Solved Problems #5

ASSIGNMENT No. 6 TITLE: SOLVED PROBLEMS No. 5

NAME: Mancenido, Jerwin P.SECTION: 1D

DATE: February 11, 2014

Specific Heat

1. Find the specific heat of 350 g of an unknown material when 34,700 Joules of heat are applied, and the temperature rises from 22ºC to 173ºC with no phase change.

Given:

m = 350 g

Q = 34,700 Joules

ΔT = 173ºC - 22ºC = 151ºC

Cp = unknown

Solution:

Original equation: Cp = Q/mΔT

c = 34,700 J/(350 g x 151ºC)

Final answer:

0.65657521286 J/(g x ºC).

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DATE: February 11, 2014RATING:




2.How many liters of water can 6.7 L of ethane boil? The initial temperature and pressure of the ethane and water is .95 bar and 25˚C.

First we must find the amount of heat released by the ethane. To do this, we calculate the number of moles of ethane gas using the ideal gas equation and multiply the molar heat of combustion by the number of moles.

ΔHcombustion= 1437.17 kJ/moln=PV/RTn=[.95*6.7]/[.08314*298]n=.2569molHeat released by ethane: (1437.17 kJ/mol)(0.2569mol) = 369.21 kJ

Then using the heat equation q=mCsΔT, we can find the mass of water that would be raised to boiling with the given amount of heat. First, the kJ must be converted to J to match the units of the specific heat.

369210= (m) (4.184)(373-298)

Using basic algebra we solve for the mass, and since water has a density of 1.0g/cm3, the mass will be equal to the volume.

m=1176.58g

W = -Δ(PV) W = Work P = Pressure V = Volume

Final Answer:

Volume of water: 1.177 L

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3.There is a house hold heater that operates at 4 V and at 35 Ω and is used to heat up 15g of copper wire. The specific heat capacity of copper is 24.440 J/mol/K. How much time is required to increase the temperature from 25˚C to 69˚C?

It is important to know the equation in circuitry that calculates power: P=V2/R, which is derived from the equation V=IR. We will also be using q=mCsΔT.

P=(4)2/35=.457 J/sq=(15)(24.440)(69-25)= 16130.4 J

We now know how many joules of heat must be added to the copper wire to increase the temperature and we know how many joules of energy are given off by the heater per second. We divide to find the number of seconds.

Ln(P1/P2)=[ΔHVap/R]*[(1/T2)-(1/T1)] P1 and P2 refer to the pressures in any unit (bar, atmosphere, Pascal) R is the gas constant that correlates with the pressure and temperature units used. T1 and T2 refer to the temperatures in Kelvin

Final Answer:

(16130.4 J)/(.457 J/s) = 35296.3 seconds

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Entropy

1. Calculate the entropy for an irreversible compression ofoxygen gas. The initial pressure of the gas is 1 bar in a volume of 100 L. The final pressure of the gas is10 bar and the temperature is 400 K.

Solution: Note that you need to obtain the number of Moles. The problem does not ask you for molar entropy.

Write down the expression for the entropy:

∆ S=nR ln (V 2

V 1)=nR ln (

P1

P2

)

We need either the ratio of volumes or the ratio of pressures.We are given the pressures so we can use those.P2/P1= 10 bar/ 1 bar = 10.

We obtain the number of moles using the ideal gas law:

n=P1 V 1

RT=

(105 Pa)(0.1 m3)

(8.31J

mol . K)(400 K )

=3.00 moles

Final Answer:

Now we can substitute into the entropy expression:

∆ S=nR ln (P1

P2

)=(3.0 mol )(8.31J

mol . K ) ln (10 )=249JK

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2. Calculate entropy change when 36.0 g of ice melts at 273 K and 1 atm.

If you look up the enthalpy of fusion for ice in a table, you would get a molar enthalpy of 6.01 kJ/mol.

When a system receives an amount of energy q at a constant temperature, T, the entropy increase S is defined by the following equation.

Final Answer:

S = q / T. , Therefore,

∆ S=(6.01 kJ mo l−1)

272 Kx

36 g18 g mo l−1

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3. Evaluate the entropy change for the reaction:CO + 3 H2 -> CH4 + H2O ,in which all reactants and products are gaseous.

