size. 5.The first law relates changes in a system’s internal E isothermal expansion energy to changes in a system’s heat and work, Δq and Δw, respectively. True or False?______H______6.The ______J____ ‘Law’ provides an explanation of F adiabatic expansion the asymmetry of ‘natural’ processes. 7. Total entropy change for an irreversible adiabatic G true process must be zero. True or False?______H______8. The stages of a Carnot Cycle are ___E, F,I,M H false ______9. The Joule-Thomson Expansion process is an I adiabatic compression ______D_____ process. 10. Ideal gas ______A, N______is(are) function(s) J second of temperature only. K equations of state L intensive M isothermal compression N O isentropic cooling P zero Q extensive R first order

2. (15 Points) For a fluid described by the ideal gas equation of state, show that the following relation "∂U % holds: $ ' = 0 . Begin with the following relation for the total differential of U(T,V(T,P)), # ∂P &T ) "∂P % , dU (T,V(T, P)) = CV (T) dT ++T $ ' − P.dV where P is pressure, T is temperature, V is volume, CV(T) is * #∂T &V - the constant volume heat capacity, and U is internal energy.

Solution:

In the general case, where U=U(T,V(T,P)), we can write:

# ∂U & # ∂U & # ∂T & # ∂U & # ∂V & % ( = % ( % ( +% ( % ( $ ∂P ' $ ∂T ' $ ∂P' $ ∂V ' $ ∂P ' T V T T T # ∂T & # ∂U & # ∂V & = CV (T)% ( +% ( % ( $ ∂P 'T $ ∂V 'T $ ∂P 'T

# ∂U & The first partial is zero since it is taken at constant temperature. In order to obtain % ( , we can use the $ ∂V 'T € given relation for the total differential of U, dU, under constant temperature conditions:

* # ∂P & - € dU(T,V (T,P)) = 0 + ,T % ( − P/ dV + $ ∂T 'V .

# ∂U & * # ∂P& - % ( = ,T % ( − P/ $ ∂V 'T + $ ∂T 'V .

Thus, we have:

€ # ∂U & * # ∂P& -# ∂V & % ( = 0 + ,T % ( − P/% ( $ ∂P 'T + $ ∂T 'V .$ ∂P 'T

* # ∂P & -# ∂V & = ,T % ( − P/% ( + $ ∂T 'V .$ ∂P 'T

Use the ideal gas EOS to evaluate the partial derivatives needed: €

# ∂U & *# nRT & # nRT & -# ∂V & % ( = % ( −% ( % ( , / $ ∂P 'T +$ V ' $ V ' .$ ∂P 'T = 0

€

3A. (25 Points) It can be shown that the Joule-Thomson coefficient can be expressed as:

1 ) "∂V % , µJT = +T $ ' −V . CP,molar * # ∂T &P -

Where P is pressure, T temperature, and V molar volume. The constant pressure specific heat is given as a molar property in this equation. For a fluid described by the of state, derive an equation for the inversion temperature. In the final stages of your analysis, you will need to consider the above equation under limiting conditions of V → ∞ .

Solution: We are given an expression for the Joule-Thomson coefficient, which should be a function of temperature at a particular pressure. We need to find the temperature where this value is zero, the inversion temperature. We must first evaluate the expression using the Van der Waals EOS. We first apply the cyclic ‘rule’ to obtain an expression in terms of partial derivatives that are easier to work with as follows:

1 * # ∂V & - µJT = ,T % ( −V / CP,molar + $ ∂T ' P .

* # ∂P& - , % ( / 1 $ ∂T ' = , −T V −V / CP,molar , # ∂P & / , % ( / + $ ∂V 'T .

Evaluate the partial derivatives for the VdW EOS: €

VdW : RT a P = − V − b V 2 $ ∂P ' R $ ∂P ' −RT & ) = → - T& ) = % ∂T (V V − b % ∂T (V V − b $ ∂P ' −RT 2a & ) = + V 2 V 3 % ∂ (T (V − b) + . + . - −RT 0 - 0 1 1 - −RT 0 µ = - V − b −V 0 = −V JT - −RT 2a 0 - 0 CP,molar CP,molar −RT 2a(V − b) 2 + 3 - 0 - V 0 + 3 , (V − b) / ,- (V − b) V /0

+ 3 . + 3 . 1 −RT(V − b)V 1 −RT(V − b)V = - −V 0 = - −V (1)0 C 3 2 C 3 2 P,molar ,- −RTV + 2a(V − b) /0 P,molar ,- −RTV + 2a(V − b) /0

2 + 3 3 2 . + 3 $ 3 ' . 1 −RT(V − b)V −RTV + 2a(V − b) 1 - −RT(V − b)V −V −RTV + 2a(V − b) 0 = - −V & ) 0 = ( ) 3 2 & 3 2 ) - 3 2 0 CP,molar - −RTV + 2a V − b −RTV + 2a V − b 0 CP,molar −RTV + 2a V − b , ( ) % ( ) ( / ,- ( ) /0

+ 4 3 4 3 2 2 . 1 −RTV + RTbV + RTV − 2aV + 4abV − 2ab V = - 0 C 3 2 P,molar ,- −RTV + 2a(V − b) /0

+ 3 3 2 2 . 1 RTbV − 2aV + 4abV − 2ab V = - 0 C 3 2 P,molar ,- −RTV + 2a(V − b) /0

€ Now, we consider that in the limit of molar volume going to infinity (as stated in the problem), the terms that remain are the cubic terms; we thus are left with:

1 # RTbV 3 − 2aV 3 & 1 # RTb − 2a&

= % 3 ( = % ( = µJT CP,molar $ −RTV ' CP,molar $ −RT '

Setting equal to zero gives: €

1 # RTb − 2a& 0 = % ( CP,molar $ −RT ' RTb − 2a = 0 −RT RTb − 2a = 0 2a T = inversion Rb

1 ) "∂V % , € 3B (5 Points). Using the relation for the Joule-Thomson Coefficient, µJT = +T $ ' −V ., determine CP,molar * # ∂T &P - the numerical value for Ideal Gas . Present your work in a logical, systematic manner, and show all steps µ JT explicitly.

