termodinamika Chapter 3

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Chapter 3 Problems

Problem 3.1:

Express the volume expansivity β and isothermal compressibility κ as functions of density ρ and its partial derivatives. The isothermal compressibility coefficient () of water at 50 o_{C and 1 bar is} _44.18

∗10−6

bar-1_{. To what pressure must water be compressed at 50}o_{C to change its density by 1%? Assume that  is} independent of P.

Given Data:

Volume expansivity=β=1

V

(

∂V ∂ T

)

P



Or

β= 1

V

(

dV dT

)

P

→(1)

Isothermal Compressibilty=κ=−1

V

(

∂ V ∂ P

)

T



Or

κ=−1

V

(

dV dP

)

T

→(2)

Temperature=T=50 0

_C

Pressure=P1=1¯¿

¯ ¿−1

κ=44.18∗10−6

¿ Density of water=ρ1=1

kg

m3 ρ2=(1+1)

kg

m3 ρ2=1.01

kg m3

P₂=?

Solution:

(a)

 We know that

ρ=1

V V=

1

(2)

 Put in (1) & (2)

 Integrating on both sides

κ

_∫

 Putting values

¯

Generally, volume expansivity β and isothermal compressibility κ depend on T and P. Prove that

(

∂ β

V

(

∂V ∂ T

)

P

(3)

V= 1

The Tait equation for liquids is written for an isotherm as:

V=V₀

(

1− AP

B+P

)

Where V is specific or molar volume, Vo is the hypothetical molar or specific volume at P = 0 and A & B are positive constant. Find an expression for the isothermal compressibility consistent with this equation.

Solution:

 We Know That,

Isothermal Compressibilty=κ=−1

V

(

∂ V ∂ P

)

T

→(1)



Given that

V=V₀

(

1− AP

B+P

)

 Where

 V0 = Hypothetical molar/specific volume at zero pressure, so it is constant

 V = Molar/specific volume

 Now,



Differentiate w.r.t Pressure

(4)

−1

V_o

(

∂V ∂ P

)

=

AB+AP−AP

(B+P)2

−1

V_o

(

∂V ∂ P

)

=

AB

(B+P)2  Since, Temperature is constant

 Therefore,

−1

V_o

(

∂V ∂ P

)

T

= AB (B+P)2

 Or, From (1)

κ= AB

(B+P)2Proved

Problem 3.4:

For liquid water the isothermal compressibility is given by:

κ= c

V(P+b)

Where c & b are functions of temperature only if 1 kg of water is compressed isothermally & reversibly from 1 bar to 500 bars at 60 o_{C, how much work is required?}

At 60 o_{C, b=2700 bars and c = 0.125 cm}3_g-1 Given Data:

Isothermal compressibility=κ= c

V(P+b) Mass of water=m=1kg Pressure=P1=1¯¿

P₂=500bars

Temperature=T=60 0_C

b=2700bars c=0.125cm3

/g Work=W=?

Solution:

 We know that

W=−

∫

PdV →(1) κ= c

V(P+b)→(2)  Also

κ=−1

V

(

dV dP

)

T

(5)

−1

 Putting values

W=0.125cm

Calculate the reversible work done in compressing 1 ft3_{of mercury at a constant temperature of 32F from} 1(atm) to 3,000(atm). The isothermal compressibility of mercury at 32F is:

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 Term, 3.9*10-6 _{has unit of atm}-1 _{& 0.1*10}-9 _{has units of atm}-2

Solution:

 We know that, work done for a reversible process is

W=−

∫

PdV →(1)

Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state at 1 bar during which the temperature change from 0o_{C to 20}o_{C. Determine ΔV}t_{, W, Q, and ΔU}t_{. The} properties for liquid carbon tetrachloride at 1 bar & 0o_{C may be assumed independent of temperature: β =} 1.2 x 10-3_K-1_{Cp = 0.84 kJ kg}-1_K-1,_{ρ = 1590 kg m}-3

Given Data:

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T₁=273.15K

V

(

dV dT

)

P

βdT=1

V dV

 Integrating on both sides,

β

_∫



Putting values

1.2∗10−3

 Now, for total volume,

∆ Vt

(8)

Work done=W=−P ∆ Vt

W=−1¯¿7.638∗10−5

m3∗101325N

1.01325¯¿m2 ∗J Nm ∗1kJ

1000J

W=−7.638∗10−3

kJ Answer

 Now,

 For a reversible process at constant pressure,we have

Q=∆ H Q=m C_P∆ T Q=5kg∗0.84 kJ

kg∗K∗(293.15−273.15)K Q=84kJ Answer

 Now,

 According to first law of thermodynamics,

∆ Ut=Q+W ∆ Ut=(84−7.368∗10−3

₎

kJ ∆ Ut=83.99kJ Answer

Problem 3.7:

A substance for which k is a constant undergoes an isothermal, mechanically reversible process from initial state (P1, V1) to (P2, V2), where V is a molar volume. a) Starting with the definition of k, show that the path of the process is described by

V=A(T)exp(−κP)

b) Determine an exact expression which gives the isothermal work done on 1 mol of this constant-k substance.

Solution:

(a)

 We know that

Isothermal compressibilty=κ=−1

V

(

dV dP

)

T

dV

V =−κdP

∫

dV_V =−κ

_∫

dP

lnV=−κP+lnA(T)

 Where ln A (T) is constant of integration & A depends on T only lnV−lnAT=−κP

ln V

A(T)=−κP  Taking anti log on both sides,

V

=e−κP

(9)

 Or

V=A(T)exp(−κP)Proved

(b)

Work done=W=?