SolutionStandard entropies of reaction, So

reaction, equal the entropy of products minus the entropy of reactants.

The standard entropies of the reactants and products have been given above, and for clarity, the entropies are given below the formula of the reactants and products: (data are given in 3 significant digits)

CO + 3 H2 -> CH4 + H2O198 131 186 189

Therefore, use:

∆ S=[ (186+189 )−(198+3 x131 ) ] JK . mol

Final Answer:

S = -216 J/K.mol

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Enthalpy of an Ideal Gas

1.) Using standard thermodynamic values, calculate the change in enthalpy of reaction (ΔHrxn) in the formation of liquid water from Hydrogen and Oxygen gas.

Chemical Equation:H2(g) + ½O2(g) => H2O(l) + heat

Product:

ΔHf H2O(l) = -285.83 kJ/mol

Reactants:ΔHf H2(g) = 0 kJ/mol (the ΔHf of elements in their standard state is defined to be 0 kJ)ΔHf O2(g) = 0 kJ/mol x 2

Use ΔH of formation (Hf) for each of the chemicals involved in the reaction found in a standard table or reference book.

[(ΔHf H2O = -285.83 kJ/mol)] - [(½) (ΔHf O2 = 0 kJ/mol) + (ΔHf H2 = 0 kJ/mol)]

Final Answer:

ΔHrxn = SUM [(-285.83 kJ) – ((½) (0 kJ + 0 kJ)] = -285.83 kJ/mol

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2) Using standard thermodynamic values, calculate the enthalpy of the reaction of the combustion of methane gas with oxygen gas to form carbon dioxide and liquid water.

Chemical Equation:CH4(g) + 2 O2(g) => CO2(g) + 2 H2O(l) + heat

Products:ΔHf H2O(l) = -285.83 kJ/mol x 2ΔHf CO2(g) = -393.51 kJ/molReactants:ΔHf CH4(g) = -74.87 kJ/molΔHf O2(g) = 0 kJ/mol x 2

Use values found in a standard table or reference book

[2 (ΔHf H2O(l) = -285.83 kJ/mol) + ΔHf CO2(g) = -393.51 kJ/mol]- [2 (ΔHf O2 = 0 kJ/mol) + (ΔHf CH4 = -74.87 kJ/mol)] =ΔHrxn = [2 (-285.83 kJ) + (-393.51 kJ)] – [(2) (0 kJ) + (-74.87 kJ)]

Final Answer:

-890.3 kJ/mol

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3. How much heat is released when burning 0.5 kg of liquid rubbing alcohol (2-propanol)? Products are carbon dioxide and liquid water. Assume an excess of oxygen.

Chemical Reaction:

2 C3H8O (l) + 9 O2 (g) => 6 CO2 (g) + 8 H2O (l) + heat0.5 kg propanol (1 mol / 60.084 g) = 8.3 mols propanol

ΔHrxn = [products - reactants] = [6 (-393.51 kJ) + 8 (-285.83 kJ)] - [2 (-318.2 kJ) + 9 (0 kJ)]= -4011.3 kJ / (2 mols propanol) = -2005.65 kJ/mol

Final Answer:

-2005.65 kJ/mol (8.3 mols) = -16646 kJ released as heat

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Constant Pressure Specific Heat

1. 300 grams of ethanol at 10 °C is heated with 14640 Joules of energy. What is the final temperature of the ethanol?

Useful information:The specific heat of ethanol is 2.44 J/g·°C.

Solution:

Use the formula

q = mcΔTwhere:q = heat energym = massc = specific heatΔT = change in temperature

14640 J = (300 g)(2.44 J/g·°C)ΔT

Solve for ΔT:ΔT = 14640 J/(300 g)(2.44 J/g·°C)ΔT = 20 °C

ΔT = Tfinal - Tinitial

Tfinal = Tinital + ΔTTfinal = 10 °C + 20 °CTfinal = 30 °C

Final Answer:

The final temperature of the ethanol is 30 °C.

