**Chapter 3 **

**Chapter 3**

**Problems**

**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 (*

*−6*

**) of water at 50****o**_{C and 1 bar is 44.18∗10}

**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}

_{C}

*Pressure=P*1=1 ¯¿ ¯ ¿−1

*κ=44.18∗10*−6¿

*Density of water=ρ*1=1

*kg*

*m*3

*ρ*2=(1+1 )

*kg*

*m*3

*ρ*2=1.01

*kg*

*m*3

*P*

_{2}=

*?*

### Solution:

* We know that*

**(a)***ρ=*1
*V* *V =*
1
*ρ*
Put in (1) & (2)
*β=ρ*

### (

*d*

*dT*1

*ρ*

### )

*P*

*β=*−

*ρ*

*ρ*2

### (

*dρ*

*dT*

### )

*P*

*β=*−1

*ρ*

### (

*dρ*

*dT*

### )

*P*

*Proved* Now,

*κ=−ρ*

### (

*d*

*dP*1

*ρ*

### )

*T*

*κ=*

*ρ*

*ρ*2

### (

*dρ*

*dP*

### )

*T*

*κ=*1

*ρ*

### (

*dρ*

*dP*

### )

*T*

*Proved*

_{ }

_{ }

* As*

**(b)***κ=*1

*ρ*

### (

*dρ*

*dP*

### )

*T*

*κdP=dρ*

*ρ*

Integrating on both sides

*κ*

### ∫

*P*1

*P*2

*dP=*

### ∫

*ρ*1

*ρ*2

*dρ*

*ρ*

*κ*|

*P*|

*P*1

*P*2 =|

*ln ρ*|

*ρ*1

*ρ*2

*κ*

### (

*P*

_{2}−

*P*

_{1}

_{)}

=_{(}

*ln ρ*

_{2}−ln ρ

_{1}

_{)}

*κ*

_{(}

*P*

_{2}−

*P*

_{1}

_{)}

=ln*ρ*2

*ρ*1 Putting values ¯ ¿

_{(}

*P*

_{2}−1

### )

=ln1.01 1 44.18∗10−6 ¿ ¯¿ 44.18∗10−6*P*

_{2}−1=0.00995∗¿

*P*

_{2}=(225.22+1)¯¿

*P*2=226.22 ¯¿

_{ Answer}

### Problem 3.2:

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

### (

*∂ β*

### Solution:

We know that*Volume expansivity=β=*1

*V*

### (

*∂V*

*∂ T*

### )

*P*

###

*Since β is very small*

*V =* 1
*∂ β*

### (

*∂ V*

*∂ T*

### )

*P*

*→ (1)*

*Isothermal Compressibilty=κ=*−1

*V*

### (

*∂ V*

*∂ P*

### )

*T*

###

*Since κ is very small*

*V =*−1
*∂κ*

### (

*∂ V*

*∂ P*

### )

*T*1

*∂ β*

### (

*∂V*

*∂ T*

### )

*P*=−1

*∂ κ*

### (

*∂V*

*∂ P*

### )

*T*1

*∂ β*

### (

1*∂T*

### )

*P*

*∂V =*−1

*∂ κ*

### (

1*∂ P*

### )

*T*

*∂V*−

### (

*∂ κ*

*∂T*

### )

*P*=

### (

*∂ β*

*∂ P*

### )

*T*

### (

*∂ β*

*∂ P*

### )

*T*=−

### (

*∂κ*

*∂T*

### )

*P*

*Proved*

### Problem 3.3:

**The Tait equation for liquids is written for an isotherm as:***V =V*_{0}

### (

1−*AP*

*B+P*

### )

**Where V is specific or molar volume, V****o**** 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 =V*_{0}

### (

1−*AP*

*B+P*

### )

Where

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

V = Molar/specific volume
Now,
*V =V _{o}*−

*AP*

*B+PVo*

*V −Vo*= −

*AP*

*B+PVo*

*V −V*

_{o}*Vo*=−

*AP*

*B+ P*

###

Differentiate w.r.t Pressure 1*V*

_{o}*∂*

*∂ P*

### (

*V −Vo*

### )

= −*∂*

*∂ P*

### (

*AP*

*B +P*

### )

1*V*

_{o}### (

*∂ V*

*∂ P*−0

### )

=−### [

*A (B+P )− AP(1)*(

*B+P )*2

### ]

−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 )*2

*Proved*

### 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? }

**Given Data:**

*Isothermal compressibility=κ=* *c*

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

*P*_{2}=500 bars
*Temperature=T =60* *0 _{C}*

*b=2700 bars*

*3 /*

_{c=0.125 cm}*g*

*Work=W =?*

### Solution:

We know that*W =−*

_{∫}

*PdV →(1)*

*κ=*

*c*

*V ( P+b )→(2)* Also

*κ=*−1

*V*

### (

*dV*

*dP*

### )

*T*

*→ (3)* Comparing (2) & (3) −1

*V*

*dV*

*dP*=

*c*

*V (P+b)*−

*dV =*

*c dP*

*P+b* Put in (1)

*W=−*

_{∫}

−*P*

*c dP*

*P+b*

*W =c*

### ∫

*1*

_{P}*P*2

*P*

*P+bdP*

*W =c*

### ∫

*1*

_{P}*P*2

*P+b−b*

*P+b*

*dP*

*W =c*

### ∫

*1*

_{P}*P*2

*P+b*

*P+bdP−c*

### ∫

*1*

_{P}*P*2

*b*

*P+bdP*

*W=c*

### ∫

*P*1

*P*2

*dP−b c*

### ∫

*P*1

*P*2 1

*P+bdP*

*W =c*|

*P*|

*P*1

*P*2 −

*bc*

### |

*ln(P+b)*

### |

*P*1

*P*2

*W=c*

### (

*P*2−

*P*1

### )

−*bc*

### [

ln### (

*P*2+

*b*

### )

−ln### (

*P*1+

*b*

### )

### ]

*W=c*

### (

*P*

_{2}−

*P*

_{1}

### )

−*bc*

### [

ln*P*2+

*b*

*P*

_{1}+

*b*

### ]

Putting values*W=0.125cm*
3
*g* ∗(500−1 )−2700¯¯ ¿0.125
*cm*3
*g* ∗ln
500+2700
1+2700
*cm*3∗¯¿
*g*
*cm*3∗¯¿
*g* −57.216¿
*W =62.375*¿
*cm*3∗¯¿
*g*
*W=5.16*¿
*cm*3
∗¯¿
*g* ∗1 m
3
1003*cm*3 ∗101325 N
1.01325 ¯¿*m*2 ∗*J*
*Nm*
*W =5.16*¿
*W=0.516J*
*g* *Answer*

### Problem 3.5:

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

**κ/(atm)****-1**_{ = 3.9 x 10}**-6**_{ - 0.1 x10}**-9**_{P(atm)}**Given Data:**

*Work done=W=?* *Volume=V =1 ft*3 *Temperature=T =32 F* *Pressure=P*1=1 atm

*Pressure=P*_{2}=3000 atm * _{κ /atm}*−1

_{=3.9∗10}−6

_{−0.1∗10}−9

_{P (atm)} Where

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)* Also

*κ=*−1

*V*

### (

*dV*

*dP*

### )

*dV =−κVdP*

Put in (1),
*W=−*

_{∫}

*P (−κVdp)*

*W=V*

### ∫

*P*1

*P*2

*κ P dP*

*W=V*

### ∫

*P*1

*P*2

### (

3.9∗10−6 −0.1∗10−9*P*

### )

*P dP*

*W=V*

### ∫

*P*1

*P*2 3.9∗10−6

*P dP−V*

### ∫

*P*1

*P*2 0.1∗10−9

*P*2

*dP*

*W=3.9∗10*−6

*V*

### ∫

1 3000*P dP−0.1∗10*−9

*V*

### ∫

1 3000*P*2

*dP*

*W=3.9∗10*−6

*V*

### |

*P*2 2

### |

1 3000 −0 .1∗10−9*V*

### |

*P*3 3

### |

1 3000*W=*1.95∗10 −6 ∗1 ft3

*atm*

### (

3000 2 −12### )