 For a mechanically reversible process, we have, dW=−PdV →(1)

 Using,

d(PV)=PdV+VdP −PdV=VdP−d(PV)

 Put in (1)

dW=VdP−d(PV)→(2)

 We know that

Isothermal compressibilty=κ=−1

V

(

dV dP

)

T

−dV κ =VdP  Put in (2)

dW=−dV

κ −d(PV)

∫

dW=−1

κ

∫

dV−

∫

d(PV) W=−1

κ ΔV−Δ(PV)

 Since volume changes from V1 to V2 & pressure changes from P1 to P2 ,  Therefore,

W=−1

κ

(

V2−V1

)

−

(

P2V2−P1V1

)

W=

(

V₁−V₂

₎

κ +P1V1−P2V2Proved

Problem 3.8:

One mole of an ideal gas with CV = 5/2 R, CP = 7/2 R expands from P1 = 8 bars & T1= 600 K to P2 = 1 bar by each of the following path:

a) Constant volume b) Constant temperature c) Adiabatically

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Given Data:

 According to first law of thermodynamics,

∆ U=Q+W →(1)

 For a constant volume process,

W=0 ∆ U=C_V∆T

 For a mechanically reversible process we have,

∆ H=CP∆ T ∆ H=7₂R

(

T2−T1

)

∆ H=

 For a constant temperature process,

∆ U=0 ∆ H=0

 We know that at constant temperature, work done is

(11)

∆ U=Q+W 0=Q+W Q=−W

 Or

Q=10.373 kJ

mol Answer

(c)

 We know that for an adiabatic process

Q=0

 Now, according to first law of thermodynamics,

∆ U=Q+W ∆ U=W →(1)

∆ U=C_V∆T

 Put in (1)

W=∆ U=C_V∆ T W=∆ U=C_V

₍

T₂−T₁

₎

→(2)  For T2 , We know that for an adiabatic process

T₁P₁

(1−γ)

γ ₌_T

2P2 (1−γ)

γ T₂=T₁

(

P1

P₂

)

(1−γ)

γ _T

2=600K

(

8 1

)

(1−1.4)

1.4 T

2=331.23K

 Put in (2)

W=∆ U=5

2R∗(331.23−600)K W=∆ U=

−5

2 ∗8.314

J

mol∗K∗268.77K W=∆ U=−5586.4

J mol

W=∆ U=5.5864 J

mol Answer

 For a mechanically reversible adiabatic process we have

∆ H=C_P∆ T ∆ H=7

2R

(

T2−T1

)

∆ H= 7

2∗8.314

J

mol∗K∗(331.23−600)K

∆ H=−7.821 J

mol Answer

Problem 3.9:

An ideal gas initially at 600k and 10 bar undergoes a four-step mechanically reversible cycle in a closed system. In step 12, pressure decreases isothermally to 3 bars; in step 23, pressure decreases at constant volume to 2 bars; in step 34, volume decreases at constant pressure; and in step 41, the gas returns adiabatically to its initial state. Take CP = (7/2) R and CV = (5/2) R.

a) Sketch the cycle on a PV diagram.

(12)

Given Data:

Initial Temperature=T₁=600K Initial Pressure=P₁=10¯¿ C_P=7

2R CV=

5 2R

Solution:

(b)

Step 12, an Isothermal process,

 Since

 For an isothermal process, temperature is constant

 Therefore,

T₂=T₁=600K P₂=3¯_¿

 We know that, for an ideal gas

P₂V₂=R T₂ V₂=RT2

P2

mol∗K∗3

¯

¿∗1.01325¯_¿_m2

101325N ∗N∗m

J V₂=8.314∗J∗600K

¿

V2=0.0166 m 3

mol

Step 23, an Isochoric process,

 Since

 For an isochoric process, Volume is constant

 Therefore,

V3=V2=0.0166 m 3

mol P3=2¯¿

P₃V₃=RT₃ T₃=P3V3

R _T

3=

2¯_¿_0.0166_m3

∗mol∗K

mol∗8.314J ∗J

N∗m ∗101325N

1.01325¯_¿_m2

T₃=400K

Step 34, an Isobaric process,

 Since

 For an isobaric process, pressure is constant

(13)

P₄=P₃=2¯¿

 For T4 , we will use an adiabatic relation of temperature and pressure

 As

T4

T₁=

(

P4

P₁

)

R CP T

4=T1

(

P4

P₁

)

R CP T

4=600K∗

(

2 10

)

2∗R

7R _T

4=378.83K

P₄V₄=RT₄ V₄=R T4

P4

mol∗K∗2

¯_¿_∗_1.01325¯_¿_m2

101325N ∗Nm

J V₄=8.314J∗378.83K

¿

V4=0.0157 m 3

mol

Step 41, an adiabatic process,

 Since

 Gas returns to its initial state adiabatically

 Therefore,

T₁=600K P₁=10¯¿

P₁V₁=R T₁ V₁=RT1

P1

mol∗K∗10

¯_¿_∗_1.01325¯_¿_m2

101325N ∗Nm

J V₁=8.314J∗600K

¿

V1=4.988∗10−3 m 3

mol

(c)

Step 12, an Isothermal process,

 Since

 For an isothermal process, temperature is constant

 Therefore

∆ U₁₂=0 ∆ H₁₂=0

(14)

Q=−R T₁lnP2

P1

Q=−8.314 J

mol∗K∗600K∗ln

3

10 Q=6006

J

mol Q=6.006∗10

3 J

mol Answer

 According to first law of thermodynamics

∆ U₁₂=Q₁₂+W₁₂ 0=Q₁₂+W₁₂ W₁₂=−Q₁₂ W₁₂=−6.006∗103 J

mol Answer

Step 23, an Isochoric process,

 Since

 For an isochoric process, Volume is constant

 Therefore,

W₂₃=0

 At constant volume we have

Q23=∆ U23=CV∆ T Q23=∆ U23=CV

(

T3−T2

)

Q23=∆ U23= 5

2R(400−600)K

Q₂₃=∆ U₂₃=

5

2∗8.314J

mol∗K ∗(−200)K

Q₂₃=∆ U₂₃=−4157 J

mol Q23=∆ U23=−4.157∗10 3 J

mol Answer

 We know that

∆ H₂₃=C_P∆ T ∆ H₂₃=C_P

₍

T₃−T₂

₎

∆ H₂₃=7

2R(400−600)K ∆ H23= 7

2∗8.314

J

mol∗K∗(−200)K ∆ H₂₃=−5820 J

mol ∆ H23=−5.82∗10 3 J

mol Answer

Step 34, an Isobaric process,

 Since

 For an isobaric process, pressure is constant

 Therefore, at constant pressure we have,

Q₃₄=∆ H₃₄=C_P∆ T Q₃₄=∆ H₃₄=7

2R

(

T4−T3

)