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2.If 150 g of lead at 100°C were placed in a calorimeter with 50 g of water at 28.8°C and the resulting

temperature of the mixture was 22°C, what are the values of qlead, qwater and qcal? (Knowing that the

specific heat of water is 4.184 J/g °C and the specific heat of lead is 0.128 J/g °C)

For lead, we know that:

m(mass) = 150 g,

Ti (initial temperature) = 100°C

Tf = 28.8°C

csp (specific heat of lead) = 0.128 J/g °C

For water:

m= 50 g,

Ti = 22°C

Tf = 28.8°C

csp = 4.184 J/g °C

q =m x csp x ΔT

qlead = 0.128 J/g °C x 150g x (28.8°C - 100°C) = -1.37 E3 J

qwater = 4.184 J/g °C x 50g x (28.8°C - 22°C) = 1.42 E3 J

qcal = -(qlead + qwater)

Final Answer:

qcal = -(qlead + qwater) = -(1.42E3 +-1.37 E3) = -50.0 J

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3.A 28.2 gram sample of Copper (Csp Cu= 0.685 J/g K) is placed in a coffee cup calorimeter containing

100 grams of water that just stopped boiling. After some time the temperature of the water becomes

constant at 92.3 oC. Assuming the atmospheric pressure is 1 atmosphere, calculate the initial

temperature of the Copper block. Assume no heat is lost to the surroundings. Csp H2O(l) = 4.18 J/g oC)

qH2O =−qCu

(mH2O)(CspH2O)(ΔT)=(mCu)(CspCu)(ΔT)

(100g)(4.18 J/goC)(−7.7oC)=−(28.2g)(0.685J/goC)(92.3oC−TfCu)

Solve for TfCu:

−3218.6=−19.317(92.3−TfCu)

1435.64=19.32TfCu

Final Answer:

TfCu=74.32=74.3oC

Note: Notice the difference in temperature units given by the two Csp values. The Csp Cu = 0.685 J/g K is

given in Kelvins and the CspH2O = 4.18 J/g oC is given in Celcius. There is no need for

conversions because when Csp values are given, the change in temperature will remain constant. In

other words, no matter what the units are, the change will still occur and be exactly the same as would

naturally.

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Constant Volume Specific Heat

A tank initially containing air at 30 atm and 1000oF is connected to a small turbine. Air discharges from the tank through the turbine, which produces work in the amount of 500 Btu. The pressure in the tank falls to 5 atm during the process and the turbine exhausts to the surroundings at 1 atm. Assume the air behaves as an ideal gas and ignore any irreversibilities within the tank and the turbine. Determine the volume of the tank, in ft3. Heat exchange with the atmosphere and changes in kinetic and potential energy are negligible.

The key to this process is that it is entirely isentropic. This will let us determine the initial and final properties of the air in the tank, as well as the properties of the turbine exhaust. The best part is that the properties of the turbine exhaust do not change during the process.

Give:P1 30 atm P2 5 atmT1 1000 oF Pout 1 atmWS 500 Btu

Find:V ? ft3

Diagram :

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Assumptions:- The system is shown in the diagram.- For the system heat exchange with the atmosphere is negligible.- Changes in kinetic and potential energies are negligible.- The process is reversible.- The air behaves as an ideal gas. This is a shakey assumption at these pressures, but the problem statement instructed us to make it !

Equations / Data / Solve :

We want to evaluate the volume of the tank in the absence of irreversibilities.We can begin by applying the 1st Law to this system.

Eqn 1

We can simplify Eqn 1 because the process is adiabatic and we have assumed that changes in kinetic and potential energies are negligible and because there is no mass flow into the system.

Eqn 2

The mass conservation equation for this process is : Eqn 3

Combining Eqns 2 & 3 yields :

Eqn 4

Because the entire process is reversible and adiabatic, the process is isentropic. Therefore,

can be determined and does not change during the process. Because two intensive properties, and Pout, are constant, we can conclude that the state of the turbine exhaust is constant and,

therefore, Tout and are constant. Therefore, Eqn 4becomes :

Eqn 5

The initial and final moles of air in the tank can be determined from the ideal gas EOS:

Eqn 6Eqn

7

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Apply Eqn 7 to both the initial and final state

Eqn 8

Now, we can solve Eqn 8 for theunknown volume of the tank :

Eqn 9

Method 1: Use the Ideal Gas Entropy Function.