*atm*2−3.333∗10 −11 ∗1 ft3

*atm*2 ∗

### (

3000 3 −13### )

*atm*3

*W =(17.55−0.8991) atm∗ft*3

*W=16.65atm∗ft*3

*Answer*

### Problem 3.6:

**Five kilograms of liquid carbon tetrachloride undergo a mechanically reversible, isobaric change of state****at 1 bar during which the temperature change from 0****o**_{C to 20}**o**_{C. Determine ΔV}**t**_{, W, Q, and ΔU}**t**_{. The}**properties for liquid carbon tetrachloride at 1 bar & 0****o**_{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:**

*Mass=m=5 kg* *Pressure=P=1 ¯*¿ *Temperature=T*1=0 0

### C

*Temperature=T*_{2}=20 0

_{C}

*T*

_{2}=(

*20+273.15) K*

*T*

_{2}=293.15 K

*−3*

_{β=1.2∗10}*K*−1

_{C}*P*=0.84

*kJ*

*kg∗K*

*ρ=1590*

*kg*

*m*3

*∆ V*

*t*=

*?*

*W=?*

*Q=?*

*∆ Ut*=

*?*

### Solution:

As*V =*1

*ρ*

*V*1= 1

*ρ*

_{1}

*V*1= 1 1590

*m*3

*kg* Also, we know that

*Volume expansivity=β=*1

*V*

### (

*dV*

*dT*

### )

*P*

*βdT =*1

*V*

*dV*

Integrating on both sides,

*β*

### ∫

*T*1

*T*2

*dT =*

### ∫

*V*1

*V*2

*dV*

*V*

*β*|

*T*|

*T*

_{1}

*T*2 =|

*lnV*|

*V*

_{1}

*V*2

*β*

### (

*T*2−

*T*1

### )

=### (

*lnV*2−ln V1

### )

*β*

### (

*T*2−

*T*1

### )

=ln*V*

_{2}

*V*1

###

Putting values 1.2∗10−3*K*∗(

*293.15−273.15 ) K =ln*

*V*

_{2}∗1590 kg

*m*3

*e*0.024 =

*V*2∗1590 kg

*m*3

*V*2= 1.024 1590∗

*m*3

*kg*

*V*

_{2}=0.000644

*m*3

*kg* Now,

*∆ V =V*

_{2}−

*V*

_{1}

*∆ V =*

### (

0.000644− 1 1590### )

*m*3

*kg*

*∆ V =15.28∗10*−6

*m*3

*kg*

*∆ Vt*=*∆ V ∗m* *∆ Vt*=

### (

15.28∗10−6∗5### )

*m*3

*kg*∗*kg* *∆ V*

*t*

=7.638∗10−5*m*3*Answer*

Now,

We know that for a reversible process,

*Work done=W=−P ∆ Vt*
*W=−1 ¯*¿7.638∗10−5
*m*3∗101325 N
1.01325 ¯¿*m*2 ∗*J*
*Nm* ∗1 kJ
*1000 J*
*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=5 kg∗0.84*

*kJ*

*kg∗K*∗(*293.15−273.15) K* *Q=84 kJ Answer*

Now,

According to first law of thermodynamics,

*∆ Ut*

=*Q+W* *∆ Ut*

=

### (

84−7.368∗10−3_{)}

_{kJ}

_{∆ U}t_{=83.99 kJ Answer}

### Problem 3.7:

**A substance for which k is a constant undergoes an isothermal, mechanically reversible process from****initial state (P****1****, V****1****) to (P****2****, V****2****), 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:

* We know that*

**(a)***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*
*AT*=*e*
−κP
*V = A (T ) e*−κP
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*

### (

*V*2−

*V*1

### )

−### (

*P*2

*V*2−

*P*1

*V*1

### )

*W=*

### (

*V*

_{1}−

*V*

_{2}

### )

*κ*+

*P*1

*V*1−

*P*2

*V*2

*Proved*

### Problem 3.8:

**One mole of an ideal gas with C****V**** = 5/2 R, C****P**** = 7/2 R expands from P****1**** = 8 bars & T****1****= 600 K to P****2**** = 1 bar****by each of the following path:**

**a) Constant volume****b) Constant temperature****c) Adiabatically**

**Assuming mechanical reversibility, calculate W, Q, ***∆* **U, and ***∆* **H for each of the three processes.****Sketch each path in a single PV diagram. **

**Given Data:***C _{V}*=5
2

*R*

*CP*= 7 2

*R*

*P*1=8 ¯¿

*T*1=600 K

*P*2=1 ¯¿

*W=?*

*Q=?*

*∆ U =?*

*∆ H=?*

### Solution:

**(a)** According to first law of thermodynamics,

*∆ U =Q+W →(1)*

For a constant volume process,

*W=0* *∆ U =C _{V}∆T*

Put in (1)

For T2 , We know that for an ideal gas
*T*_{1}
*P*1
=*T*2
*P*2
*T*2=
*T*_{1}
*P*1
∗*P*2
8 ¯¿∗1¯¿
*T*_{2}=*600 K*_{¿} *T*2=75 K
Put in (2),
*Q=∆U =*5
2*R (75−600 ) K* *Q=∆U =*
−5
2 ∗8.314
*J*
*mol∗K*∗525 K *Q=∆U =−10912*
*J*
*mol*
*Q=∆U =−10.912* *kJ*
*mol* *Answer*
Also

For a mechanically reversible process we have,

*∆ H=C _{P}∆ T*

*∆ H=*7 2

*R*

### (

*T*2−

*T*1

### )

*∆ H=*7 2∗8.314

*J*

*mol∗K*∗(75−600 ) K

*∆ H=−15277*

*J*

*mol*

*∆ H=−15.277*

*kJ*

*molAnswer*

**(b)** For a constant temperature process,

*∆ U =0* *∆ H=0*

We know that at constant temperature, work done is

*W=R T*_{1}ln*P*2
*P*1
*W=8.314* *J*
*mol∗K*∗600 K∗ln
1
8 *W=−10373*
*J*
*mol* *W=−10.373*
*kJ*
*molAnswer*

Now, according to first law of thermodynamics,

*∆ U =Q+W* *0=Q+W* *Q=−W*

Or

*Q=10.373* *kJ*

*mol* *Answer*
**(c)**

*Q=0*

Now, according to first law of thermodynamics,

*∆ U =Q+W* *∆ U =W →(1)*

*∆ U =C _{V}∆T*

Put in (1)

*W=∆ U=CV∆ T* *W=∆ U=CV*

### (

*T*2−

*T*1

### )

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

*T*1*P*1
(*1−γ*)
*γ* _{=}* _{T}*
2

*P*2 (

*1−γ*)

*γ*

*T*

_{2}=

*T*

_{1}

### (

*P*1

*P*

_{2}

### )

(*1−γ*)

*γ*

*2=600 K*

_{T}### (

8 1### )

(1−1.4) 1.4*T*2=331.23 K Put in (2)

*W=∆ U=*5 2

*R∗(331.23−600) K*

*W=∆ U=*−5 2 ∗8.314

*J*

*mol∗K*∗268.77 K

*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 2

*R*

### (

*T*2−

*T*1

### )

*∆ 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 C****P**** = (7/2) R and C****V**** = (5/2) R.**

**a) Sketch the cycle on a PV diagram.**

**b) Determine (where unknown) both T and P for states 1, 2, 3, and 4.****c) Calculate Q, W, ***∆* **U, and ***∆* **H for each step of the cycle.****Given Data:**

*Initial Temperature=T*_{1}=600 K *Initial Pressure=P*_{1}=10 ¯¿ *CP*=
7
2*R* *CV*=
5
2*R*