Q34=∆ H34= 7

2∗8.314

J

mol∗K∗(378.83−400)K Q₃₄=∆ H₃₄=−616 J

mol Answer

 For an Isobaric process we have

W₃₄=−R ∆ T W₃₄=−R

(

T₄−T₃

)

W₃₄=−8.314 J

mol∗K∗(378.83−400)K W34=176

J

(15)

 We know that,

∆ U₃₄=C_V∆T ∆ U₃₄=5

2R

(

T4−T3

)

∆ U34= 5

2∗8.314

J

mol∗K∗(378.83−400)K ∆ U₃₄=−440 J

mol Answer

Step 41, an adiabatic process,

 Since

 For an adiabatic process there is no exchange of heat

 Therefore,

Q₄₁=0

 We know that,

∆ U41=CV∆ T ∆ U41=5₂R

(

T1−T4

)

∆ U41=5₂∗8.314_molJ

∗K∗(600−378.83)K ∆ U41=4597_molJ

∆ U₄₁=4.597∗103 J

mol Answer

 We know that

∆ H₄₁=C_P∆T ∆ H₄₁=7

2R

(

T1−T4

)

∆ H41= 7

2∗8.314

J

mol∗K∗(600−378.83)K ∆ H41=6435.8

J mol

∆ H₄₁=6.4358∗103 J

mol Answer

∆ U₄₁=Q₄₁+W₄₁ ∆ U₄₁=W₄₁ W₄₁=4.597∗103 J

molAnswer

Problem 3.10:

An ideal gas, CP= (5/2) R and CV= (3/2) R is changed from P1 = 1bar and V1t = 12m3 to P2 = 12 bar and

V2t = 1 m3 by the following mechanically reversible processes:

a) Isothermal compression

b) Adiabatic compression followed by cooling at constant pressure. c) Adiabatic compression followed by cooling at constant volume. d) Heating at constant volume followed by cooling at constant pressure. e) Cooling at constant pressure followed by heating at constant volume.

Calculate Q, W, change in U, and change in H for each of these processes, and sketch the paths of all processes on a single PV diagram.

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C_P=5

2R CV=

3

2R Initial pressure=P1=1¯¿ Vt1=12m3 Final pressure=P2=12¯¿

V2t=1m3 Q=? W=? ∆ H=? ∆ U=?

Solution:

 Since

Temperature=constant

 Therefore, for all parts of the problem,

∆ H=0 ∆ U=0

(a)

Isothermal compression,

 For an isothermal process, we have

Q=−R T₁lnP2

P1

 Since

 For an ideal gas, we have

P₁V₁=R T₁

 Therefore,

Q=−P₁V₁lnP2

P1

Q=−1¯¿12m3

∗ln 12

1 ∗101325N 1.01325¯_¿_m2 ∗J

Nm ∗1kJ

1000J

Q=−2981.88kJ Answer

∆ U=Q+W 0=Q+W W=−Q W₁₂=2981.88kJ Answer

(b)

Adiabatic compression followed by cooling at constant pressure

 Since

 For an adiabatic process, there is no exchange of heat

 Therefore,

Q=0Answer

 The process completes in two steps

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P₂

(

V'

₎

γ

=P₁V₁ V'=V₁

(

P1 P2

)

1

γ

 For mono atomic gas, we have

γ=1.67

V'=12m3∗

(

1

12

)

1 1.67 _V'

=2.71m3

 We know that,

W₁=P2V '

−P1V1

γ−1

W₁=

(12∗2.71−1∗12)¯¿m3

1.67−1 ∗101325N

1.01325¯¿m2 ∗J

Nm ∗1kJ

1000J

W1=3063kJ →(1)

 Second step, cooling at constant pressure P2

 We know that, for a mechanically reversible process

W2=−P2

(

V2−V

'

)

W₂=−12(1−¯2.71)

m3

∗101325N

1.01325¯¿m2 ∗J

Nm ∗1kJ

1000J

W2=2052kJ →(2)

 Now

W=W₁+W₂ W=(3063+2052)kJ W=5115kJ Answer

(c)

Adiabatic compression followed by cooling at constant volume

 Since

 For an adiabatic process, there is no exchange of heat

 Therefore,

Q=0Answer

 First step, an adiabatic compression to volume V2 , intermediate pressure can be given as

P'V2

γ

=P1V1 P

'

=P₁

(

V1

V₂

)

γ

(18)

γ=1.67

P'=1

(

12¯ 1

)

1.67

P'=63.42¯_¿  We know that,

W₁=P '

V2−P1V1

γ−1

W₁=

(63.42∗1−1∗12)¯¿m3

1.67−1 ∗101325N

1.01325¯¿m2 ∗J

Nm ∗1kJ

1000J

W₁=7674.76kJ

 Second step, cooling at constant Volume,

 Therefore, No work will be done

W₂=0

 Now

W=W₁+W₂ W=(7674.76+0)kJ W=7674.76kJ Answer

(d)

Heating at constant volume followed by cooling at constant pressure

 The process completes in two steps

 Step 1, Heating at constant volume to P2

 Therefore no work will be done

W₁=0

 Step 2, Cooling at constant pressure P2 To V2

 We know that, for a mechanically reversible process

W₂=−P₂∆ V W₂=−P₂

(

V₂−V₁

)

W₂=−12(1−¯12)

m3

∗101325N

1.01325¯_¿_m2 ∗J

Nm ∗1kJ

1000J

W₂=13200kJ

 Now

 According to first law of thermodynamics

∆ U=Q+W 0=Q+W Q=−W Q=−13200kJ Answer

(19)

Cooling at constant pressure followed by heating at constant volume

 The process completes in two steps

 Step 1, Cooling at constant Pressure P1 to V2

 Therefore, for a mechanically reversible process

W₁=−P₁∆ V W₁=−P₁

(

V₂−V₁

)

W₁=−1(1−¯12)

m3

∗101325N

1.01325¯_¿_m2 ∗J

Nm ∗1kJ

1000J

W₁=1100kJ

 Step 1, Heating at constant Volume V2 to pressure P2

 Therefore no work will be done

W₂=0

 Now

∆ U=Q+W 0=Q+W Q=−W Q=−1100kJ Answer

Problem 3.11:

The environmental lapse rate dT_dz characterizes the local variation of temperature with elevation in the

earth's atmosphere. Atmospheric pressure varies with elevation according to the hydrostatic formula,

dP

dz =−M ρg

Where M is a molar mass, ρ is molar density and g is the local acceleration of gravity. Assume that the atmosphere is an ideal gas, with T related to P by the polytropic formula equation (3.35 c). Develop an expression for the environmental lapse rate in relation to M, g, R, and δ.

