Células B-Ibru estimuladas con LPS o CpG no presentan alteraciones

 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∗101325 N 1.01325 ¯¿m² ∗J

Nm ∗1 kJ 1000 J

W₁=1100 kJ

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

 Therefore no work will be done W₂=0

 Now

W=W₁+W₂ 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)

 The polytropic relation is

T P

 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

 R=Specific gas constant=R^'/M

 Put (a) in above equation

ρ= 1

RT∗P_o

(

^TTo

)

^{δ −1δ}

 Put in (1)

dP

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

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 m_tank dt → (2)

 Combining equation (1) & (2) d(mU)_tank

dt −H^'d m_tank dt =0

1

dt

{

^{d (mU )}tank−H^'d m_tank

}

⁼⁰ d (mU )_tank=H^'d m_tank

 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^'=C_PT^'→(b )

 Put (a) & (b) in (3)

C_VT =C_PT^' T =C_P CV

T^'

 Since heat capacities are constant, therefore

γ=C_P

CV T =γ T^'Proved

Problem 3.14:

A tank of 0.1-m³ volume contains air at 25 ^oC and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at the constant conditions of 45^oC 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 ^oC, 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 m³ T₁=25 C^o_. =298 K P₁=101.33 kPa T₂=45 C^o_. =318 K P₂=1500 kPa

Heat lost =Q=? C_P=7

2R C_V=5 2R

Solution:

 According to first law of thermodynamics

∆ U =Q+W →(1)

 Since

 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

 Initial number of moles of gas can be obtained as,

P₁V =n₁R T₁ n₁=P₁V R T1

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

P₂V =n₂R T₁ n₂=P₂V RT₁

 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 C

P

(

^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∗0.1 m³

298 K ∗7

2 ∗(318−298) K −0.1 m³(1500−101.33 )kPa=Q

Q=−172.717 m³

kPa∗1 kN 1 kPa∗m²∗1 kJ

1 kNm

Q=−172.717 kJ 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 ^oC, 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 CP = (7/2) R & CV = (5/2) R Given Data:

Volume of thetank=V1=4 m³ V₂=V₁∗1 3 =4

3m³ Pressure=P₂=6 ¯¿ Temperature=T1=100 Co.

V₃=V₁∗2 3 =8

3m³

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} 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=−P₂V₂lnV₂ V1

W=−6 ¯¿4

3m³ln 4

3∗4 W =8.788¯¿m³∗101325 N

1.01325¯¿m² ∗1 kJ 1000 Nm W=878.8 kJ Answer

Problem 3.18:

An ideal gas initially at 30 ⁰C 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 ⁰C 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 Cp = (7/2) R and CV = (5/2) R

Given Data:

T1=30 C0. T₁=303.15 K P₁=100 kPa Q=? W=? ∆ U =? ∆ H=? C_P=7

2R C_V=5

2R

So lution:

(a)

P₂=500 kPa

1) Adiabatic Compression from point 1 to point 2 Q₁₂=0

 Now, from first law of thermodynamics,

∆ U₁₂=Q₁₂+W₁₂ ∆ U₁₂=W₁₂

W₁₂=∆U₁₂=C_V∆ T₁₂ W₁₂=∆U₁₂=5

2R

(

^T2−T₁

)

^{→ (1)}

 For ‘T2’

 We know that

T₂

T₁=

(

^PP21

)

^{γ −1γ T2=T1}

(

^PP21

)

^{γ −1γ T2=303.15 K}

⁽

⁵⁰⁰¹⁰⁰

⁾

^{1.4−11.4 T2=480.13 K}

 Put in (1)

W₁₂=∆U₁₂=5

2∗8.314 J

mol∗K (480.13−303.15) K∗1 kJ

1000 J W₁₂=∆ U₁₂=3.679 kJ mol

 Also, we have

∆ H₁₂=C_P

(

^T2−T₁

)

^{∆ H}12=7

2∗8.314 J

mol∗K (480.13−303.15) K∗1 kJ

1000 J ∆ H₁₂=5.15 kJ mol 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

2R

(

^T3−T₂

)