### Solution:

**(b)****Step 12, an Isothermal process,**

**Step 12, an Isothermal process,**

Since

For an isothermal process, temperature is constant Therefore,

*T*_{2}=*T*_{1}=600 K *P*_{2}=3 ¯¿

We know that, for an ideal gas

*P*_{2}*V*_{2}=*R T*_{2} *V*_{2}=*RT*2
*P*2
*mol∗K∗3*
¯
¿∗1.01325 ¯¿*m*2
*101325 N* ∗*N∗m*
*J*
*V*_{2}=*8.314∗J∗600 K*
¿
*V*_{2}=0.0166 *m*
3
*mol*

**Step 23, an Isochoric process,**

**Step 23, an Isochoric process,**

Since

For an isochoric process, Volume is constant Therefore,

*V*_{3}=*V*_{2}=0.0166 *m*3

*mol* *P*3=2 ¯¿

We know that, for an ideal gas

*P*_{3}*V*_{3}=*RT*_{3} _{T}_{3}=*P*3*V*3
*R* * _{T}*
3=
2¯¿

*0.0166 m*3 ∗

*mol∗K*

*mol∗8.314 J*∗

*J*

*N∗m*∗101325 N 1.01325 ¯¿

*m*2

*T*

_{3}=400 K

**Step 34, an Isobaric process,**

**Step 34, an Isobaric process,**

Therefore,

*P*_{4}=*P*_{3}=2 ¯¿

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

As
*T*4
*T*_{1}=

### (

*P*4

*P*

_{1}

### )

*R*

*CP*

*T*4=

*T*1

### (

*P*4

*P*

_{1}

### )

*R*

*CP*

*T*4=600 K∗

### (

2 10### )

*2∗R*

*7R*

*4=378.83 K*

_{T} We know that, for an ideal gas

*P*_{4}*V*_{4}=*RT*_{4} *V*_{4}=*R T*4
*P*4
*mol∗K∗2*
¯_{¿}_{∗1.01325 ¯}_{¿}* _{m}*2

*101325 N*∗

*Nm*

*J*

*V*4=

*8.314 J∗378.83 K*¿

*V*

_{4}=0.0157

*m*3

*mol*

**Step 41, an adiabatic process,**

**Step 41, an adiabatic process,**

Since

Gas returns to its initial state adiabatically Therefore,

*T*_{1}=600 K *P*_{1}=10 ¯¿

We know that, for an ideal gas

*P*_{1}*V*_{1}=*R T*_{1} *V*_{1}=*RT*1
*P*1
*mol∗K∗10*
¯_{¿}_{∗1.01325 ¯}_{¿}* _{m}*2

*101325 N*∗

*Nm*

*J*

*V*

_{1}=

*8.314 J∗600 K*¿

*V*1=4.988∗10 −3

*m*3

*mol*

**(c)****Step 12, an Isothermal process,**

**Step 12, an Isothermal process,**

Since

For an isothermal process, temperature is constant Therefore

*∆ U*_{12}=0 *∆ H*_{12}=0

For an isothermal process, we have

*Q=−R T*_{1}ln*P*2
*P*1
*Q=−8.314* *J*
*mol∗K*∗600 K∗ln
3
10 *Q=6006*
*J*
*mol* *Q=6.006∗10*
3 *J*
*mol* *Answer*

According to first law of thermodynamics

*∆ U*_{12}=*Q*_{12}+*W*_{12} *0=Q*_{12}+*W*_{12} *W*_{12}=−*Q*_{12} *W*12=−6.006∗10
3 *J*

*mol* *Answer*

**Step 23, an Isochoric process,**

**Step 23, an Isochoric process,**

Since

For an isochoric process, Volume is constant Therefore,

*W*_{23}=0

At constant volume we have

*Q*_{23}=*∆ U*_{23}=*C _{V}∆ T*

*Q*

_{23}=

*∆ U*

_{23}=

*C*

_{V}### (

*T*

_{3}−

*T*

_{2}

### )

*Q*

_{23}=

*∆ U*

_{23}=5 2

*R ( 400−600 ) K*

*Q*

_{23}=

*∆ U*

_{23}= 5 2∗8.314 J

*mol∗K*∗(−200) K

*Q*

_{23}=

*∆ U*

_{23}=−4157

*J*

*mol*

*Q*23=

*∆ U*23=−4.157∗10 3

*J*

*molAnswer* We know that

*∆ H*23=

*CP∆ T*

*∆ H*23=

*CP*

### (

*T*3−

*T*2

### )

*∆ H*23= 7 2

*R ( 400−600) K*

*∆ H*23= 7 2∗8.314

*J*

*mol∗K*∗(−200) K

*∆ H*23=−5820

*J*

*mol*

*∆ H*23=−5.82∗10 3

*J*

*molAnswer*

**Step 34, an Isobaric process,**

**Step 34, an Isobaric process,**

Since

*Q*_{34}=*∆ H*_{34}=*C _{P}∆ T*

*Q*34=

*∆ H*34= 7 2

*R*

### (

*T*4−

*T*3

### )

*Q*34=

*∆ H*34= 7 2∗8.314

*J*

*mol∗K*∗(

*378.83−400) K*

*Q*34=

*∆ H*34=−616

*J*

*molAnswer*

For an Isobaric process we have

*W*_{34}=−*R ∆ T* *W*_{34}=−*R*

### (

*T*

_{4}−

*T*

_{3}

### )

*W*34=−8.314

*J*

*mol∗K*∗(378.83−400) K

*W*34=176

*J*

*mol*

*Answer* We know that,

*∆ U*

_{34}=

*C*

_{V}∆ T*∆ U*

_{34}=5 2

*R*

### (

*T*4−

*T*3

### )

*∆ U*34= 5 2∗8.314

*J*

*mol∗K*∗(378.83−400) K

*∆ U*

_{34}=−440

*J*

*mol*

*Answer*

**Step 41, an adiabatic process,**

**Step 41, an adiabatic process,**

Since

For an adiabatic process there is no exchange of heat
Therefore,
*Q*_{41}=0
We know that,
*∆ U*_{41}=*C _{V}∆ T*

*∆ U*

_{41}=5 2

*R*

### (

*T*1−

*T*4

### )

*∆ U*41= 5 2∗8.314

*J*

*mol∗K*∗(

*600−378.83) K*

*∆ U*41=4597

*J*

*mol*

*∆ U*

_{41}=4.597∗103

*J*

*mol*

*Answer* We know that

*∆ H*

_{41}=

*C*

_{P}∆T*∆ H*

_{41}=7 2

*R*

### (

*T*1−

*T*4

### )

*∆ H*41= 7 2∗8.314

*J*

*mol∗K*∗(

*600−378.83) K*

*∆ H*41=6435.8

*J*

*mol*

*∆ H*

_{41}=6.4358∗103

*J*

*mol*

*Answer*

*∆ U*_{41}=*Q*_{41}+*W*_{41} *∆ U*_{41}=*W*_{41} *W*41=4.597∗10
3 *J*

*molAnswer*

### Problem 3.10:

* An ideal gas, CP= (5/2) R and CV= (3/2) R is changed from P1 = 1bar and V*1

*t*

**= 12m****3**_{ to P}

**2 ****= 12 bar and***V*2*t* _{= 1 m}**3**_{ 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.**

**Given Data:**

*C _{P}*=5

2*R* *CV*=

3

2*R* *Initial pressure=P*1=1 ¯¿ *V*1*t*=12 m3 *Final pressure=P*2=12¯¿

*V*2
*t*
=1 m3 *Q=?* *W=?* *∆ H=?* *∆ U =?*

### Solution:

Since*Temperature=constant*

Therefore, for all parts of the problem,

*∆ H=0* *∆ U =0*
**(a)**

**Isothermal compression,**

**Isothermal compression,**

For an isothermal process, we have

*Q=−R T*_{1}ln*P*2

*P*1
Since

*P*_{1}*V*_{1}=*R T*_{1}
Therefore,
*Q=−P*_{1}*V*_{1}ln*P*2
*P*1
*Q=−1¯*¿*12 m*3∗ln
12
1 ∗101325 N
1.01325 ¯¿*m*2 ∗*J*
*Nm* ∗1 kJ
*1000 J*
*Q=−2981.88 kJ Answer*