Solution:

 Given that

dP

dz=−M ρg →(1)

 The polytropic relation is

T P

1−δ

(20)

 Or

 To =Temperature at sea level, so it is constant

 Po = Pressure at sea level, so it is constant

 Differentiate w.r.t to Temperature on both sides

dP  Put (a) in above equation

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P_o

An evacuated tank is filled with gas from a constant pressure line. Develop an expression relating the temperature of the gas in the tank to temperature T’ of the gas in line. Assume that gas is ideal with constant heat capacities, and ignore heat transfer between the gas and the tank. Mass and energy balances for this problem are treated in Ex. 2.13.

Solution:

 Choose the tank as the control volume. There is no work, no heat transfer & kinetic & potential energy changes are assumed negligible.

 Therefore, applying energy balance

d(mU)_tank

 Tank is filled with gas from an entrance line, but no gas is being escaped out,

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 Where prime (‘) denotes the entrance stream

 Applying mass balance

m'=d mtank

dt →(2)

 Combining equation (1) & (2)

d(mU)_tank

dt −H

'd mtank

dt =0

1

dt

{

d(mU)tank−H '_{d m}

tank

}

=0 d(mU)tank=H'd mtank  Integrating on both sides

∫

m1

m2

d(mU)_tank=H'

∫

m1

m2

d m_tank ∆(mU)_tank=H'

(

m2−m1

)

m2U2−m1U1=H'

(

m2−m1

)

 Because mass in the tank initially is zero, therefore

m₁=0

m2U2=H'm2

U2=H'→(3)

 We know that

U=C_VT U₂=C_VT₂→(a)

 Also

H'

=CPT'→(b)  Put (a) & (b) in (3)

CVT=CPT' T=

C_P CV

T'

 Since heat capacities are constant, therefore

γ=CP

CV T=γ T '

Proved

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A tank of 0.1-m3_{volume contains air at 25}o_{C and 101.33 kPa. The tank is connected to a compressed-air} line which supplies air at the constant conditions of 45o_{C and 1,500 kPa. A valve in the line is cracked so} that air flows slowly into the tank until the pressure equals the line pressure. If the process occurs slowly enough that the temperature in the tank remains at 25 o_{C, how much heat is lost from the tank? Assume} air to be an ideal gas for which CP = (7/2) R and CV = (5/2) R

Given Data:

Volume=V=0.1m3 T1=25o.C=298K P1=101.33kPa T2=45o.C=318K P2=1500kPa

Heat lost=Q=? C_P=7

2R CV=

5 2R

Solution:

∆ U=Q+W →(1)

 Since

∆ H=∆ U+∆(PV) ∆ U=∆ H−∆(PV) ∆ U=∆ H−P ∆ V−V ∆ P→(a)  Also, we know that

W=−P ∆ V →(b)  Put (a) & (b) in (1)

∆ H−P ∆V−V ∆ P=Q−P ∆ V ∆ H−V ∆ P=Q →(2)  Also, we have

∆ H=nC_P∆T ∆ H=nC_P

₍

T₂−T₁

₎

 Put in (2)

n C_P

(

T₂−T₁

)

−V ∆ P=Q →(3)

 For “n”

 We know that for an ideal gas,

PV=nRT

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P₁V=n₁R T₁ n₁=P1V

R T1

 The final number of moles of gas at temperature T1 are

P₂V=n₂R T₁ n₂=P2V

R T1

 Now, Applying molar balance

n=n₁−n₂ n=P1V

R T1

−P2V

R T1

n=

(

P1−P2

)

V

R T₁  Put in (3)

(

P₁−P₂

₎

V R T₁ CP

(

T₂−T₁

)

−V ∆ P=Q

(

P₁−P₂

)

V R T₁ ∗7

2 R∗

(

T2−T1

)

−V ∆ P=Q

(

P1−P2

)

V

T1

∗7

2 ∗

(

T2−T1

)

−V

(

P2−P1

)

=Q

(101.33−1500)kPa∗0.1m3

298K ∗7

2 ∗(318−298)K−0.1m

3

(1500−101.33)kPa=Q

Q=−172.717m3

kPa∗1kN

1kPa∗m2∗1kJ

1kNm

Q=−172.717kJ Answer

Problem 3.17:

A rigid, no conducting tank with a volume of 4 m3_{is divided into two unequal parts by a thin membrane.} One side of the membrane, representing 1/3 of the tank, contains nitrogen gas at 6 bars and 100 o_{C, and} the other side, representing 2/3 of the tank, is evacuated. The membrane ruptures and the gas fills the tank.

a) What is the final, temperature of the gas? How much work is done? Is the process reversible? b) Describe a reversible process by which the gas can be returned to its initial state, How much work

is done

(25)

Volume of the tank=V1=4m3 V2=

V₁∗1

3 =

4

3m

3 _Pressure₌_P

2=6¯¿ Temperature=T1=100o.C

V₃=V1∗2

3 =

8

3m

3

Solution:

(a)

Finaltemperature=T₂=?