The Gibbs Equation for Ideal Gases in terms of the Ideal Gas Entropy Function is :

Eqn 10

We can apply Eqn 10 to the process that the gas inside the tank undergoes AND to the process that the gas undergoes as it passes through the turbine:

Eqn 11

We can solve Eqns 10 & 11 for the unknowns SoT2 and So

Tout :

Eqn 12Eqn

13

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We can look up SoT1 in the Ideal Gas Property Tables and use it with the known pressures in Eqn

13 to determine SoT2 and So

Tout :

T1 810.93 K T (K)So

(kJ/kg-K) SoT1 1.046

kJ/kg-K

R 8.314 J/mol-K 810 1.045 SoT2 0.5325

kJ/kg-K

MW 29.00 g/mol 820 1.058 SoTout 0.07112

kJ/kg-K

Now, we can use SoT2 and So

Tout and the Ideal Gas Property Tables to determine both T2and Tout and then U1, U2 and Uout by interpolation :

T (K)Uo

(kJ/kg) T (K) So(kJ/kg-K) Uo(kJ/kg)810 392.73 500 0.5270 148.23820 400.92 510 0.5477 155.81 T2 = 502.67 K

Uo1 393.49 kJ/kg Uo

2 150.25 kJ/kg11411.2 J/mol 4357.37 J/mol

And at the turbine outlet : T (K) So(kJ/kg-K) Uo(kJ/kg)310 0.0707 107.41320 0.1016 117.45 Tout = 310.14 K

Hoout 107.55 kJ/kg

3118.83 J/mol

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We can plug all of the given and determined values back into Eqns 3, 7, 8 & 9 to evaluaten1, n2, Δn, WS and, finally,V :

WS 527550 J P1 3039750 PaR 8.314 J/mol-K P2 506625 Pa

Conversion Factors : V 0.14701 m3

1 Btu = 1055.1 J V 5.192 ft3

1 lbm = 0.453593 kg1 atm = 101325 Pa1 m3 = 35.315 ft3

MW 28.97 g/mol

n1 = 66.28 moles m1 = 1.922 kg m1 = 4.238 lbm

n2 = 17.82 moles m2 = 0.517 kg m2 = 1.139 lbm

Δn = -48.46 moles Δm = -1.405 kg Δm = -3.098 lbm

Method 2: Use the Ideal Gas Relative Pressure.

When an ideal gas undergoes an isentropic process :Eqn

14

Where Pr is the Ideal Gas Relative Pressure, which is a function of T only and we can look up in the Ideal Gas Property Table for air.

We can solve Eqn 14 For Pr(T2), as follows :Eqn

15

Look-up Pr(T1) and use it in Eqn 15 To determine Pr(T2) :

T (K) Pr T1 810.93 K810 38.108 Pr(T1) 38.279820 39.954 Pr(T2) 6.380

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Once we know Pr(T2) we can determine T2 by interpolation on the the Ideal Gas Property Table.We can then use T1 and T2 to determin U1 and U2 from the Ideal Gas Property Tables.

T (K) U (kJ/kg) T (K) Pr U (J/mol)810 392.73 500 6.2732 148.23820 400.92 510 6.7424 155.81 T2 = 502.27 K

U1 393.49 kJ/kg U2 149.95 kJ/kg11399.4 J/mol 4344.1 J/mol

Because the turbine is also an isentropic process, we can determine the relative pressure of the turbine effluent:

Eqn 16 Rearranging: Eqn 17

Pr(Tout) 1.276

Now, we can use Pr(Tout) to determine T2 and then Hout using the Ideal Gas Property Tables :

T (K) Pr H (J/mol)310 1.1452 97.40 Tout 319.74 K320 1.2794 107.41 Hout 107.15 kJ/kg

3104.27 J/mol

We can plug all of the given and determined values back into Eqns 3, 7, 8 & 9 to evaluaten1, n2, Δn, WS and, finally,V :

WS 527550J V 0.14697 m3

R 8.314J/mol-K V 5.190 ft3

P1 3039750PaP2 506625Pa

n1 = 66.26 moles m1 = 1.920 kg m1 = 4.232 lbm

n2 = 17.83 moles m2 = 0.517 kg m2 = 1.139 lbm

Δn = -48.43 moles Δm = -1.403 kg Δm = -3.093 lbm

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Answers : Method 1 Method 2The volume of the tank is: V 5.19 5.19 ft3




Internal Energy of an Ideal Gas

Given the temperature, pressure and volume of neon in an ideal gas state, we can calculate the mass of neon in the system using the Ideal Gas EOS. This allows us to convert the enthalpy into specific enthalpy. We can use the definition of enthalpy or specific enthalpy to relate U to H and PV and then eliminate PV using the Ideal Gas EOS again. The units may get tricky.