According to first law of thermodynamics

*∆ U =Q+W* *0=Q+W* *W=−Q* *W*_{12}=2981.88 kJ Answer

**(b)**

**Adiabatic compression followed by cooling at constant pressure**

**Adiabatic compression followed by cooling at constant pressure**

Since

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

*Q=0 Answer*

The process completes in two steps

First step, an adiabatic compression to final pressure P2 , intermediate volume can be given as

*P*_{2}

### (

*V'*

### )

*γ*=

*P*

_{1}

*V*

_{1}

*V'*=

*V*

_{1}

### (

*P*1

*P*

_{2}

### )

1
*γ*

For mono atomic gas, we have

*γ=1.67*
*V'*=12 m3∗

### (

1 12### )

1 1.67*=2.71 m3 We know that,*

_{V}'*W*1=

*P*2

*V'*−

*P*1

*V*1

*γ−1*

*W*_{1}=
(12∗2.71−1∗12)¯¿*m*3
1.67−1 ∗101325 N
1.01325 ¯¿*m*2 ∗*J*
*Nm* ∗1 kJ
*1000 J*
*W*1=3063 kJ →(1)

Second step, cooling at constant pressure P2

We know that, for a mechanically reversible process

*W*2=−*P*2

### (

*V*2−

*V*

*'*

### )

*W*

_{2}=−12(1−2.71)¯

*m*3∗101325 N 1.01325 ¯¿

*m*2 ∗

*J*

*Nm*∗1 kJ

*1000 J*

*W*2=2052 kJ →(2) Now

*W=W*

_{1}+

*W*

_{2}

*W =(3063+ 2052)kJ*

*W=5115kJ Answer*

**(c)****Adiabatic compression followed by cooling at constant volume**

**Adiabatic compression followed by cooling at constant volume**

Since

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

*Q=0 Answer*

First step, an adiabatic compression to volume V2 , intermediate pressure can be given as
*P'V*2
*γ*
=*P*1*V*1 *P*
*'*
=*P*_{1}

### (

*V*1

*V*

_{2}

### )

*γ*

For mono atomic gas, we have

*γ=1.67*
*P'*=1

### (

12¯ 1### )

1.67*P'*=63.42¯¿ We know that,

*W*_{1}=*P*
*'*
*V*2−*P*1*V*1
*γ−1*
*W*_{1}=
(63.42∗1−1∗12)¯¿*m*3
1.67−1 ∗101325 N
1.01325 ¯¿*m*2 ∗*J*
*Nm* ∗1 kJ
*1000 J*
*W*_{1}=7674.76 kJ

Second step, cooling at constant Volume, Therefore, No work will be done

*W*_{2}=0

Now

*W=W*_{1}+*W*_{2} *W=(7674.76+0) kJ* *W=7674.76 kJ Answer*
**(d)**

**Heating at constant volume followed by cooling at constant pressure**

**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

*W*_{1}=0

Step 2, Cooling at constant pressure P2 To V2 We know that, for a mechanically reversible process

*W*_{2}=−*P*_{2}*∆ V* *W*_{2}=−*P*_{2}

### (

*V*

_{2}−

*V*

_{1}

### )

*W*

_{2}=−12(1−12)¯

*m*3

_{∗101325 N}1.01325 ¯¿

*m*2 ∗

*J*

*Nm*∗1 kJ

*1000 J*

*W*

_{2}=13200 kJ Now

*W=W*

_{1}+

*W*

_{2}

*W=(0+13200) kJ*

*W=13200 kJ Answer*

According to first law of thermodynamics

*∆ U =Q+W* *0=Q+W* *Q=−W* *Q=−13200 kJ Answer*

**(e)**

The process completes in two steps

Step 1, Cooling at constant Pressure P1 to V2 Therefore, for a mechanically reversible process

*W*_{1}=−*P*_{1}*∆ V* *W*_{1}=−*P*_{1}

### (

*V*

_{2}−

*V*

_{1}

### )

*W*

_{1}=−1(1−12)¯

*m*3∗101325 N 1.01325 ¯¿

*m*2 ∗

*J*

*Nm*∗1 kJ

*1000 J*

*W*

_{1}=1100 kJ

Step 1, Heating at constant Volume V2 to pressure P2 Therefore no work will be done

*W*_{2}=0

Now

*W=W*_{1}+*W*_{2} *W=(1100+0) kJ* *W=1100kJ Answer*

According to first law of thermodynamics

*∆ 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)*

*T P*
*1−δ*
*δ* _{=}* _{Constant}*
Or

*T P*

*1−δ*

*δ*

_{=}

_{T}*oPo*

*1−δ*

*δ* Where

To =Temperature at sea level, so it is constant
Po = Pressure at sea level, so it is constant
*T*
*P*
*δ−1*
*δ*
= *To*
*P _{o}*

*δ−1*

*δ*

*T*

*To*=

### (

*P*

*Po*

### )

*δ−1*

*δ*

### (

*T*

*To*

### )

*δ*

*δ−1*

_{=}

*P*

*Po*

*P=P*

_{o}### (

*T*

*To*

### )

*δ*

*δ −1*

_{→(a)}*P=*

*Po*

*To*

*δ*

*δ−1*∗

*T*

*δ*

*δ −1*

Differentiate w.r.t to Temperature on both sides

*dP*
*dT*=
*Po*
*T _{o}*

*δ*

*δ −1*∗

*δ*

*δ−1*∗

*T*1

*δ−1*

_{dP=}*Po*

*T*

_{o}*δ*

*δ−1*∗

*δ*

*δ−1*∗

*T*1

*δ−1*

_{dT →(2)} We know that, for an ideal gas

*ρ=* *P*
*RT*

Where

*R=Specific gas constant=R'*/*M*
Put (a) in above equation

*ρ=* 1
*RT*∗*Po*

### (

*T*

*To*

### )

*δ*

*δ −1* Put in (1)

*dP*
*dz*=−*M*
*g∗1*
*RT* *Po*

### (

*T*

*To*

### )

*δ*

*δ −1*

_{dP=−M}g∗1*RT*

*Po*

### (

*T*

*To*

### )

*δ*

*δ−1*

_{∗}

* Put (2) in above*

_{dz}*P*

_{o}*T*

_{o}*δ*

*δ−1*∗

*δ*

*δ−1*∗

*T*1

*δ −1*

_{dT =−M}g∗1*RT*

*Po*

### (

*T*

*T*

_{o}### )

*δ*

*δ−1*

_{∗}

_{dz}*dT*

*dz*= −

*δ−1*

*δ*∗

*M g*

*R*∗

*Po*

*P*∗

_{o}*To*

*δ*

*δ−1*

*T*

_{o}*δ*

*δ−1*∗

*T*

*δ*

*δ −1*

*T∗T*1

*δ−1*

*dT*

*dz*= −

*δ−1*

*δ*∗

*M g*

*R*∗

*Po*

*P*∗

_{o}*To*

*δ*

*δ−1*

*T*

_{o}*δ*

*δ−1*∗

*T*

*δ*

*δ −1*

*T*

*δ*

*δ−1*

*dT*

*dz*= −

*δ*

*δ−1*∗

*M g*

*R*

*Proved*

### Problem 3.12:

**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.

*d (mU ) _{tank}*

*dt*+

*∆ ( Hm)=0*

*d (mU )*

_{tank}*dt*+

*H*

*''*

*m''*−

*H'm'*=0 Since

Tank is filled with gas from an entrance line, but no gas is being escaped out,
Therefore,
*d (mU ) _{tank}*