 Here T₃=303.15 K

Q₂₃=∆ H₂₃=7

2∗8.314 J

mol∗K (303.15−480.13) K∗1 kJ

1000 J 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 1000 J

∆ 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=500 kPa

 For an Isothermal process we have

W₃₁=−R T₃lnP₃

P₁W₃₁=−8.314 J

mol∗K∗303.15 K∗ln 500 100∗1 kJ

1000 J

W₃₁=−4.056 kJ mol

 According to first law of thermodynamics

∆ U₃₁=Q₃₁+W₃₁

0=Q₃₁+W₃₁ Q₃₁=−W₃₁ Q₃₁=4.056 kJ mol

 For the complete cycle,

Q=Q₁₂+Q₂₃+Q₃₁ Q=0−5.15+4.056 Q=−1.094 kJ

molAnswer

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 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 W₁₂=W₁₂

0.8 W₁₂=3.679

0.8 W₁₂=4.598 kJ mol

 According to first law of thermodynamics

∆ 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

W₂₃=W₂₃

0.8 W₂₃=1.471

0.8 W₂₃=1.839 kJ mol

 According to first law of thermodynamics

∆ 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

 According to first law of thermodynamics

∆ U₃₁=Q₃₁+W₃₁ Q₃₁=−W₃₁+0

Q₃₁=3.245 kJ mol

 For the complete cycle,

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

mol K C_V=? 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 C_V=12.686 J mol∗K

 As

γ=C_P

C_V ^γ=1.6554

(a)

 Since, for an isothermal process

 Temperature remains constant, therefore

T₂=T₁=600 K Answer

 For an ideal gas we have

P₁V₁

T₁ =P₂V₂

T₂ P₂= P₁V₁

T1

∗T₂

V₂ P₂=

1000 kPa∗1 m³

600 K ∗600 K 5 m³

P₂=200 kPa Answer

 We know that, for an isothermal process

W=−R T₁lnV₂ V₁

 Since P₁V₁=R T₁

 Therefore,

W=−P₁V₁lnV₂ V₁

W=−1000 kPa∗1 m³ln 5 1∗N Pa∗m²∗J

Nm

W=−1609.43 kJ Answer

(b)

 We know that, for an adiabatic process

P1V1γ

=P2V2γ P₂=P₁

(

^VV12

)

^{γ P2=1000 kPa∗}

⁽

¹⁵

⁾

^{1.6554 P2}=69.65 kPa Answer

 For an ideal gas we have P₁V₁

T₁ =P₂V₂

T₂ T₂=P₂V₂

P₁V₁∗T₁ T₂=69.65 kPa∗5 m³

1000 kPa∗1 m³∗600 K T₂=208.95 K Answer

 For an adiabatic process work done is

W=P₂V₂−P₁V₁

γ−1 W=(69.65∗5−1000∗1) kPa∗m³ 1.6554−1

N Pa∗m²∗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_rdV ∆ U =W=−P_r

(

^V2−V₁

)

∆ U =W=−100(5−1)

kPa∗m³∗N Pa∗m² ∗J

Nm

∆ U =−400 kJ n C_V∆T =−400 kJ n C_V

(

^T2−T₁

)

^{=−400 kJ}

T₂=−400 kJ

n C_V +T₁→(1)

 For an ideal gas we have,

P₁V₁=nR T₁ n=P₁V₁ R T₁

n=

1000 kPa∗1 m³∗mol∗K 8.314 J∗600 K ∗kN

kPa∗m² ∗kJ

kNm

n=0.2005 mol

 Put in (1)

T₂=

−400 kJ∗mol∗K

0.2005 mol∗12.686 J∗1000 J

1 kJ +600 K T₂=−157.26 K +600 K T₂=442.74 K Answer

 For an ideal gas we have

P₁V₁

T₁ =P₂V₂

T₂ P₂= P₁V₁

T1

∗T₂

V₂ P₂=

1000 kPa∗1 m³

600 K ∗442.74 K 5 m³

P₂=147.58 kPa Answe r

Problem 3.20:

One mole of air, initially at 150 ⁰C 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 ⁰C 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 C⁰_. =423.15 K Initial pressure=P₁=8 ¯¿

Finaltemperature=T₃=50 C⁰_. =323.15 K Final pressure=P₃=3 ¯¿ C_P=7

2R C_V=5 2R

Solution:

 Since process is reversible

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

Step 12:

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=

8.314 J∗423.15 K mol∗K ∗1 kJ

1000 J ∗ln3∗423.15 8∗323.15 W₁₂=−2.502 kJ

mol

 According to first law of thermodynamics

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

Step 23:

 For step 23 volume is constant,

 Therefore,

W₂₃=0

 According to first law of thermodynamics

∆ 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

❑_❑ ¿❑_❑ ¿❑

❑ ¿8.314❑

❑

¿❑

❑ ⁽⁾ K❑_❑ ¿❑_❑

¿−2.0785❑

❑

 W e know that

∆ H₂₃=C_P∆ T ∆ H₂₃=C_P

(

^T3−T₂

)

_{∆ H}

23=7

2∗8.314 J

mol∗K∗1 kJ

1000 J (423.15−323.15) K

∆ H₂₃=2.91 kJ mol

 For the complete cycle,

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 ⁰C 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=150 C⁰_. =423.15 K Velocity=u₁=2.5 m

sec T₂=? u₂=50 m

sec C_P=7 2R Molecualr weight of Nitrogen=28 g

mol

Solution:

 Applying energy balance for steady state flow process

∆ H +∆ u²

2 +g ∆ z=Q+W_S

 Since

∆ z=W_S=Q=0

 Therefore,

∆ H +∆ u²

2 =0 C_P∆ T =−∆u²

2 C_P

(

^T2−T₁

)

^=−u2

2−u₁²

2 T₂=−u22

−u12

2C_P +T₁

T₂=

−

(

50²−2.5²

)

∗2∗m²∗mol∗K

2∗7∗8.314 J∗sec² ∗28 g Nitrogen 1 mol Nitrogen ∗J

N∗m ∗N∗sec²

kg∗m ∗1 kg

1000 g +423.15 K

T₂=−1.199 K +423.15 K

T₂=421.95 K T₂=(421.95−273.15) C⁰_. T₂=148.8 C⁰_. Answe r

Problem 3.22:

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

a) The gas is first heated at constant volume until its temperature is 130 ⁰C; 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 ⁰C; 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 ⁰C

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=30 C⁰_. T₁=(30+273.15) K T₁=303.15 K P₁=1 ¯¿ T₂=130 C⁰_. T₃=(130+273.15) K T₃=403.15 K P₃=10 ¯¿ Q=? W=? ∆ U =? ∆ H=?

Solution:

C_P=7

2R C_V=5 2R

 Each part consist of two steps, 12 & 23

 For the overall processes

∆ U =∆ U₁₂=∆U₂₃=C_V∆T ∆ U =∆ U₁₂=∆U₂₃=5

2R

(

^T3−T₁

)

∆ U =∆ U₁₂=∆U₂₃=5

2∗8.314 J

mol∗K(403.15−303.15)K∗1 kJ 1000 J

∆ 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 1000 J

∆ H=∆ H₁₂=∆ H₂₃=2.91 kJ

mol→ (b) Answe r

(a)

Step 12:

 For step “12” volume is constant

 Therefore

W₁₂=0

 Here

T₂=T₃

 According to first law of thermodynamics

∆ U₁₂=Q₁₂+W₁₂ ∆ U₁₂=Q₁₂ Q₁₂=∆ U₁₂=C_V∆ 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

 Here

T₂=T₃

 Now, intermediate pressure can be calculated as

P₁ T₁=P₂

T₂ P₂=P₁

T₁∗T₂ 1 ¯¿

303.15 K∗403.15 K P₂=¿

P₂=1.329 b ar

 For an isothermal process we have W₂₃=R T₂lnP₃

P2

W₂₃=8.314 J

mol∗K∗403.15K∗1 kJ

1000 J∗ln 10 1.329

W₂₃=6.764 kJ mol

 According to first law of thermodynamics

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

 For the complete cycle,

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,

∆ U₁₂=2.079 kJ

mol

[

¿(a)

]

 According to first law of thermodynamics

∆ 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∗403.15K∗1 kJ

1000 J ∗ln10 1

W₂₃=7.718 kJ mol

 According to first law of thermodynamics

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

 For the complete cycle,

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) kJ

mol Q=−4.808 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

(c)

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

 Therefore

∆ U₁₂=∆ H₁₂=0

 Here

P₂=P₃

 For an isothermal process we have W₁₂=R T₁lnP₂

P₁ W₁₂=8.314 J

mol∗K∗303.15K∗1 kJ

1000 J ∗ln10 1

W₁₂=5.8034 kJ mol

 According to first law of thermodynamics

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

Step 23:

 For step “23” volume is constant

 Therefore, at constant pressure we have

Q₂₃=∆ H₂₃=2.91 kJ

mol

[

^¿(b)

]

 Here T₂=T₃

 Now

∆ U₂₃=2.079 kJ

mol

[

¿(a)

]

 According to first law of thermodynamics

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

mol W₂₃=−0.831 kJ mol

 For the complete cycle,

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

mol Answe r

Solution:

C_P=5

2R C_V=3 2R

 Each part consist of two steps, 12 & 23

 For the overall processes

∆ U =∆ U₁₂=∆U₂₃=C_V∆T ∆ U =∆ U₁₂=∆U₂₃=3

2R

(

^T3−T₁

)

∆ U =∆ U₁₂=∆U₂₃=3

2∗8.314 J

mol∗K(403.15−303.15)K∗1 kJ 1000 J

∆ U =∆ U₁₂=∆U₂₃=1.247 kJ

mol→(a) Answe r

 Now

∆ H=∆ H₁₂=∆ H₂₃=C_P∆ 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 1000 J

∆ H=∆ H₁₂=∆ H₂₃=2.079 kJ

mol→ (b) Answer

(a)

Step 12:

 For step “12” volume is constant

 Therefore

W₁₂=0

 Here

T₂=T₃

 According to first law of thermodynamics

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

[

^¿(a)

]

 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 ¯¿

303.15 K∗403.15 K P₂=¿

P₂=1.329 b ar

 For an isothermal process we have

W₂₃=R T₂lnP₃ P2

W₂₃=8.314 J

mol∗K∗403.15K∗1 kJ

1000 J∗ln 10 1.329

W₂₃=6.764 kJ mol

 According to first law of thermodynamics

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

 For the complete cycle,

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

mol Answe r

(b)

Step 12:

 For step “12” volume is constant

 Therefore, at constant pressure we have

Q₁₂¿∆ H₁₂=2.079 kJ

mol

[

^¿(b)

]

 Also,

∆ U₁₂=1.247 kJ

mol

[

¿(a)

]

 According to first law of thermodynamics

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

mol W₁₂=−0.832 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∗403.15K∗1 kJ

1000 J ∗ln10 1

W₂₃=7.718 kJ mol

 According to first law of thermodynamics

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

 For the complete cycle,

Work=W =W₁₂+W₂₃ W=(−0.832+7.718) kJ

mol W=6.886 kJ

mol Answer

Q=Q₁₂+Q₂₃ Q=(2.079−7.718) kJ

mol Q=−5.639 kJ

mol Answer

∆ 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)

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

 Therefore

∆ U₁₂=∆ H₁₂=0

 Here P₂=P₃

 For an isothermal process we have W₁₂=R T₁lnP₂

P₁ W₁₂=8.314 J

mol∗K∗303.15K∗1 kJ

1000 J ∗ln10 1

W₁₂=5.8034 kJ mol

 According to first law of thermodynamics

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

Step 23:

 For step “23” volume is constant

 Therefore, at constant pressure we have

Q₂₃=∆ H₂₃=2.079 kJ

mol

[

¿(b)

]

 Here T₂=T₃

 Now

∆ U₂₃=1.247 kJ mol

[

¿(a)

]

 According to first law of thermodynamics

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

mol W₂₃=−0.832 kJ mol

 For the complete cycle,

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

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

2R C_V=5 2R

Solution:

 The process consist of two steps, 12 & 23

Step 12:

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

 Therefore

∆ U₁₂=∆ H₁₂=0

P₂

393.15 K∗303.15 K P₂=¿

P₂=9.25 b ar

 For an isothermal process we have W₁₂=R T₁lnP₂

P₁ W₁₂=8.314 J

mol∗K∗303.15 .15K∗1 kJ

1000 J∗ln9.25 1

W₁₂=5.607 kJ mol

 According to first law of thermodynamics

∆ 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

 According to first law of thermodynamics

∆ U₂₃=Q₂₃+W₂₃ ∆ U₂₃=Q₂₃

Q₂₃=∆ U₂₃=C_V∆ T Q₂₃=∆ U₂₃=5

2R

(

^T3−T₁

)