∆ U=Q+W

 Since

 No work is done & no heat is transferred  Therefore

Q=W=0

∆ U=0 mCV∆ T=0 ∆ T=0 T2−T1=0 T2=T1 T2=100℃Answer

 No, process is not reversible

(b)

 Since

 Therefore, the process is isothermal

 For an isothermal process we have

W=−R T₂lnV2

V1

 As, for an ideal gas

P₂V₂=R T₂ W=−P2V2ln

V₂ V1

W=−6¯_¿4

3m

3

ln 4

3∗4 W=8.788¯¿

m3∗101325N

1.01325¯¿m2 ∗1

kJ

1000Nm

W=878.8kJ Answer

Problem 3.18:

An ideal gas initially at 30 0_{C and 100 kPa undergoes the following cyclic processes in a closed system:} a) In mechanically reversible processes, it is first compressed adiabatically to 500 kPa then cooled at

(26)

b) The cycle traverses exactly the same changes of state but each step is irreversible with an efficiency of 80% compared with the corresponding mechanically reversible process NOTE: the initial step can no longer be adiabatic

Find Q W ∆ U and ∆ H for each step of the process and for the cycle Take Cp = (7/2) R and CV =

1) Adiabatic Compression from point 1 to point 2

Q₁₂=0

 Now, from first law of thermodynamics,

∆ U₁₂=Q₁₂+W₁₂ ∆ U₁₂=W₁₂

2) Cooling at constant pressure from point 2 to point 3  Therefore at constant pressure we have,

Q₂₃=∆ H₂₃=C_P∆T₂₃ Q₂₃=∆ H₂₃=7

(27)

 Here

T₃=303.15K

Q₂₃=∆ H₂₃=7

2∗8.314

J

mol∗K (303.15−480.13)K∗1 kJ

1000J Q23=∆ H23=−5.15

kJ mol

 Also, we have

∆ U₂₃=C_V

(

T₃−T₂

)

∆ U₂₃=5

2∗8.314

J

mol∗K (303.15−480.13)K∗1 kJ

1000J ∆ U₂₃=−3.679 kJ

mol

 Now, from first law of thermodynamics,

∆ U₂₃=Q₂₃+W₂₃ W₂₃=∆U₂₃−Q₂₃ W₂₃=−3.679+5.15 W₂₃=1.471 kJ

mol

3) Isothermal expansion from point 3 to point 1

 Since for an isothermal process temperature remains constant

 Therefore,

∆ U₃₁=∆ H₃₁=0

 Here

P₃=P₂=500kPa

 For an Isothermal process we have

W₃₁=−RT₃ln P3

P1

W₃₁=−8.314 J

mol∗K∗303.15K∗ln

500 100∗1kJ

1000J

W₃₁=−4.056 kJ

mol

∆ U₃₁=Q₃₁+W₃₁

0=Q₃₁+W₃₁ Q₃₁=−W₃₁ Q₃₁=4.056 kJ

mol

 For the complete cycle,

(28)

W=W12+W23+W31 W=3.679+1.471−4.056

W=1.094 kJ

mol Answer

∆ H=∆ H₁₂+∆ H₂₃+∆ H₃₁ ∆ H=5.15−5.15+0 ∆ H=0 kJ

mol Answer

∆ U=∆ U12+∆ U23+∆ U31 ∆ U=3.679−3.679+0 ∆ U=0_molkJ Answer

(b)

 If each step that is 80% accomplishes the same change of state then values of ∆ U & ∆ H will remain same as in part (a) but values of Q & W will change.

1. Adiabatic Compression from point 1 to point 2

W₁₂=W12

0.8 W12=

3.679

0.8 W12=4.598

kJ mol

∆ U₁₂=Q₁₂+W₁₂

3.679 kJ

mol=Q12+4.598

kJ mol

Q₁₂=3.679 kJ

mol−4.598 kJ

mol Q12=−0.92

kJ mol

2. Cooling at constant pressure from point 2 to point 3

W₂₃=W23

0.8 W23=

1.471

0.8 W23=1.839

kJ mol

∆ U₂₃=Q₂₃+W₂₃ −3.679 kJ

mol=Q23+1.839

kJ

mol Q23=−3.679

kJ

mol−1.839 kJ

mol Q23=−5.518

kJ mol

3. Isothermal expansion from point 3 to point 1  Since initial step can no longer be adiabatic , therefore

W₃₁=W₃₁∗0.8 W31=−4.056

kJ

mol∗0.8 W31=3.245

kJ mol

(29)

Q31=3.245

kJ mol

Q=Q₁₂+Q₂₃+Q₃₁ Q=−0.92−5.518+3.245 Q=−3.193 kJ

mol Answer

W=W₁₂+W₂₃+W₃₁ W=4.598+1.839−3.245 W=3.192 kJ

mol Answer

Problem 3.19:

One cubic meter of an ideal gas at 600 K and 1,000 kPa expands to five times its initial volume as follows: a) By a mechanically reversible, isothermal process

b) By a mechanically reversible adiabatic process

c) By adiabatic irreversible process in which expansion is against a restraining pressure of 100 kPa For each case calculate the final temperature, pressure and the work done by the gas, Cp=21 J mol-1K-1.

Given Data:

V1=1m3 T1=600K P1=1000kPa V2=5V1 V2=5m3 CP=21

J

mol K CV=? T2=? P2=?

W=?

Solution:

 We know that,

C_P−C_V=R C_V=C_P−R C_V=(21−8.314) J

mol∗K CV=12.686

J mol∗K  As

γ=CP

CV

γ=1.6554

(a)

 Since, for an isothermal process

 Temperature remains constant, therefore

T₂=T₁=600K Answer

(30)

P₁V₁

 We know that, for an isothermal process

W=−R T₁lnV2

W=−1609.43kJ Answer

(b)

 We know that, for an adiabatic process

P1V1

P₂=69.65kPa Answer

 For an ideal gas we have

 For an adiabatic process work done is

W=P2V2−P1V1

W=−994.43kJ Answer

(c)

P_r=100kPa

 Since, for an adiabatic process

Q=0

(31)

∆ U=Q+W ∆ U=W ∆ U=W=−P_rdV ∆ U=W=−P_r

₍

V₂−V₁

₎

∆ U=W=−100(5−1)

kPa∗m3

∗N Pa∗m2 ∗J

Nm

∆ U=−400kJ n C_V∆T=−400kJ n C_V

(

T₂−T₁

)

=−400kJ

T₂=−400kJ

n C_V +T1→(1)