*dt*+

*0−H*

*'*

_{m}'_{=0}

*d (mU )tank*

*dt*−

*H*

*'*

_{m}'_{=0→(1)}

**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

### ∫

*m*1

*m*2

*d (mU )tank*=

*H*

*'*

### ∫

*m*1

*m*2

*d mtank* *∆ (mU )tank*=*H*
*'*

### (

*m*2−

*m*1

### )

*m*2*U*2−*m*1*U*1=*H'*

### (

*m*2−

*m*1

### )

Because mass in the tank initially is zero, therefore

*m*_{1}=0

*m*2*U*2=*H'm*2

*U*2=*H'→(3)*
We know that

*U=C _{V}T*

*U*

_{2}=

*C*

_{V}T_{2}

*→( a)*

Also

*H'*

=*CPT'→(b )*
Put (a) & (b) in (3)

*CVT =CPT'* *T =*

*CP*

*CV*

*T'*

Since heat capacities are constant, therefore

*γ=CP*

*CV* *T =γ T*
*'*

*Proved*

### Problem 3.14:

**A tank of 0.1-m****3**_{ 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 45****o**_{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 =0.1 m*3 *T*1=25 C*.*
*o*
=298 K *P*_{1}=101.33 kPa *T*2=45 C*.*
*o*
=318 K *P*_{2}=1500 kPa
*Heat lost =Q=?* *C _{P}*=7
2

*R*

*CV*= 5 2

*R*

### Solution:

According to first law of thermodynamics

*∆ U =Q+W →(1)*

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}_{(}

*T*

_{2}−

*T*

_{1}

_{)}

Put in (2)
*n C*

_{P}### (

*T*

_{2}−

*T*

_{1}

### )

−*V ∆ P=Q →(3)* For “n”

We know that for an ideal gas,

*PV =nRT*

Initial number of moles of gas can be obtained as,

*P*_{1}*V =n*_{1}*R T*_{1} *n*_{1}=*P*1*V*

*R T*1

The final number of moles of gas at temperature T1 are
*P*_{2}*V =n*_{2}*R T*_{1} *n*2=

*P*_{2}*V*
*RT*1

Now, Applying molar balance

*n=n*_{1}−*n*_{2} *n=P*1*V*
*R T*1
−*P*2*V*
*R T*1
*n=*

### (

*P*1−

*P*2

### )

*V*

*R T*1 Put in (3)

### (

*P*

_{1}−

*P*

_{2}

### )

*V*

*R T*

_{1}

*CP*

### (

*T*

_{2}−

*T*

_{1}

_{)}

−*V ∆ P=Q*

### (

*P*

_{1}−

*P*

_{2}

_{)}

*V*

*R T*

_{1}∗7 2

*R∗*

### (

*T*2−

*T*1

### )

−*V ∆ P=Q*

### (

*P*

_{1}−

*P*

_{2}

### )

*V*

*T*

_{1}∗7 2 ∗

### (

*T*2−

*T*1

### )

−*V*

### (

*P*2−

*P*1

### )

=*Q*(

*101.33−1500 )kPa∗0.1 m*3

*298 K*∗7 2 ∗(

*318−298) K −0.1 m*3

_{(}

_{1500−101.33 )kPa=Q}*Q=−172.717 m*3

*kPa∗1 kN*

*1 kPa∗m*2∗1 kJ

*1 kNm*

*Q=−172.717 kJ Answer*

### Problem 3.17:

**A rigid, no conducting tank with a volume of 4 m****3**_{ 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**

**Assume nitrogen is an ideal gas for which C****P**** = (7/2) R & C****V**** = (5/2) R****Given Data:***Volume of thetank=V*1=4 m3 *V*2=
*V*_{1}∗1
3 =
4
3*m*
3 * _{Pressure=P}*
2=6 ¯¿

*Temperature=T*1=100 C

*o.*

*V*3=

*V*

_{1}∗2 3 = 8 3

*m*3

### Solution:

**(a)** According to first law of thermodynamics

*∆ U =Q+W*

Since

**No work is done & no heat is transferred **

Therefore

*Q=W =0*

*∆ U =0* *mC _{V}∆ T =0*

*∆ T =0*

*T*

_{2}−

*T*

_{1}=0

*T*

_{2}=

*T*

_{1}

*T*

_{2}=100

*℃ Answer*

**No, process is not reversible**

**(b)**

Since

Therefore, the process is isothermal For an isothermal process we have

*W=−R T*_{2}ln*V*2

*V*1
As, for an ideal gas

*P*_{2}*V*_{2}=*R T*_{2} *W=−P*_{2}*V*_{2}ln*V*2
*V*1
*W=−6 ¯*¿4
3*m*
3
ln 4
3∗4 *W =8.788¯*¿
*m*3_{∗101325 N}
1.01325¯¿*m*2 ∗1
*kJ*
*1000 Nm*
*W=878.8 kJ 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**

**a constant pressure of 500 kPa to 30 ****0**_{C and finally expanded isothermally to its original state }**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 C****p**** = (7/2) R and C****V**** =****(5/2) R**

**Given Data:***T*1=30 C0*.* *T*1=303.15 K *P*1=100 kPa *Q=?* *W=?* *∆ U =?* *∆ H=?* *CP*=
7
2*R* *CV*=
5
2*R*

### So

### lution:

**(a)***P*

_{2}=500 kPa

**1) Adiabatic Compression from point 1 to point 2**

*Q*_{12}=0

Now, from first law of thermodynamics,

*∆ U*_{12}=*Q*_{12}+*W*_{12} *∆ U*_{12}=*W*_{12}
*W*12=*∆U*12=*CV∆ T*12 *W*12=*∆U*12=
5
2*R*

### (

*T*2−

*T*1

### )

*→ (1)* For ‘T2’ We know that

*T*2

*T*

_{1}=

### (

*P*2

*P*

_{1}

### )

*γ −1*

*γ*

*2=*

_{T}*T*1

### (

*P*2

*P*

_{1}

### )

*γ −1*

*γ*

*2=303.15 K*

_{T}### (

500 100### )

1.4−1 1.4*2=480.13 K Put in (1)*

_{T}*W*

_{12}=

*∆U*

_{12}=5 2∗8.314

*J*

*mol∗K*(480.13−303.15) K∗1

*kJ*

*1000 J*

*W*12=

*∆ U*12=3.679

*kJ*

*mol* Also, we have

*∆ H*12=

*CP*

### (

*T*2−

*T*1

### )

*∆ H*12= 7 2∗8.314

*J*

*mol∗K*(

*480.13−303.15) K∗1*

*kJ*

*1000 J*

*∆ H*12=5.15

*kJ*

*mol*

**2) Cooling at constant pressure from point 2 to point 3**

Therefore at constant pressure we have,

*Q*23=*∆ H*23=*CP∆T*23 *Q*23=*∆ H*23=

7

Here
*T*_{3}=303.15 K
*Q*_{23}=*∆ H*_{23}=7
2∗8.314
*J*
*mol∗K* (*303.15−480.13) K∗1*
*kJ*
*1000 J* *Q*23=*∆ H*23=−5.15
*kJ*
*mol*
Also, we have
*∆ U*_{23}=*C _{V}*

_{(}

*T*

_{3}−

*T*

_{2}

_{)}

*∆ U*

_{23}=5 2∗8.314

*J*

*mol∗K*(

*303.15−480.13 ) K∗1*

*kJ*

*1000 J*

*∆ U*

_{23}=−3.679

*kJ*

*mol*

Now, from first law of thermodynamics,

*∆ U*23=*Q*23+*W*23 *W*23=*∆ U*23−*Q*23 *W*23=−3.679+5.15 *W*23=1.471

*kJ*
*mol*

**3) Isothermal expansion from point 3 to point 1**

Since for an isothermal process temperature remains constant Therefore,

*∆ U*_{31}=*∆ H*_{31}=0

Here

*P*_{3}=*P*_{2}=500 kPa

For an Isothermal process we have

*W*_{31}=−*R T*_{3}ln*P*3
*P*_{1}*W*31=−8.314
*J*
*mol∗K*∗303.15 K∗ln
500
100∗1 kJ
*1000 J*
*W*_{31}=−4.056 *kJ*
*mol*