IG : RT V = P # ∂V & R # ∂V & RT % ( = →T% ( = $ ∂T ' P P $ ∂T ' P P 1 + # ∂V & . 1 + RT RT . µJT = -T % ( −V 0 = - − 0 = 0 CP,molar , $ ∂T ' P / CP,molar , P P /

€

4 (10 Points) An ideal gas is originally confined in a volume V1 within an insulated container whose total volume is (V1+V2). The remaining container volume is evacuated. The partition separating the volumes is removed allowing the gas to expand to fill the entire container. If the initial temperature of the gas is T, what is the final temperature? Use equations to justify your answer.

Solution:

Since the container is insulated, we consider an adiabatic process. We consider as our system the gas. Thus, since the expansion occurs against vacuum, we have:

dU = dq + dw = 0 − pext dV = 0

However, for an ideal gas: €

dU = CV dT

Thus, since dU=0 and Cv>0, we have dT=0. Integrating for the process, we have ΔT=0. Thus the final temperature is T as well. €

5A (7.5 Points) The standard reaction enthalpy for the following reaction,

TiO2(solid) + 2C(graphite) + 2Cl2 (gas) →2CO(gas) + TiCl4 (liquid)

at 298K is -80.0 kJ/mol. Use the following data to estimate the standard enthalpy at 135.8 Celsius.

€ TiO2(solid) Cl2(gas) C(graphite) CO(gas) TiCl4(liquid) 0 ΔfH (kJ/mol) -945 -110.5 -1 -1 CP,m (J mol K ) 55.06 33.91 8.53 29.12 145.2

408.95 0  0  ΔH (135.8 C) = ΔH (25 C) + ∫ ΔCP,molardT 298K 408.95 0  0  = ΔH (25 C) + ΔCP,molar ∫ dT = ΔH (25 C) + ΔCP,molarΔT 298K

where we assume the change in specific heats is independent of temperature. Substituting values, we obtain: € 0  0  ΔH (138.5 C) = ΔH (25 C) + ΔCP,molarΔT kJ $ J J J J J ' = −80.0 +& (1)(145.2 + (2)(29.12 − (1)(55.06 ) − (2)(8.53 ) − (2)(33.91 ) ΔT mol % mol K mol K mol K mol K mol K ( kJ kJ −80.0 + (0.0635 )(408.95K − 298K) mol mol K kJ kJ = −80.0 + 7.05 mol mol kJ = −72.95 mol

5B (7.5 Points) In an experiment to estimate the caloric value of food, a sample of foodstuffs is burned (oxidized) in an oxygen atmosphere. The temperature of the calorimeter rose by 2.89 Celsius. When a € current of 1.27 Amperes from a 12.5 Volt source flowed through the same calorimeter for 157 seconds, the temperature rose by 3.88 Celsius. Give an estimate of the heat of combustion for the food. The energy associated as heat for a current of “I” Amperes due to a voltage “V” for “t” seconds is E = IVt. Keep in mind that 1 (Volt)(Ampere) = 1 Joule/second.

Considering that the system including calorimeter and foodstuffs is isolated, dUTotal=0. Thus, the heat liberated from oxidation is used to heat the calorimeter, thus causing temperature increase of the calorimeter. We need the heat capacity (calorimeter constant from Engel & Reid terminology), which is determined from the information about how much temperature rise occurs upon application of a certain amount of current. First, to determine the calorimeter constant:

J q IVt (1.27A)(12.5V )(157s) 2492.4AVs 2492.4 s 2492.4J C = = = = = s = cal ΔT ΔT 3.88 C 3.88 C 3.88 C 3.88 C J = 642.4  C Thus, oxidation of the foodstuffs giving rise to 2.80 Celsius rise in calorimeter temperature leads to

J  q = Ccal ΔT = (642.4 )(2.89 C) =1856.4J €  C

€ 6 (20 Points) For the following questions, state any assumption(s) you make. Take the specific heat at cal cal constant volume as C = 5 and R = 2 V mol C mol C

A. How much heat is required to raise the temperature of 1000 grams of nitrogen from -20C to 100C at constant pressure?

Considering nitrogen to follow ideal gas behavior over this temperature range, the heat required is:

# & 1000gr # cal cal &  qP = n(CV + R)ΔT = % (% 5 + 2 (( 120 C) = 30000cal = 30kcal 28gr /mole mole C mole C $ '$ '

B. What is the change in internal energy? € # & 1000gr # cal &  ΔU = n(CV )ΔT = % (% 5 (( 120 C) = 21429cal = 21.429kcal 28gr /mole mole C $ '$ '

C. How much external work was done? €

w = q − ΔU = n(CV )ΔT = 30kcal − 21.429kcal = 8.6kcal

€ D. How much heat is required if volume is kept constant?

# & 1000gr # cal &  qV = n(CV )ΔT = % (% 5 (( 120 C) = 21429cal = 21.429kcal 28gr /mole mole C $ '$ '

€