Given: V = 98 L P = 2 MPaT = 400 oC H = 1200 kJ

Find: U = ??? kJ/kg

Assumptions: - Equilibrium conditions- Neon is an ideal gas at this T and P

Solution :Since neon behaves as an ideal gas, the definition of specific enthalpy can be modified as follows:

Eqn 1 Eqn 2Eqn

3

But : Eqn 4Eqn

5

For ideal gases : Eqn 6Eqn

7

Molecular weight of neon :( NIST WebBook )

MW 20.18 g / mol

Universal Gas Constant values : R 0.08205 atm L/gmol KR 8.314 J/mol K or Pa m3/mol K

Note: To convert °C to K, add 273.15 to °C. T = 673.15 K

m = 0.7067 kgRT / MW 277.3 kJ/kgH = 1698.0 kJ/kg

Final Answer: U = 1421 kJ/kg1. Determine the specific internal energy of 98 L of neon gas at 400°C and 2 MPa. The neon gas

has a total enthalpy of 1200 kJ . Assume that the neon gas is an ideal gas.

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2. Which has a higher molar internal energy: the superheated water vapor stored in Tank A or the superheated water vapor stored in Tank B? What is the difference in molar internal energy between the two tanks? Use data from the NIST WebBook.

Because the water vapor is superheated, it has 2 degrees of freedom. In this case both the T and P must be specified to completely determine the state. Because the state is completely determined, we can use the given T and P values to look up properties like Uand H in the Superheated Tables in the Steam Tables.

Given: PA = 1.5 atm PB = 0 atmTA = 200 °C TB = 150 °C

Find: ΔU = UA - UB = ??? kJ/mol

Solution: The internal energy of a substance is the sum of the kinetic energies stored in the vibrational, rotational, and translational motion of the molecules. Tank A has more energy by virtue of its higher temperature. Therefore, it must have the higher intern

Verify:We must look up the isobaric properties of superheated water in the NIST Webbook. Use the ASHRAE convention. A portion of the thermodynamic table used in this problem is given below.

Temperature (°C)Pressure

(atm)

Internal Energy (kJ/mol)

Phase

140 1.5 46.2 vapor150 1.5 46.479 vapor160 1.5 46.756 vapor170 1.5 47.032 vapor180 1.5 47.306 vapor190 1.5 47.581 vapor200 1.5 47.855 vapor210 1.5 48.129 vapor

The internal energies at the two given temperatures are:

T = 200°C T = 150°CUA = 47.855 KJ/mol UB = 46.479 KJ/mol

As we predicted, the internal energy of the water vapor in Tank A is greater than in Tank B.The U of Tank A is greater by:

Final Answer: ΔU = UA - UB = 1.376 KJ/mol

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3. Superheated water vapor at 300°C expands isothermally in a piston-and-cylinder device from 10 atm to 5 atm. Calculate the change in the molar enthalpy and molar internal energy in units of KJ / mol.

Because the water vapor is superheated, it has 2 degrees of freedom. In this case both the T and P must be specified to completely determine the state. Because the state is completely determined, we can use the given T and P values to look up properties like Uand H in the Superheated Tables in the Steam Tables.

Given: T1 = 300 oC P1 = 10 atm

T2 = 300 oC P2 = 5 atm

Find: ΔU ??? kJ/molΔH ??? kJ/mol

Solution : Use the NIST Webbook to determine the properties of superheated water vapor at the initial and final pressures. As always, use the ASHRAE convention. A portion of the thermodynamic table used in this problem is provided below.