*∆ U*_{31}=*Q*_{31}+*W*_{31}

*0=Q*_{31}+*W*_{31} *Q*_{31}=−*W*_{31} *Q*_{31}=4.056 *kJ*

*mol*

For the complete cycle,

*Q=Q*12+*Q*23+*Q*31 *Q=0−5.15+4.056* *Q=−1.094 _{mol}kJ*

*Answer*

*W=W*

_{12}+

*W*

_{23}+

*W*

_{31}

*W=3.679+1.471−4.056*

*W=1.094*

*kJ*

*mol*

*Answer*

*∆ H=∆ H*

_{12}+

*∆ H*

_{23}+

*∆ H*

_{31}

*∆ H=5.15−5.15+0*

*∆ H=0*

*kJ*

*mol*

*Answer*

*∆ U =∆ U*

_{12}+

*∆ U*

_{23}+

*∆ U*

_{31}

*∆ U =3.679−3.679+0*

*∆ U =0*

*kJ*

*mol*

*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**

*W*_{12}=*W*12
0.8 *W*12=
3.679
0.8 *W*12=4.598
*kJ*
*mol*

According to first law of thermodynamics

*∆ U*_{12}=*Q*_{12}+*W*_{12}
3.679 *kJ*
*mol*=*Q*12+4.598
*kJ*
*mol*
*Q*_{12}=3.679 *kJ*
*mol*−4.598
*kJ*
*mol* *Q*12=−0.92
*kJ*
*mol*

*W*_{23}=*W*23
0.8 *W*23=
1.471
0.8 *W*23=1.839
*kJ*
*mol*

According to first law of thermodynamics

*∆ U*_{23}=*Q*_{23}+*W*_{23} −3.679 *kJ*
*mol*=*Q*23+1.839
*kJ*
*mol* *Q*23=−3.679
*kJ*
*mol*−1.839
*kJ*
*mol* *Q*23=−5.518
*kJ*
*mol*

**3. Isothermal expansion from point 3 to point 1**

Since initial step can no longer be adiabatic , therefore

*W*_{31}=*W*_{31}∗0.8 *W*_{31}=−4.056 *kJ*

*mol*∗0.8 *W*31=3.245

*kJ*
*mol*

According to first law of thermodynamics

*∆ U*31=*Q*31+*W*31 *Q*31=−*W*31+0

*Q*_{31}=3.245 *kJ*

*mol*

For the complete cycle,

*Q=Q*_{12}+*Q*_{23}+*Q*_{31} *Q=−0.92−5.518+3.245* *Q=−3.193* *kJ*

*mol* *Answer*
*W=W*_{12}+*W*_{23}+*W*_{31} *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, C****p****=21 J mol****-1****K****-1****.**

*V*1=1 m
3 * _{T}*
1=600 K

*P*1=1000 kPa

*V*2=5 V1

*V*2=5 m 3

_{C}*P*=21

*J*

*mol K*

*CV*=

*?*

*T*2=

*?*

*P*2=

*?*

*W=?*

### Solution:

We know that,*C*−

_{P}*C*=

_{V}*R*

*C*=

_{V}*C*−

_{P}*R*

*CV*=(21−8.314)

_{mol∗K}J*CV*=12.686

* As*

_{mol∗K}J*γ=CP*

*CV*

*γ=1.6554*

**(a)** Since, for an isothermal process

Temperature remains constant, therefore

*T*_{2}=*T*_{1}=*600 K Answer*

For an ideal gas we have

*P*_{1}*V*_{1}
*T*1
=*P*2*V*2
*T*2 *P*2=
*P*_{1}*V*_{1}
*T*1
∗*T*_{2}
*V*_{2} *P*2=
*1000 kPa∗1 m*3
*600 K* ∗600 K
*5 m*3
*P*_{2}=200 kPa Answer

We know that, for an isothermal process

*W=−R T*_{1}ln*V*2

*V*1
Since

*P*_{1}*V*_{1}=*R T*_{1}

*W=−P*_{1}*V*_{1}ln*V*2
*V*1 * _{W=−1000 kPa∗1 m}*3
ln
5
1∗

*N*

*Pa∗m*2∗

*J*

*Nm*

*W=−1609.43 kJ Answer*

**(b)** We know that, for an adiabatic process

*P*1*V*1*γ*=*P*2*V*2*γ* *P*2=*P*1

### (

*V*1

*V*

_{2}

### )

*γ*

*P*

_{2}=1000 kPa∗

### (

1 5### )

1.6554*P*

_{2}=69.65 kPa Answer

For an ideal gas we have

*P*_{1}*V*_{1}
*T*1
=*P*2*V*2
*T*2
*T*_{2}=*P*2*V*2
*P*1*V*1
∗*T*_{1} *T*_{2}=*69.65 kPa∗5 m*
3
*1000 kPa∗1 m*3∗600 K *T*2=208.95 K Answer
For an adiabatic process work done is

*W=P*2*V*2−*P*1*V*1
*γ−1* *W=*(*69.65∗5−1000∗1) kPa∗m*
3
1.6554−1
*N*
*Pa∗m*2∗*J*
*Nm*
*W=−994.43 kJ Answer*
**(c)***P _{r}*=

*100 kPa*

Since, for an adiabatic process

*Q=0*

According to first law of thermodynamics

*∆ U =Q+W* *∆ U =W* *∆ U =W=−P _{r}dV*

*∆ U =W=−P*

_{r}### (

*V*

_{2}−

*V*

_{1}

### )

*∆ U =W=−100*(5−1)

*kPa∗m*3∗

*N*

*Pa∗m*2 ∗

*J*

*Nm*

*∆ U =−400 kJ*

*n C*

_{V}∆T =−400 kJ*n C*

_{V}### (

*T*

_{2}−

*T*

_{1}

### )

=−400 kJ*T*

_{2}=−400 kJ

*n C*+

_{V}*T*1

*→(1)*

For an ideal gas we have,
*P*_{1}*V*_{1}=*nR T*_{1} *n=P*1*V*1
*R T*1
*n=*
*1000 kPa∗1 m*3
∗*mol∗K*
*8.314 J∗600 K* ∗*kN*
*kPa∗m*2 ∗*kJ*
*kNm*
*n=0.2005 mol*
Put in (1)
*T*_{2}=
−400 kJ∗mol∗K
*0.2005 mol∗12.686 J*∗1000 J
*1 kJ* +*600 K*
*T*_{2}=−157.26 K +600 K *T*_{2}=442.74 K Answer

For an ideal gas we have

*P*1*V*1
*T*1
=*P*2*V*2
*T*2 *P*2=
*P*_{1}*V*_{1}
*T*1
∗*T*2
*V*2
*P*_{2}=
*1000 kPa∗1 m*3
*600 K* ∗442.74 K
*5 m*3
*P*_{2}=147.58 kPa 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 C****P**** = (7/2) R and C****V ****= (5/2) R, calculate W, Q,**

**∆ U , and ∆ H****Given Data:**

*Mole of air=n=1mol* *Initial Temperature=T*1=150 C*.*
0
=*423.15 K* *Initial pressure=P*_{1}=8 ¯¿
*Finaltemperature=T*3=50 C*.*
0
=323.15 K *Final pressure=P*_{3}=3 ¯¿ *C _{P}*=7
2

*R*

*CV*= 5 2

*R*

### Solution:

Since process is reversible

Two different steps are used in this case to reach final state of the air.

