Physical chemistry 1 (Hóa lý 1) exercise

Physical Chemistry 1: Thermodynamics, Chemical Equilibrium and Phase Equilibrium Questions Ho Chi Minh University of Technology - VNU-HMC (Bach Khoa) Faculty of Chemical Engineering Bài tập lớn Hóa lý 1

TABLE OF CONTENT

Question 1 3 Question 2 6 Question 3 .10

Question 1

Assuming that a heat pump operates as an ideal vapor-compression cycle using R-134a as a refrigerant Its working pressure is 180 kPa and 1.0 MPa The flow rate is 0.25 kg/s

a/ Determine specific volume, temperature, pressure, specific enthalpy, specific internal energy, specific entropy at the inlet port of each equipment in this cycle?

b/ Calculate the powers of each equipment, and the coefficient of performance (COPHP)?

c/ Show the obtained results on T-s diagram?

d/ What is the difference between ideal and actual vapor-compression cycle? Explain?

Solution:

The ideal vapor-compression cycle consists of four process:

Process Description

1-2 Isentropic compression in a compressor

2-3 Constant-pressure heat rejection in a condenser

3-4 Throttling in an expansion device

4-1 Constant-pressure heat absorption in an evaporator

a/ R-134a enters the compressor at state 1 as saturated vapor

P1 = 180 kPa ⟹

{

T1= −12.73℃

v1 = 0.11049 m3/kg

u1 = 223.01 kJ/kg

h1 = 242.90 kJ/kg

s1 = 0.93979 kJ/kg K Then the refrigerant is isentropically to the condenser pressure (state 2)

{P2 = 1.0 MPa

s2 = s1 = 0.93979 kJ/kg K

Obviously, the entropy s2 is between 0.9180 and 0.9526 We can use the linear interpolation formula to determined T2, v2, u2, h2

P2 = 1.0 MPa

m3/kg

u kJ/kg

h kJ/kg

s kJ/kg K

The linear interpolation formula:

y = yA+yB− yA

xB− xA(x − xA)

In this case, x is the entropy s and y can be T, v, u, h By linear interpolation they are determined

to be:

{

T2 = 46.3℃

v2 = 0.021281 m3/kg

u2 = 257.39 kJ/kg

h2 = 278.68 kJ/kg

At state 2, the refrigerant enters the condenser and leaves as saturated liquid at state 3 with the constant-pressure, so:

P3 = P2 = 1.0 MPa ⟹

{

T3 = 39.37℃

v3 = 0.00087 m3/kg

u3 = 106.47 kJ/kg

h3 = 107.34 kJ/kg

s3 = 0.39196 kJ/kg K

At state 3, the refrigerant is throttled to the evaporator pressure (state 4) by passing it through an expansion valve or capilarry tube as adiabatic process Then the refrigerant enters the evaporator

as a low-quality saturated mixture and completely evaporates and leaves as saturated vapor (state 1) with the constant-pressure, so:

{P4 = P1 = 180 kPa

h4 = h3 = 107.34 kJ/kg ⟹ {

T4 = −12.73℃

hf= 34.94 kJ/kg

hg = 242.90 kJ/kg Because the refrigerant at state 4 is the saturated mixture, so we have the equation h = x hg+ (1 − x) hf with x is the ratio of the mass of vapor to the total mass of the mixture

x =h4− hf

hg− hf=

107.34 − 34.94 242.90 − 34.94= 0.35

We apply similarly this equation to v4, u4, s4:

{

v4 = 0.03895 m3/kg

u4 = 100.33 kJ/kg

s4 = 0.41930 kJ/kg K b/

Evaporator: QL= m(h1− h4) = (0.25 kg/s) [(242.90 − 107.34) kJ/kg] = 33.89 kW Compressor: Win = m(h2− h1) = (0.25 kg/s) [(278.68 − 242.90) kJ/kg] = 8.95 kW Condenser: QH = m(h2− h3) = (0.25 kg/s) [(278.68 − 107.34) kJ/kg] = 42.84 kW

The coefficient of performance: COPHP = QH

Win =

42.84 kW 8.95 kW = 4.79 c/

Figure 2: T-s diagram for the obtained result of ideal vapor-compression cycle

d/ The two main differences between the ideal and actual vapor-compression cycle are the fluid frictions, that causes pressure drop and the heat transfer to or from surroundings