 For an ideal gas we have,

P₁V₁=nR T₁ n=P1V1

R T1

n=

1000kPa∗1m3∗mol∗K

8.314J∗600K ∗kN

kPa∗m2 ∗kJ kNm

n=0.2005mol

 Put in (1)

T₂=

−400kJ∗mol∗K

0.2005mol∗12.686J∗1000J

1kJ +600K

T₂=−157.26K+600K T₂=442.74K Answer

 For an ideal gas we have

P₁V₁ T1

=P2V2

T2 P2=

P₁V₁ T1

∗T₂

V₂ P2=

1000kPa∗1m3

600K ∗442.74K

5m3

P₂=147.58kPa Answe r

Problem 3.20:

One mole of air, initially at 150 0_{C and 8 bars undergoes the following mechanically reversible changes. It} expands isothermally to a pressure such that when it is cooled at constant volume to 50 0_{C its final} pressure is 3 bars. Assuming air is an ideal gas for which CP = (7/2) R and CV = (5/2) R, calculate W, Q,

∆ U , and ∆ H

Given Data:

Mole of air=n=1mol Initial Temperature=T1=150.C

0

=423.15K Initial pressure=P₁=8¯¿

Finaltemperature=T3=500.C=323.15K Final pressure=P3=3¯¿ CP=

7

2R CV=

5 2R

Solution:

 Since process is reversible

(32)

Step 12:

 For step 12 temperatures is constant,

T₁=T₂

 Therefore

∆ U₁₂=∆ H₁₂=0

 For an isothermal process we have

W₁₂=R T₁lnV1

V2

 As

V2=V3 W12=R T1ln

V₁ V3

→(1)

 We know that

P₁V₁ T1

=P3V3

T3

V₁ V3

=P3¿T1

T3∗P1

W₁₂=R T₁lnP1¿T3

T1∗P3 W12=

8.314J∗423.15K mol∗K ∗1kJ

1000J ∗ln

3∗423.15 8∗323.15

W12=−2.502

kJ mol

∆ U₁₂=Q₁₂+W₁₂ 0=Q₁₂+W₁₂ Q₁₂=−W₁₂ Q₁₂=2.502 kJ

mol

Step 23:

 For step 23 volume is constant,

 Therefore,

W₂₃=0

∆ U₂₃=Q₂₃+W₂₃ ∆ U₂₃=Q₂₃+0 Q₂₃=∆ U₂₃ Q₂₃=∆ U₂₃=C_V∆ T Q₂₃=∆ U₂₃=C_V

(

T₃−T₂

)

Q₂₃=∆ U₂₃=5

2R(323.15−423.15)K

Q₂₃=∆ U₂₃=5

2∗8.314

J

mol∗K∗1kJ

1000J (−100)K Q23=∆U23=−2.0785

kJ mol

(33)

∆ H₂₃=C_P∆ T ∆ H₂₃=C_P

(

T₃−T₂

)

_{∆ H} 23=

7

2∗8.314

J

mol∗K∗1kJ

1000J (423.15−323.15)K ∆ H₂₃=2.91 kJ

mol

Work=W=W₁₂+W₂₃ W=(−2.502+0) kJ

mol W=−2.502

kJ

mol Answe r

Q=Q₁₂+Q₂₃ Q=(2.502−2.0785) kJ

mol Q=0.424

kJ

mol Answer

∆ U=∆ U₁₂+∆ U₂₃ ∆ U=(0−2.0785) kJ

mol ∆ U=−2.0785

kJ

mol Answe r

∆ H=∆ H₁₂+∆ H₂₃ ∆ H=(0−2.91) kJ

mol ∆ H=−2.91 kJ

mol Answe r

Problem 3.21:

An ideal gas flows through a horizontal tube at steady state. No heat is added and no shaft work is done. The cross-sectional area of the tube changes with length, and this causes the velocity to change. Derive an equation relating the temperature to the velocity of the gas. If nitrogen at 150 0_{C flows past one section of} the tube with a velocity of 2.5 m/s, what is the temperature at another section where its velocity is 50 m/s? Let CP = (7/2) R

Given Data:

Temperature=T1=1500.C=423.15K Velocity=u1=2.5

m

sec T2=? u2=50

m

sec CP=

7 2R

Molecualr weight of Nitrogen=28 g

mol

Solution:

 Applying energy balance for steady state flow process

∆ H+∆ u

2

2 +g ∆ z=Q+WS

 Since

(34)

 Therefore,

∆ H+∆ u

2

2 =0 CP∆ T=

−∆u2

2 CP

(

T2−T1

)

=

−u2 2

−u1 2

2 T2=

−u2 2

−u1 2

2C_P +T1

T₂=

−(502−2.52

)∗

2∗m2∗mol∗K

2∗7∗8.314J∗sec2 ∗28g Nitrogen

1mol Nitrogen ∗J

N∗m ∗N∗sec

2

kg∗m ∗1kg

1000g +423.15K

T₂=−1.199K+423.15K

T₂=421.95K T2=(421.95−273.15)0.C T2=148.80.C Answe r

Problem 3.22:

One mole of an ideal gas, initially at 30 0_{C and 1 bar, is changed to 130}0_{C and 10 bars by three different} mechanically reversible processes:

a) The gas is first heated at constant volume until its temperature is 130 0_{C; then it is compressed} isothermally until its pressure is 10 bar

b) The gas is first heated at constant pressure until its temperature is 130 0_{C; then it is compressed} isothermally to 10 bar

c) The gas is first compressed isothermally to 10 bar; then it is heated at constant pressure to 130 0_C Calculate Q, W, ∆ U∧∆ H in each case. Take CP = (7/2) R and CV = (5/2) R. alternatively, take CP = (5/2) R and CV = (3/2) R

Given Data:

T1=300.C T1=(30+273.15)K T1=303.15K P1=1¯¿ T2=1300.C T3=(130+273.15)K

T₃=403.15K P₃=10¯¿ Q=? W=? ∆ U=? ∆ H=?