Q₂₃=∆ U₂₃=5

2∗8.314 J

mol∗K (393.15−303.15)K∗1 kJ

1000 J Q₂₃=∆ U₂₃=1.871 kJ

mol∗K (393.15−303.15) K∗1 kJ

1000 J ∆ H₂₃=2.619 kJ mol

 For the complete cycle,

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

molAnswe r

∆ U =∆ U₁₂+∆ U₂₃ ∆ U =(0+1.871) kJ

mol ∆ U =1.871 kJ

molAnswe r

∆ H=∆ H₁₂+∆ H₂₃ ∆ H=(0+2.691) kJ

mol ∆ H=2.691 kJ

mol Answe r

Problem 3.24:

A process consists of two steps: (1) One mole of air at T = 800 K and P = 4 bars are cooled at constant volume to T = 350 K. (2) The air is then heated air constant pressure until its temperature reaches 800 K.

If this two step process is replaced by a single isothermal expansion of the air from 800 K and 4 bar to some final pressure P, what is the value of P that makes the work of two step processes the same? Assume mechanical reversibility and treat air as an ideal gas with CP = (7/2) R and CV = (5/2) T.

Given Data:

T₁=800 K P₁=4 ¯¿ T₂=350 K P=?

Solution:

 For the first step volume is constant

 Therefore,

W₁₂=0

 For the work done is W =W₂₃=−P₂∆V →(1)

 For one mole of an ideal gas we have,

P ∆ V =R ∆ T P ∆ V =R ∆ T

 Put in (1)

W=−R ∆ T W=−R

(

^T3−T₂

)

 Since

T₃=T₁

 Therefore W=−R

(

^T1−T₂

)

^→(2)

 For an isothermal process we have

W=R T₁ln P P₁→ (3)

 Compare (2) and (3)

−R

(

^T1−T₂

)

^{=R T}1ln P

P₁ T₂−T₁=T₁ln P P₁

T₂−T₁

T₁ =ln P P₁

P 4 ¯¿ (350−800) K

800 K =ln¿

4 ¯¿ e^−0.5625=P

¿

4 ¯¿0.5698=P P=2.279Answer¯

Problem 3.25:

A scheme for finding the internal volume V^Bt of the gas cylinder consists of the following steps. The cylinder is filled with a gas to low pressure P1, and connected through a small line and valve to an evacuated reference tank of known volume V^tA . The valve is opened, and the gas flows through the line into the reference tank. After the system returns to its initial temperature, a sensitive pressure transducer provides a valve for the pressure change ∆ P in the cylinder. Determine the cylinder volume V^Bt from the following data:

a) V^t_A=256 cm³ b) ∆ P/ P₁=−0.0639

Given Data:

V_B^t=? V^t_A=256 cm³ ∆ P

 When gas flows through the line into the tank then tank’s total volume becomes VtA

+VtB

 Now, by applying condition for an ideal gas

P₁V^t_B=P₂

(

^VA

A closed, non-conducting, horizontal cylinder is fitted with non-conducting, frictionless, floating piston which divides the cylinder in two Sections A & B. The two sections contains equal masses of air, initially at the same conditions, T1 = 300 K and

P1 = 1 atm. An electrical heating element in section A is activated, and the air temperature slowly increases: TA in section A because of heat transfer, and TB in section B because of adiabatic compression by slowly moving piston. Treat air as an ideal gas with CP = (7/2) R and let nA be the number of moles of air in section A. For the process as described, evaluate one of the following sets of quantities:

a) TA, TB, and Q/ nA, if P (final) = 1.25 atm b) TB, Q/ nA, and P (final), if TA = 425 K c) TA, Q/nA, and P (final), if TB = 325 K d) TA, TB, and P (final), if Q/nA = 3 kJ mol^-1.