*T*_{1}=*T*_{2}

Therefore

*∆ U*_{12}=*∆ H*_{12}=0

For an isothermal process we have

*W*_{12}=*R T*_{1}ln*V*1
*V*2
As
*V*_{2}=*V*_{3} *W*_{12}=*R T*_{1}ln *V*1
*V*3
*→(1)*
We know that
*P*_{1}*V*_{1}
*T*1
=*P*3*V*3
*T*3
*V*_{1}
*V*3
=*P*3¿*T*1
*T*3∗*P*1
*W*_{12}=*R T*_{1}ln*P*1¿*T*3
*T*1∗*P*3 *W*12=
*8.314 J∗423.15 K*
*mol∗K* ∗1 kJ
*1000 J* ∗ln
3∗423.15
8∗323.15
*W*12=−2.502
*kJ*
*mol*

According to first law of thermodynamics

*∆ U*_{12}=*Q*_{12}+*W*_{12} *0=Q*_{12}+*W*_{12} *Q*_{12}=−*W*_{12} *Q*_{12}=2.502 *kJ*

*mol*

**Step 23:**

**Step 23:**

For step 23 volume is constant, Therefore,

*W*_{23}=0

According to first law of thermodynamics

*∆ U*_{23}=*Q*_{23}+*W*_{23} *∆ U*_{23}=*Q*_{23}+0 *Q*_{23}=*∆ U*_{23} *Q*_{23}=*∆ U*_{23}=*C _{V}∆ T*

*Q*

_{23}=

*∆ U*

_{23}=

*C*

_{V}_{(}

*T*

_{3}−

*T*

_{2}

_{)}

*Q*

_{23}=

*∆ U*

_{23}=5

❑_{❑} ¿❑_{❑} ¿❑
❑ ¿8.314❑❑
¿❑
❑ () *K*❑❑ ¿❑❑
¿−2.0785❑
❑
W e know that
*∆ H*_{23}=*C _{P}∆ T*

*∆ H*

_{23}=

*C*

_{P}### (

*T*

_{3}−

*T*

_{2}

### )

*23= 7 2∗8.314*

_{∆ H}*J*

*mol∗K*∗1 kJ

*1000 J*(423.15−323.15) K

*∆ H*

_{23}=2.91

*kJ*

*mol*

For the complete cycle,

*Work=W =W*_{12}+*W*_{23} *W=(−2.502+0 )* *kJ*
*mol* *W=−2.502*
*kJ*
*mol* *Answe r*
*Q=Q*_{12}+*Q*_{23} *Q=(2.502−2.0785)* *kJ*
*mol* *Q=0.424*
*kJ*
*mol* *Answer*

*∆ U =∆ U*12+*∆ U*23 *∆ U =(0−2.0785 ) _{mol}kJ*

*∆ U =−2.0785*

_{mol}kJ*Answe r*

*∆ H=∆ H*_{12}+*∆ H*_{23} *∆ 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 C****P**** = (7/2) R**

*Temperature=T*1=150 C0*.* =*423.15 K* *Velocity=u*1=2.5
*m*
*sec* *T*2=*?* *u*2=50
*m*
*sec* *CP*=
7
2*R*

*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
*∆ z=W _{S}*=

*Q=0* Therefore,

*∆ H +∆ u*2 2 =0

*CP∆ T =*−

*∆u*2 2

*CP*

### (

*T*2−

*T*1

### )

= −*u*2 2 −

*u*1 2 2

*T*2= −

*u*22−

*u*12

*2C*+

_{P}*T*1

*T*

_{2}= −

### (

502−2.52### )

∗2∗m2∗*mol∗K*

*2∗7∗8.314 J∗sec*2 ∗28 g Nitrogen

*1 mol Nitrogen*∗

*J*

*N∗m*∗

*N∗sec*2

*kg∗m*∗1 kg

*1000 g*+

*423.15 K*

*T*

_{2}=−1.199 K +423.15 K

*T*

_{2}=421.95 K

*T*2=(421.95−273.15) C

*.*0

*T*2=148.8 C

*.*0

_{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**

**Calculate Q, W, ∆ U ∧∆ H in each case. Take C****P**** = (7/2) R and C****V**** = (5/2) R. alternatively, take C****P**** =****(5/2) R and C****V**** = (3/2) R****Given Data:***T*1=30 C*.*
0 * _{T}*
1=(30+273.15) K

*T*1=303.15 K

*P*1=1 ¯¿

*T*2=130 C

*.*0

*3=(130+273.15) K*

_{T}*T*

_{3}=403.15 K

*P*

_{3}=10 ¯¿

*Q=?*

*W=?*

*∆ U =?*

*∆ H=?*

### Solution:

*CP*= 7 2

*R*

*CV*= 5 2

*R*

Each part consist of two steps, 12 & 23 For the overall processes

*∆ U =∆ U*_{12}=*∆U*_{23}=*C _{V}∆T*

*∆ U =∆ U*

_{12}=

*∆U*

_{23}=5

2*R*

### (

*T*3−

*T*1

### )

*∆ U =∆ U*12=*∆U*23=5_{2}∗8.314* _{mol∗K}J* (403.15−303.15)

*K∗1 kJ*

_{1000 J}*∆ U =∆ U*_{12}=*∆U*_{23}=2.079 *kJ*

*mol→(a) Answe r*

Now
*∆ H=∆ H*_{12}=*∆ H*_{23}=*C _{P}∆ T*

*∆ H=∆ H*

_{12}=

*∆ H*

_{23}=7 2

*R*

### (

*T*2−

*T*1

### )

*∆ H=∆ H*

_{12}=

*∆ H*

_{23}=7 2∗8.314

*J*

*mol∗K*(

*403.15−303.15) K∗1*

*kJ*

*1000 J*

*∆ H=∆ H*

_{12}=

*∆ H*

_{23}=2.91

*kJ*

*mol→ (b) Answe r*

**(a)**

**Step 12:**

**Step 12:**

For step “12” volume is constant Therefore

*W*_{12}=0

Here

*T*_{2}=*T*_{3}

According to first law of thermodynamics

*∆ U*_{12}=*Q*_{12}+*W*_{12} *∆ U*_{12}=*Q*_{12} *Q*_{12}=*∆ U*_{12}=*C _{V}∆ T*

*Q*

_{12}=

*∆ U*

_{12}=2.079

*kJ*

*mol*

### [

¿(*a)*

### ]

Also we have *∆ H*

_{12}=2.91

*kJ*

*mol*

### [

¿(*b)*

### ]

**Step 23:**

**Step 23:**

Since for step “23” process is isothermal Therefore

*∆ U*_{23}=*∆ H*_{23}=0

Here

*T*_{2}=*T*_{3}

Now, intermediate pressure can be calculated as

*P*_{1}
*T*1
=*P*2
*T*2
*P*_{2}=*P*1
*T*1
∗*T*_{2} 1
¯_{¿}
*303.15 K*∗403.15 K
*P*_{2}=¿
*P*_{2}=1.329 b ar

For an isothermal process we have

*W*_{23}=*R T*_{2}ln*P*3
*P*2
*W*23=8.314
*J*
*mol∗K*∗403.15
*K∗1 kJ*
*1000 J*∗ln
10
1.329

*W*23=6.764

*kJ*
*mol*

According to first law of thermodynamics

*∆ U*_{23}=*Q*_{23}+*W*_{23} *0=Q*_{23}+*W*_{23} *Q*_{23}=−*W*_{23} *Q*_{23}=−6.764 *kJ*

*mol*

For the complete cycle,

*Work=W =W*_{12}+*W*_{23} *W=(0+6.764 )* *kJ*
*mol* *W=6.764*
*kJ*
*mol* *Answe r*
*Q=Q*_{12}+*Q*_{23} *Q=(2.079−6.764)* *kJ*
*mol* *Q=−4.685*
*kJ*
*mol* *Answe r*
*∆ U =∆ U*_{12}+*∆ U*_{23} *∆ U =(2.079+0)* *kJ*
*mol* *∆ U =2.079*
*kJ*
*molAnswe r*
*∆ H=∆ H*_{12}+*∆ H*_{23} *∆ H=(2.91+0)* *kJ*
*mol* *∆ H=2.91*
*kJ*
*mol* *Answe r*
**(b)**