Figure 3: T-s diagram for the actual vapor-compression cycle Process Description

1-2 Irreversible and non-adiabatic compression of refrigerant Heat transfer from

surroundings to refrigerant and entropy increases

1-2’ Heat transfer from refrigerant to surroundings and entropy decreases

2-3 Temperature (and pressure) drop due to fluid friction and heat transfer

3-4 Pressure drops in the condenser because fluid friction

4-5 Temperature and pressure drop (as in process 2-3)

5-6 Throttling process

6-7 The throttle valve and evaporator are usually located very close to each other, so

pressure drop in connecting line is negligible

Question 2

Decompose excess NH4Cl as follows: NH4Cl (s) = NH3 (g) + HCl (g) Using the given tables

in this subject to take thermodynamic properties, then:

a/ Find the equation of ∆rxnCp as a function of temperature?

b/ Find the equation of ∆rxnH as a function of temperature?

c/ Find the equation of Kp as a function of temperature?

d/ Estimate the total pressure in a range of 100 – 800 oC (increase each 50 oC, 100, 150, 200, ,

800 oC)? Draw a diagram to show total pressure as a function of temperature?

Solution

a/ The heat capacity as a function of temperature of NH4Cl (solid), we read from NIST Chemistry WebBook that:

Cp(NH4Cl) = 2101,47 − 11,194T + 23,342 10−3T2− 1,7 × 10−5T3− 268,353 ×

105T−2 J K−1mol−1

The heat capacities as a function of temperature of NH3 and HCl (gas), we read from TABLE A.2.2 that:

Cp(NH3) = 29,746 + 25,108 × 10−3T − 1,546 × 105T−2 J K−1mol−1

Cp(HCl) = 26,239 + 5,18 × 10−3T + 1,255 × 105T−2 J K−1mol−1

The reaction heat capacity of NH4Cl (s) → NH3 (g) + HCl (g) is calculated as follows:

∆rCp = Cp(HCl)+ Cp(HCl)− Cp(NH4Cl)

∆rCp = −2045,485 + 11,224T − 23,342 × 10−3T2+ 1,7 × 10−5T3+ 268,602 ×

105T−2 J K−1mol−1

b/ The standard enthalpy of formation of NH3 (g), HCl (g), NH4Cl (s) are:

∆fHNHo 3 = −46,11 kJ mol−1

∆fHHClo = −92,31 kJ mol−1

∆fHNHo 4Cl = −314,43 kJ mol−1

The standard reaction enthalpy of NH4Cl (s) → NH3 (g) + HCl (g) is:

⇒ ∆rH298K = ∆fHNHo 3+ ∆fHHClo − ∆fHNHo 4Cl

∆rH298K= −46,11 + −92,31 − −314,43 = 176,01 kJ mol−1 = 176010 J mol−1

The enthalpy of reation at temperature T (K) is calculated as follows:

∆rHT = ∆rH298K + ∫298T ∆rCpdT

∆rHT = 176010 + ∫ (−2045,485 + 11,224T − 23,342 × 10298T −3T2+ 1,7 × 10−5T3+ 268,602 × 105T−2)dT

∆rHT = 176010 − 2045,485T + 5,612T2− 7,781 × 10−3T3+ 0,425 × 10−5T4− 268,602 × 105T−1+ 373718,451

∆rHT = 549728,451 − 2045,485T + 5,612T2− 7,781 × 10−3T3+ 0,425 × 10−5T4− 268,602 × 105T−1 J mol−1

c/ The standard entropy of NH3 (g), HCl (g), NH4Cl (s) are:

Sm(NHo 3)= 192,45 J K−1mol−1

Sm(HCl)o = 186,91 J K−1mol−1

Sm(NHo 4Cl) = 94,6 J K−1mol−1

The standard reaction entropy of NH4Cl (s) → NH3 (g) + HCl (g) is:

⇒ ∆rS298K = Sm(NHo 3)+ Sm(HCl)o − Sm(NHo 4Cl)

∆rS298K = 192,45 + 186,91 − 94,6 = 284,76 J K−1mol−1

The entropy of reation at temperature T (K) is calculated as follows:

∆rST = ∆rS298K + ∫ ∆rCp

T dT

T 298

∆rST = 284,76 + ∫ −2045,485+11,224T−23,342×10−3T2+1,7×10−5T3+268,602×105T−2

T 298

∆rST = 284,76 + ∫ (−2045,485T298T −1+ 11,224 − 23,342 × 10−3T + 1,7 × 10−5T2+ 268,602 × 105T−3)dT

∆rST = 284,76 − 2045,485lnT + 11,224T − 11,671 × 10−3T2+ 0,567 × 10−5T3− 134,301 × 105T−2+ 9346,183