Given Data:

T₁=300 K P₁=1 atm C_P=7 2R

Solution:

 According to ideal gas equation,

PV =nRT

 Applying ideal gas equation for initial conditions

 On section “A”

P₁V_A=n_AR T₁ V_A=n_AR T₁ P₁

 On section “B”

P₁V_B=n_BRT₁

 Since

n_A=n_B

 Therefore,

P₁V_B=n_AR T₁ V_B=n_ART₁ P₁

 Total initial volume can be given as

V_i=V_A+V_B V_i=n_AR T₁

P₁ +n_AR T₁

P₁ V_i=2.n_ART₁ P₁

 Let P2 be the final pressure & TA & TB are the final temperatures of section A & section B respectively

 Applying ideal gas equation for final conditions

 On section “A”

P₂V_A=n_AR T_A

V_A=n_AR T_A P₂

 On section “B”

P₂V_B=n_AR T_B V_B=n_ART_B P2

 Total Final volume can be given as

V_f=V_A+V_B V_f=n_AR T_A

P₂ +n_AR T_B

P₂ V_f=n_AR

(

^TA+T_B

)

P₂

 Since the total volume is constant, therefore V_i=V_f

2.n_AR T₁

P₁ =n_AR

(

^TA+T_B

)

P₂

2. T₁

P₁ =

(

^TA+T_B

)

P₂ →(1)

(a)

P₂=1.25 atm

 Since the process occurring in section B is reversible adiabatic compression

 Therefore, for an adiabatic compression we have

T₁

(

P₁

)

1−γ

γ =T_B

(

P₂

)

1−γ

γ T_B=T₁

(

P₁

)

1−γ γ

(

^P2

)

1− γ γ

T_B=T₁

(

^PP21

)

^{γ−1γ →(2)}

 We know that,

C_P−C_V=R C_V=C_P−R C_V=7

2R−R C_V=7 R−2 R

2 C_V=5 2R

 As

γ=C_P CV

γ=7∗R∗2

2∗5∗R γ=1.4

 Put in (2)

T_B=300 K

(

^1.251

)

^{1.4 −11.4 TB}=300 K∗1.0658 T_B=319.74 K Answe r

 Put in (1)

2∗300 K

1 atm =

(

^TA+319.74 K

)

1.25 atm

600 K∗1.25=T_A+319.74 K T_A=750 K−319.74 K T_A=430.26 K Answe r

 According to first law of thermodynamics

∆ U =Q+W

 Since volume is constant, therefore

∆ U =Q→(a)

 For section A & B

∆ U =∆ U_A+∆ U_B

 Put in (a)

Q=∆U_A+∆U_B Q=n_AC_V∆ T +n_AC_V∆ T Q=n_AC_V

(

^TA−T₁

)

⁺ⁿAC_V

(

^TB−T₁

)

^Q=nAC_V

[

^TA−T₁+T_B−T₁

]

Q=n_AC_V

[

^TA+T_B−2T₁

]

_n^Q

A

=C_V

[

^TA+T_B−2 T₁

]

^→(3) _n^Q

A

=5

2R (430.26+319.74−2∗300) K Q

n_A=5

2∗8.314 J

mol∗K∗150.02K∗1 kJ 1000 J

Q

n_A=3.118 kJ

mol Answe r

(b)

T_A=425 K

 From equation (1)

2. T₁

P₁ =

(

^TA+T_B

)

P₂

P₂

P₁=

(

^TA+T_B

)

2.T₁

 Put in (2)

T_B=T₁

(

^{T2. TA+T}1^B

)

^{γ −1γ}

 Assume T_B=319 K

T_B=300 K

(

^425+3192∗300

)

^1.4−11.4

T_B=300 K∗(1.0634) T_B=319.02 K

 Since 319 ≈ 319.02 , therefore

T_B=319.02 K Answe r

 Put in (1)

2∗300 K

1 atm =(425+319.02) K

P₂ P₂=744.02

600 atm P₂=1.24 atm Answe r

 From equation (3)

Q

n_A=C_V

[

^TA+T_B−2 T₁

]

Q n_A=5

2R (425+319.02−2∗300 ) K Q n_A=5

2∗8.314 J

mol∗K∗144.02K∗1 kJ 1000 J

Q

n_A=2.993 kJ

mol Answer

(c)

T_B=325 K

 Put in (2)

T_B=T₁

(

^PP21

)

^γ−1γ 325 K=300 K

(

1 atm^P2

)

^{γ −1γ 325}300=

(

1 atm^P2

)

^{γ −1γ}

⁽

³²⁵³⁰⁰

⁾

^{γ −1γ =1 atmP2 P2=}

⁽

³²⁵³⁰⁰

⁾

^{1.4−11.4 atm}

P₂=1.323 atm Answe r

 Put in (1)

2. T₁

P₁ =

(

^TA+T_B

)