Solution:

C_P=7

2R CV=

5 2R

 Each part consist of two steps, 12 & 23

 For the overall processes

∆ U=∆ U12=∆U23=CV∆T ∆ U=∆ U12=∆U23= 5

(35)

∆ U=∆ U12=∆U23= 5

2∗8.314

J

mol∗K(403.15−303.15)

K∗1kJ

1000J

∆ U=∆ U₁₂=∆U₂₃=2.079 kJ

mol→(a)Answe r

 Now

∆ H=∆ H₁₂=∆ H₂₃=C_P∆ T

∆ H=∆ H₁₂=∆ H₂₃=7

2R

(

T2−T1

)

∆ H=∆ H12=∆ H23= 7

2∗8.314

J

mol∗K (403.15−303.15)K∗1 kJ

1000J

∆ H=∆ H₁₂=∆ H₂₃=2.91 kJ

mol→(b)Answer

(a)

Step 12:

 For step “12” volume is constant

 Therefore

W₁₂=0

 Here

T₂=T₃

∆ U12=Q12+W12 ∆ U12=Q12 Q12=∆ U12=CV∆ T Q12=∆ U12=2.079

kJ mol

[

¿(a)

]

 Also we have



∆ H₁₂=2.91 kJ

mol

[

¿(b)

]

Step 23:

 Since for step “23” process is isothermal

 Therefore

∆ U₂₃=∆ H₂₃=0

(36)

T₂=T₃

 Now, intermediate pressure can be calculated as

P₁ T1

=P2

T2

P₂=P1

T1

∗T₂ 1

¯¿

303.15K∗403.15K P₂=¿

P₂=1.329b ar

W₂₃=R T₂lnP3

P2

W₂₃=8.314 J

mol∗K∗403.15

K∗1kJ

1000J ∗ln

10 1.329

W₂₃=6.764 kJ

mol

∆ U23=Q23+W23 0=Q23+W23 Q23=−W23 Q23=−6.764

kJ mol

Work=W=W₁₂+W₂₃ W=(0+6.764) kJ

mol W=6.764

kJ

mol Answe r

Q=Q₁₂+Q₂₃ Q=(2.079−6.764) kJ

mol Q=−4.685

kJ

mol Answe r

∆ U=∆ U₁₂+∆ U₂₃ ∆ U=(2.079+0) kJ

mol ∆ U=2.079

kJ

molAnswe r

∆ H=∆ H₁₂+∆ H₂₃ ∆ H=(2.91+0) kJ

mol ∆ H=2.91 kJ

mol Answe r

(b)

Step 12:

 For step “12” volume is constant

 Therefore, at constant pressure we have

Q₁₂¿∆ H₁₂=2.91 kJ

mol

[

¿(b)

]

 Also,

(37)

∆ U₁₂=Q₁₂+W₁₂ W₁₂=∆ U₁₂−Q₁₂ W₁₂=(2.079−2.91) kJ

mol W12=−0.831

kJ mol

Step 23:

 Since for step “23” process is isothermal ( T = Constant)

 Therefore

∆ U₂₃=∆ H₂₃=0

 Here

T₂=T₃∧P₁=P₂

 For an isothermal process we have W₂₃=R T₂lnP3

P2

W₂₃=8.314 J

mol∗K∗403.15

K∗1kJ

1000J ∗ln

10 1

W₂₃=7.718 kJ

mol

∆ U₂₃=Q₂₃+W₂₃ 0=Q₂₃+W₂₃ Q₂₃=−W₂₃ Q₂₃=−7.718 kJ

mol

Work=W=W₁₂+W₂₃ W=(−0.831+7.718) kJ

mol W=6.887

kJ

mol Answer

Q=Q12+Q23 Q=(2.91−7.718)_molkJ Q=−4.808_molkJ Answe r

∆ U=∆ U12+∆ U23 ∆ U=(2.079+0)_molkJ ∆ U=2.079_molkJ Answe r

∆ H=∆ H₁₂+∆ H₂₃ ∆ H=(2.91+0) kJ

mol ∆ H=2.91 kJ

mol Answe r

(c)

 Therefore

∆ U₁₂=∆ H₁₂=0

 Here

(38)

W₁₂=R T₁lnP2

P1

W₁₂=8.314 J

mol∗K∗303.15

K∗1kJ

1000J ∗ln

10 1

W12=5.8034

kJ mol

∆ U₁₂=Q₁₂+W₁₂ 0=Q₁₂+W₁₂ Q₁₂=−W₁₂ Q₁₂=−5.8034 kJ

mol

Step 23:

Q₂₃=∆ H₂₃=2.91 kJ

mol

[

¿(b)

]

 Here

T₂=T₃

 Now

∆ U₂₃=2.079 kJ

mol

[

¿(a)

]

∆ U₂₃=Q₂₃+W₂₃ W₂₃=∆U₂₃−Q₂₃ W₂₃=(2.079−2.91) kJ

mol W23=−0.831

kJ mol

Work=W=W₁₂+W₂₃ W=(5.8034−0.831) kJ

mol W=4.972

kJ

mol Answer

Q=Q₁₂+Q₂₃ Q=(−5.8034+2.91) kJ

mol Q=−2.894

kJ

molAnswe r

∆ U=∆ U₁₂+∆ U₂₃ ∆ U=(0+2.079) kJ

mol ∆ U=2.079

kJ

molAnswe r

∆ H=∆ H₁₂+∆ H₂₃ ∆ H=(0+2.91) kJ

mol ∆ H=2.91 kJ

(39)

Solution:

C_P=5

2R CV=

3 2R

 Each part consist of two steps, 12 & 23

 For the overall processes

∆ U=∆ U₁₂=∆U₂₃=C_V∆T ∆ U=∆ U12=∆U23=3₂R

(

T3−T1

)

∆ U=∆ U₁₂=∆U₂₃=3

2∗8.314

J

mol∗K(403.15−303.15)

K∗1kJ

1000J

∆ U=∆ U₁₂=∆U₂₃=1.247 kJ

mol→(a)Answe r

 Now

∆ H=∆ H12=∆ H23=CP∆ T

∆ H=∆ H₁₂=∆ H₂₃=5

2R

(

T2−T1

)

∆ H=∆ H₁₂=∆ H₂₃=5

2∗8.314

J

mol∗K(403.15−303.15)K∗1 kJ

1000J

∆ H=∆ H₁₂=∆ H₂₃=2.079 kJ

mol→(b)Answe r

(a)