Temp. (°C)

Pressure (atm)

Internal Energy (kJ/mol)

Enthalpy (kJ/mol)

Phase

300 4 50.533 55.252 vapor300 5 50.499 55.206 vapor300 6 50.465 55.159 vapor300 7 50.43 55.112 vapor300 8 50.395 55.065 vapor300 9 50.359 55.018 vapor300 10 50.324 54.97 vapor300 11 50.288 54.921 vapor

The internal energy and enthalpy at the given pressures are:

P = 10 atm P = 5 atmU1 = 50.324 KJ/mol U2 = 50.499 KJ/mol

H1 = 54.97 KJ/mol H2 = 55.206 KJ/mol

Remember that the change in any property is defined as the final state minus the initial state.

FinalAnswers: ΔU = U2 - U1 = 0.175 KJ/molΔH = H2 - H1 = 0.236 KJ/mol

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Temperature-Entropy Coordinates

A closed system consists of an ideal gas with constant specific heats. a.) The gas undergoes a process in which temperature increases from T1 to T2. Show that the entropy change for the process is greater if the change in state occurs at constant pressure than if it occurs at constant volume. Sketch the processes on PV and TS coordinates. b.) Using the results of (a), show on TS coordinates that a line of constant specific volume passing through a state has a greater slope than a line of constant pressure passing through a state. c.) The gas undergoes a process in which pressure increases from P1 to P2. Show that the ratio of the entropy change for an isothermal process to the entropy change for a constant-volume process is (1 - γ). Sketch the process on PV and TS coordinates.

Read : Sketch the process in parts (a), (b) and (c) first to get a better understanding of the processes.For part (a) use equations relating entropy to CP and CV.For part (b) recall that the slope on a TS diagram is (dT/dS).For part (c) determine ΔS for each process and determine the ratio.

Given :A closed system consisting of an ideal gas with constant specific heat ratio γ.

Find:Part(a)For the process where the T increases from T1 to T2: show that ΔS is greater if the change in state occurs at constant P than if it occurs at constant V.Sketch PV and TS diagrams for the process.

Part(b)

Show on a TS diagram that a line of constant specific V passing through a state has a greater slope than a line of constant P.

Part(c)For the process where the P increases from P1 to P2: show that the ratio of ΔSfor an isothermal process to ΔS for a constant V process is (1 - γ).Sketch PV and TS diagrams for the process.

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Diagrams :Part (a) and (b) Part (c)

Part (a) and (b) Part (c)

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Equations / Data / Solve :

Part a.)

There are two key equations for calculating the entropy change of an ideal gas.

Eqn 1Eqn

2

For Process 1-A, specific volume is constant. For Process 1-B, pressure is constant.We can apply Eqn 1 to Process 1-A and Eqn 2 to Process 1-B.

Eqn 3Eqn

4

Because the specific heats are constant, Eqns 3 & 4 can be integrated to obtain :

Eqn 5Eqn

6

Notice that both the intial and final temperatures are the same: TA = TB = T2.

Next, we can take the ratio of Eqn 2 to Eqn 1 :Eqn

7

Cancelling terms in Eqn 7 leaves us with :Eqn

8

For ideal gases :Eqn

9

Use Eqn 9 to eliminate CP from Eqn 8 :Eqn

10

Because R and CV are both positive numbers, we can conclude that : (SB - S1) > (SA - S1)

Eqn 1

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Part b.)

Here, we compare, at state 1, (dT/dS)V to (dT/dS)P.

Since (dT/dS) at fixed V (or fixed P) is : Eqn 12

In part (a), we showed that, for the same ΔT, ΔS at constant P is greater than ΔS at constant V.

Consequently : Eqn 13

On a TS Diagram, a constant specific volume line passing through State 1 has a greater slope than a constant pressure line passing through the same state.

Part c.)

For Process 1-A, temperature is constant. For Process 1-B, volume is constant.

Apply Eqn 2 to Process 1-A and Eqn 1 to Process 1-B.

Eqn 14Eqn

15

We need to consider the ratio of Eqn 14 to Eqn 15 and compare its value to 1 to determine which is greater, ΔS1-A or ΔS1-B.

Eqn 16

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But, for ideal gases undergoing a constant volume process 1-B :

Eqn 17 and :Eqn

18

Therefore :Eqn

19

Now, we can use Eqns 9 & 19to simplify Eqn 16 : Eqn

20

Verify : The assumptions made in this solution cannot be verified with the given information.

Answers: Part a.) (SB - S1) > (SA - S1)

Part b.) Part c.)

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Solved Problems #5