**Step 12:**

**Step 12:**

For step “12” volume is constant

Therefore, at constant pressure we have

*Q*_{12}¿*∆ H*_{12}=2.91 *kJ*

*mol*

### [

¿(*b)*

### ]

Also,

*∆ U*_{12}=2.079 *kJ*

*mol*

### [

¿(*a)*

### ]

According to first law of thermodynamics

*∆ U*_{12}=*Q*_{12}+*W*_{12} *W*_{12}=*∆U*_{12}−*Q*_{12} *W*12=(2.079−2.91)

*kJ*

*mol* *W*12=−0.831

*kJ*
*mol*

**Step 23:**

**Step 23:**

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

*∆ U*_{23}=*∆ H*_{23}=0

Here

*T*2=*T*3∧*P*1=*P*2

*W*_{23}=*R T*_{2}ln*P*3
*P*2
*W*_{23}=8.314 *J*
*mol∗K*∗403.15
*K∗1 kJ*
*1000 J* ∗ln
10
1
*W*_{23}=7.718 *kJ*
*mol*

According to first law of thermodynamics

*∆ U*_{23}=*Q*_{23}+*W*_{23} *0=Q*_{23}+*W*_{23} *Q*_{23}=−*W*_{23} *Q*23=−7.718

*kJ*
*mol*

For the complete cycle,

*Work=W =W*_{12}+*W*_{23} *W=(−0.831+7.718)* *kJ*
*mol* *W=6.887*
*kJ*
*mol* *Answer*
*Q=Q*_{12}+*Q*_{23} *Q=(2.91−7.718)* *kJ*
*mol* *Q=−4.808*
*kJ*
*mol* *Answe r*
*∆ U =∆ U*_{12}+*∆ U*_{23} *∆ U =(2.079+0)* *kJ*
*mol* *∆ U =2.079*
*kJ*
*molAnswe r*
*∆ H=∆ H*_{12}+*∆ H*_{23} *∆ H=(2.91+0)* *kJ*
*mol* *∆ H=2.91*
*kJ*
*mol* *Answe r*
**(c)**

Since for step “12” process is isothermal ( T = Constant) Therefore

*∆ U*12=*∆ H*12=0
Here

*P*_{2}=*P*_{3}

For an isothermal process we have

*W*_{12}=*R T*_{1}ln*P*2
*P*1
*W*_{12}=8.314 *J*
*mol∗K*∗303.15
*K∗1 kJ*
*1000 J* ∗ln
10
1
*W*_{12}=5.8034 *kJ*
*mol*

According to first law of thermodynamics

*∆ U*_{12}=*Q*_{12}+*W*_{12} *0=Q*_{12}+*W*_{12} *Q*_{12}=−*W*_{12} *Q*12=−5.8034_{mol}kJ

**Step 23:**

**Step 23:**

For step “23” volume is constant

Therefore, at constant pressure we have

*Q*_{23}=*∆ H*_{23}=2.91 *kJ*
*mol*

### [

¿(*b)*

### ]

Here*T*2=

*T*3 Now

*∆ U*

_{23}=2.079

*kJ*

*mol*

### [

¿(*a)*

### ]

According to first law of thermodynamics

*∆ U*_{23}=*Q*_{23}+*W*_{23} *W*_{23}=*∆ U*_{23}−*Q*_{23} *W*_{23}=(2.079−2.91) *kJ*

*mol* *W*23=−0.831

*kJ*
*mol*

For the complete cycle,

*Q=Q*_{12}+*Q*_{23} *Q=(−5.8034+2.91)* *kJ*
*mol* *Q=−2.894*
*kJ*
*molAnswe r*
*∆ U =∆ U*_{12}+*∆ U*_{23} *∆ U =(0+2.079)* *kJ*
*mol* *∆ U =2.079*
*kJ*
*molAnswe r*
*∆ H=∆ H*12+*∆ H*23 *∆ H=(0+2.91) _{mol}kJ*

*∆ H=2.91*

_{mol}kJ*Answe r*

### Solution:

*C _{P}*=5

2*R* *CV*=

3
2*R*

Each part consist of two steps, 12 & 23 For the overall processes

*∆ U =∆ U*_{12}=*∆U*_{23}=*C _{V}∆T*

*∆ U =∆ U*

_{12}=

*∆U*

_{23}=3 2

*R*

### (

*T*3−

*T*1

### )

*∆ U =∆ U*

_{12}=

*∆U*

_{23}=3 2∗8.314

*J*

*mol∗K*(403.15−303.15)

*K∗1 kJ*

*1000 J*

*∆ U =∆ U*

_{12}=

*∆U*

_{23}=1.247

*kJ*

*mol→(a) Answe r*

Now
*∆ H=∆ H*_{12}=*∆ H*_{23}=*C _{P}∆ T*

*∆ H=∆ H*

_{12}=

*∆ H*

_{23}=5 2

*R*

### (

*T*2−

*T*1

### )

*∆ H=∆ H*

_{12}=

*∆ H*

_{23}=5 2∗8.314

*J*

*mol∗K*(

*403.15−303.15) K∗1*

*kJ*

*1000 J*

*∆ H=∆ H*

_{12}=

*∆ H*

_{23}=2.079

*kJ*

*mol→ (b) Answer*

**(a)**

**Step 12:**

**Step 12:**

For step “12” volume is constant Therefore

*W*_{12}=0

Here

*T*_{2}=*T*_{3}

According to first law of thermodynamics

*∆ U*_{12}=*Q*_{12}+*W*_{12} *∆ U*_{12}=*Q*_{12} *Q*_{12}=*∆ U*_{12}=*C _{V}∆ T*

*Q*

_{12}=

*∆ U*

_{12}=1.247

*kJ*

*mol*

### [

¿(*a)*

### ]

Also we have*∆ H*12=2.079

*kJ*

*mol*

### [

¿(*b)*

### ]

**Step 23:**

**Step 23:**

Since for step “23” process is isothermal Therefore

*∆ U*_{23}=*∆ H*_{23}=0

Here

*T*_{2}=*T*_{3}

Now, intermediate pressure can be calculated as

*P*1
*T*1
=*P*2
*T*2
*P*_{2}=*P*1
*T*1
∗*T*_{2} 1
¯_{¿}
*303.15 K*∗403.15 K
*P*2=¿
*P*_{2}=1.329 b ar

*W*_{23}=*R T*_{2}ln*P*3

*P*2

*W*23=8.314* _{mol∗K}J* ∗403.15

*K∗1 kJ*∗ln

_{1000 J}_{1.329}10

*W*_{23}=6.764 *kJ*

*mol*

According to first law of thermodynamics

*∆ U*23=*Q*23+*W*23 *0=Q*23+*W*23 *Q*23=−*W*23 *Q*23=−6.764

*kJ*
*mol*

For the complete cycle,

*Work=W =W*_{12}+*W*_{23} *W=(0+6.764 )* *kJ*
*mol* *W=6.764*
*kJ*
*mol* *Answe r*
*Q=Q*_{12}+*Q*_{23} *Q=(1.247−6.764 )* *kJ*
*mol* *Q=−5.516*
*kJ*
*molAnswe r*
*∆ U =∆ U*_{12}+*∆ U*_{23} *∆ U =(1.247+0)* *kJ*
*mol* *∆ U =1.247*
*kJ*
*mol* *Answer*
*∆ H=∆ H*_{12}+*∆ H*_{23} *∆ H=(2.079+0)* *kJ*
*mol* *∆ H=2.079*
*kJ*
*mol* *Answe r*
**(b)**

**Step 12:**

**Step 12:**

For step “12” volume is constant

Therefore, at constant pressure we have

*Q*_{12}¿*∆ H*_{12}=2.079 *kJ*

*mol*

### [

¿(*b)*

### ]

Also,

*∆ U*_{12}=1.247 *kJ*

*mol*

### [

¿(*a)*