∆rST = 9630,943 − 2045,485lnT + 11,224T − 11,671 × 10−3T2+ 0,567 × 10−5T3− 134,301 × 105T−2 J K−1mol−1

The Gibbs energy of the reaction at temperature T (K) is calculated as follows:

∆rGT= ∆rHT− T∆rST

∆rGT= 549728,451 − 2045,485T + 5,612T2− 7,781 × 10−3T3+ 0,425 × 10−5T4− 268,602 × 105T−1− T(9630,943 − 2045,485lnT + 11,224T − 11,671 × 10−3T2+ 0,567 × 10−5T3− 134,301 × 105T−2)

∆rGT= 549728,451 − 134,301 × 105T−1+ 2045,485TlnT − 11676,428T − 5,612T2+ 3,89 × 10−3T3− 0,142 × 10−5T4 J mol−1

The equilibrium constant of the reaction at temperature T (K) is calculated as follows:

∆rGT= −RTlnKp

= lnKp =∆rGT

−RT

lnKp =

549728,451−134,301×105T−1+2045,485TlnT−11676,428T−5,612T2+3,89×10−3T3−0,142×10−5T4

−8,314T

lnKp = −66120,814T−1+ 16,154 × 105T−2− 246,029lnT + 1404,43 + 0,675T − 0,468 × 10−3T2+ 0,017 × 10−5T3

d) Because ofexcess NH4Cl, there are three constituents and two phases (one solid, one gas) The total pressure of the system is the sum of partial pressure of NH3 and partial pressure of HCl at equilibrium

PNH3 = PHCl

PTotal = PNH3+ PHCl

where, PNH3 – the partial pressure of NH3, PHCl – the partial pressure of HCl, PTotal – the total pressure Therefore

PNH3 = PHCl =PTotal

2 The equilibrium constant:

KP = PNH3× PHCl = (PTotal

2

Then, PTotal = 2√KP

In additional

lnKp = −66120,814T−1+ 16,154 × 105T−2− 246,029lnT + 1404,43 + 0,675T − 0,468 × 10−3T2+ 0,017 × 10−5T3

Put the right side of equation equal to A

Finally, PTotal = 2√eA The total pressure as a function of temperature

Question 3

Give 350 g of mixture X (the blue point) on the ternary diagram as follows:

a/ Determine the chemical compositions (by wt%) of initial X mixture and formed phases? b/ Calculate the masses of the formed phases?

c/ Determine the minimum mass of acetic acid that must be added to 350 g of mixture X to obtain a single-phase system (only one homogeneous solution)

Solution:

0 20000 40000 60000 80000 100000

120000

140000

160000

180000

200000P (barr)

T (K)

Figure 1 The diagram of total pressure

a/ From X point, draw the line which is perpendicular to the sides of the equilateral triangle (green lines) The chemical compositions (by wt%) of initial X mixture are determined by reading the

length of the line Hence, X mixture consists of 24%wt acetic acid, 53%wt of water and 23% of

chloroform. Using the Tarachenco’s empirical rule: tie lines meet each other at a point S From point S, we bring out the tie line P1P2 whose endpoints (from X) give the 2 formed phases in equilibrium (P1 and P2)

Chemical compositions of the 2 formed phases are similarly determined by above method

Thus, phase P1 consists of 27%wt of acetic acid, 65%wt of water, 8%wt of chloroform and P2 consists of 16%wt acetic acid, 8%wt of water, 76%wt of chloroform

b/ Materials balance for the system:

P1+ P2 = X = 350 (∗)

Apply lever rule for the system: P1

P2 =

0.76 − 0.24 0.24 − 0.08 (∗∗)

Solve (*) and (**), then, P1 = 267.65g và P2=82.35g

c/ To obtain the single phase, acetic acid must be added along the yellow line, so that the achieved system is the point which is on the pink curve (Z) The minimum mass of the acetic acid was found out by lever rule

mAcOH = X ×AZ

1.3 4.6= 98.91g

REFERENCES

[1] Yunus A Çengel & Michael A Boles (2015) Thermodynamics: An Engineer Approach 8 th , McGraw-Hill Education New York

[2] M W Chase (1/8/1998) NIST Chemistry WebBook NIST-JANAF Thermochemical Tables

4th Edition, American Institute of Physics

[3] Peter Atkins (2006) Physical Chemistry 8th Edition Oxford University Press, New York [4] Robert G Mortimer (2008) Physical Chemistry 3rd Edition Elsevier publisher, Canada