Step 12:

 For step “12” volume is constant

 Therefore

W₁₂=0

 Here

T₂=T₃

∆ U₁₂=Q₁₂+W₁₂ ∆ U₁₂=Q₁₂ Q₁₂=∆ U₁₂=C_V∆ T Q₁₂=∆ U₁₂=1.247 kJ

(40)

 Also we have

∆ H₁₂=2.079 kJ

mol

[

¿(b)

]

Step 23:

 Since for step “23” process is isothermal

 Therefore

∆ U₂₃=∆ H₂₃=0

 Here

T₂=T₃

 Now, intermediate pressure can be calculated as

P₁ T1

=P2

T2

P₂=P1

T1

∗T₂ 1

¯_¿

303.15K∗403.15K P2=¿

P₂=1.329b ar

 For an isothermal process we have W₂₃=R T₂lnP3

P2

W23=8.314

J

mol∗K∗403.15

K∗1kJ

1000J ∗ln

10 1.329

W₂₃=6.764 kJ

mol

∆ U₂₃=Q₂₃+W₂₃ 0=Q₂₃+W₂₃ Q₂₃=−W₂₃ Q23=−6.764

kJ mol

Work=W=W₁₂+W₂₃ W=(0+6.764) kJ

mol W=6.764

kJ

mol Answe r

Q=Q₁₂+Q₂₃ Q=(1.247−6.764) kJ

mol Q=−5.516

kJ

molAnswe r

∆ U=∆ U₁₂+∆ U₂₃ ∆ U=(1.247+0) kJ

mol ∆ U=1.247

kJ

mol Answer

∆ H=∆ H₁₂+∆ H₂₃ ∆ H=(2.079+0) kJ

mol ∆ H=2.079

kJ

(41)

(b)

Step 12:

Q₁₂¿∆ H₁₂=2.079 kJ

mol

[

¿(b)

]

 Also,

∆ U₁₂=1.247 kJ

mol

[

¿(a)

]

∆ U₁₂=Q₁₂+W₁₂ W₁₂=∆ U₁₂−Q₁₂ W12=(1.247−2.079)

kJ

mol W12=−0.832

kJ mol

Step 23:

 Therefore

∆ U23=∆ H23=0

 Here

T₂=T₃∧P₁=P₂

W₂₃=R T₂lnP3

P2

W23=8.314

J

mol∗K∗403.15

K∗1kJ

1000J ∗ln

10 1

W₂₃=7.718 kJ

mol

∆ U₂₃=Q₂₃+W₂₃ 0=Q₂₃+W₂₃ Q₂₃=−W₂₃ Q23=−7.718

kJ mol

Work=W=W₁₂+W₂₃ W=(−0.832+7.718) kJ

mol W=6.886

kJ

molAnswe r

Q=Q₁₂+Q₂₃ Q=(2.079−7.718) kJ

mol Q=−5.639

kJ

(42)

∆ U=∆ U₁₂+∆ U₂₃ ∆ U=(1.247+0) kJ

mol ∆ U=1.247

kJ

mol Answer

∆ H=∆ H₁₂+∆ H₂₃ ∆ H=(2.079+0) kJ

mol ∆ H=2.079

kJ

mol Answe r

(c)

Step 12:

 Therefore

∆ U₁₂=∆ H₁₂=0

 Here

P2=P3

 For an isothermal process we have W₁₂=R T₁lnP2

P1

W₁₂=8.314 J

mol∗K∗303.15

K∗1kJ

1000J ∗ln

10 1

W₁₂=5.8034 kJ

mol

∆ U₁₂=Q₁₂+W₁₂ 0=Q₁₂+W₁₂ Q₁₂=−W₁₂ Q₁₂=−5.8034 kJ

mol

Step 23:

Q23=∆ H23=2.079

kJ

mol

[

¿(b)

]

 Here

T₂=T₃

 Now

∆ U₂₃=1.247 kJ

mol

[

¿(a)

]

(43)

∆ U₂₃=Q₂₃+W₂₃ W₂₃=∆U₂₃−Q₂₃ W23=(1.247−2.079)

kJ

mol W23=−0.832

kJ mol

Work=W=W₁₂+W₂₃ W=(5.8034−0.832) kJ

mol W=4.9714

kJ

molAnswe r

Q=Q₁₂+Q₂₃ Q=(−5.8034+2.079) kJ

mol Q=−3.724

kJ

mol Answe r

∆ U=∆ U₁₂+∆ U₂₃ ∆ U=(0+1.247) kJ

mol ∆ U=1.247

kJ

mol Answer

∆ H=∆ H₁₂+∆ H₂₃ ∆ H=(0+2.079) kJ

mol ∆ H=2.079

kJ

mol Answe r

Problem 3.23:

One mole of an ideal gas, initially at 30 ℃ and 1 bars, undergoes the following mechanically reversible changes. It is compressed isothermally to point such that when it is heated at constant volume to 120 ℃ its final pressure is 12 bars. Calculate Q, W, ∆ U∧∆ H for the process. Take CP = (7/2) R and CV = (5/2) R.

Given Data:

T₁=30℃ T₁=(30+273.15)K T₁=303.15K P₁=1¯¿ T₃=120℃ T₃=(120+273.15)K

T₃=393.15K P₃=12¯¿ Q=? W=? ∆ U=? ∆ H=? C_P=7

2R CV=

5 2R

Solution:

 The process consist of two steps, 12 & 23

Step 12:

 Therefore

∆ U₁₂=∆ H₁₂=0

(44)

P₂

W₁₂=R T₁lnP2

∆ U₁₂=Q₁₂+W₁₂ 0=Q₁₂+W₁₂ Q₁₂=−W₁₂ Q₁₂=−5.607 kJ

mol

Step 23:

 Since for step “23” volume is constant

 Therefore

W₂₃=0

∆ U23=Q23+W23 ∆ U23=Q23

Work=W=W₁₂+W₂₃ W=(5.607+0) kJ

mol W=5.607

kJ

molAnswe r

Q=Q₁₂+Q₂₃ Q=(−5.607+1.871) kJ

mol Q=−3.736

kJ