P₂

2∗300 K

1 atm =T_A+325 K

1.323 atm T_A+325 K =600 K∗1.323 T_A=793.9 K−325 K T_A=468.9 K Answe r

 From equation (3)

Q

n_A=C_V

[

^TA+T_B−2 T₁

]

Q n_A=5

2R (468.9+325−2∗300 ) K

Q n_A=5

2∗8.314 J

mol∗K∗193.9K∗1 kJ 1000 J

Q

n_A=4.0302 kJ

mol Answe r

(d)

Q

n_A=3 kJ mol

 From equation (1)

2. T₁

P₁ =

(

^TA+T_B

)

P₂ T_A+T_B=2.T₁∗P₂ P₁ → (b )

 From equation (3) Q

n_A=C_V

[

^TA+T_B−2 T₁

]

^TA+T_B−2 T₁= Q

n_A∗C_V T_A+T_B= Q

n_A∗C_V+2T₁→(c )

 Comparing (b) and (c)

2. T₁∗P₂ P₁ = Q

n_A∗C_V+2 T₁

P₂= P₁

2.T1

[

^{nA∗CQ V+2 T1}

]

^{P2=2∗300 K1 atm}

[

^{5 R∗mol2∗3 kJ +2∗300 K}

]

^P2=1 atm

600 K

[

5∗mol∗8.314 J^{6 kJ∗mol∗K} ∗1000 J

1 kJ +600 K

]

P₂=1 atm

600 K [144.335+600]K P₂=1.2406 atm Answe r

 Put in (2)

T_B=T₁

(

^PP21

)

^γ−1γ

T_B=300 K

(

^1.24061

)

^{1.4 −11.4 TB}=300 K∗1.0635

T_B=319.06 K Answer

 Put in (1) 2. T₁

P₁ =

(

^TA+T_B

)

P₂

2∗300 K

1 atm =

(

^TA+319.06 K

)

1.2406 atm

600 K∗1.2406=T_A+319.06 K T_A=744.36 K−319.06 K T_A=425.3 K Answe r

Problem 3.27:

One mole of an ideal gas with constant heat capacities undergoes an arbitrary mechanically reversible process. Show that:

∆ U = 1

γ−1∆ ( PV )

Given Data:

Number of moles=n=1

Solution:

 We know that

∆ U =nC_V∆ T ∆ U =1.C_V∆ T ∆ U =C_V

(

^T2−T₁

)

^→(1)

 For an ideal gas we have, C_P−C_V=R C_P

C_V−C_V C_V= R

C_V

C_P

C_V−1= R C_V

 Since

γ=C_P CV

 Therefore,

γ−1= R

C_V C_V= R γ −1

 Put in (1)

∆ U = R

γ−1

(

^T2−T₁

)

^{∆ U =}_γ−1¹

(

^{R T}2−R T₁

)

^→(2)

 For one mole of an ideal gas we have

P₁V₁=R T₁→(a) P₂V₂=R T₂→(b)

 Put (a) & (b) in (2)

∆ U = 1

γ−1

(

^P2V₂−P₁V₁

)

^{∆ U =}_γ−1¹ ∆ ( PV ) Proved

Problem 3.28:

Derive an equation for the work of a mechanically reversible, isothermal compression of 1 mole of a gas from an initial pressure p1 to a final pressure p2 when the equation of state is the virial expansion truncated to:

Z =1+B ' P

How does the result compare with the corresponding equation for an ideal gas?

Solution:

 For a mechanically reversible process we have,

W =−

∫

V₁ V2

PdV →(1)

 Given that

Z =1+B ' P

 Also

Z =PV RT

 Therefore,

PV

RT=1+B^'P V =RT

P

(

1+B^'P

)

V =RT

(

P¹+B'

)

 Differentiate both sides w.r.t to pressure

dV

dP=RT

(

⁻¹P²+0

)

^{dV =−}P^RT2 dP

 Put in (1)

W=−

∫

V₁ V2

PdV W =−

∫

P₁ P2

−PRT

P² dP W =RT

∫

P₁ P2

1

PdP W =RT|ln P|_P^P₁^{2 W=RT}

(

^{ln P}2−ln P₁

)

W=RT ln P₂

P₁Proved

Problem 3.30:

For methyl chloride at 100 ^℃ the second and third virial coefficients are:

B=−242.5 cm³mol⁻¹;C=25 200 cm⁶mol⁻².

Calculate the work of mechanically reversible, isothermal compression of 1 mol of methyl chloride 1 bar to 55 bars at 100 ^℃ . Base calculations on the following forms of virial equations

a) Z =1+B V +C

V² b) Z =1+B^'P+C ' P²

Where B^'= B

RT∧C^'=C−B² (RT )² Why don’t both equations give exactly the same result?

Given Data:

Temperature=T =100℃ T =(100+273.15) K T =373.15 K B=−242.5 cm³mol⁻¹

C=25200 cm⁶mol⁻² P₁=1 ¯¿ P₂=55 ¯¿ B^'= B

RT C^'=C−B²

(RT )^{2 W=?}

Solution:

 As

B^'= B RT

m²∗1.01325 ¯¿∗1 m³ 100³cm³

B^'=

−242.5 cm³∗mol∗K mol∗8.314 J∗373.15 K∗J

N∗m ∗101325 N

¿

¯¿

B^'=−7.817∗10⁻³1

¿

 Now,

C^'=C−B²

 For a mechanically reversible process we have,

W=−

∫

 Put (a) and (b) in (4)

30780cm³

mol=31023.6cm³

mol∗

(

¹⁻30780^242.5 + 25200 30780²

)

30780cm³

mol=31023.6cm³

mol(1−0.007878+0.000026598) 30780cm³

mol=31023.6∗0.99215cm³ mol

30780cm³

mol=30780cm³ mol

 Since

L. H . S=R . H . S

 Therefore

Initial volume=V₁=30780cm³ mol

 Again using

V₂=RT

P₂

(

^1+VB2+VC2

2

)

^{→ (5 )}

 Assume that

V₂=241.33cm³ mol→(c)

mol∗K∗55

¯¿∗N∗m

J ∗1.01325 ¯¿m²

101325 N ∗100³∗cm³ 1 m³

RT

P₂ =564.067cm³ mol→(d ) RT

P₂=8.314 J∗373.15 K

¿

 Put (c) and (d) in (4)

241.33cm³

mol=564.067cm³

mol∗

(

¹⁻241.33^242.5 + 25200

241.33²

)

^241.33mol^cm3=564.067cm³

mol(1−1.0048+0.4327)

241.33cm³

mol=564.067∗0.4278cm³

mol 241.33cm³

mol=241.33cm³ mol

 Since

L. H . S=R . H . S

 Therefore

Finalvolume=V₂=241.33cm³ mol

 Now from equation (3)

W=−RT

∫

V1

V2

(

^1+VB+V^C2

)

^{V1 dV W=−RT}

[ ^∫

^{VV12 V1 dV +B}

^∫

^{VV12 V12dV +C}

^∫

^{VV12 V13dV}

]

W=−RT

[

^{|ln V|VV12−B}

^|

^V1

^|

^VV12−12C

^|

^V12

^|

^VV12

]

^W=−RT

[ ⁽

^{ln V2−ln V1}

⁾

^−B

⁽

^V12−V11

⁾

^−12C

⁽

^V122−V112

⁾ ]

W=−RT

[

^lnVV21−B

⁽

^V12−V11

⁾

^−12C

⁽

^V122−V112

⁾ ]

W=

−8.314∗J∗373.15 K mol∗K ∗1 kJ

1000 J

[

^{ln241.3330780+242.5}

⁽

^{241.331 −307801}

⁾

^−252002

⁽

^{241.331 2−307801 2}

⁾ ]

W=−3.102 kJ

mol[−4.848+0.996−0.2163]W=12.62 kJ

mol Answer

(b)

Z =1+B^'P+C^'P² PV

RT=1+B^'P+C^'P² V =RT

P

(

1+B^'P+C^'P²

)

V =RT

(

P¹+B^'+C ' P

)

dV

dP=RT

(

⁻¹P²+0+C^'

)

^{dV =−RT}

(

P¹²+C '

)

^dP

 Put in (2)

W=−

∫

P₁ P2

−PRT

(

_P^12+C'

)

^{dP W=RT}

[ ^∫

^{PP12 P1dP+C '}

^∫

^{PP12 PdP}

]

W =RT

[

^|lnP|PP12+C2^'

|

P²

|

_P^P₁²

]

W=RT

[

^lnPP21

+C^'

(

^P2

2−P₁²

) ]

¿¯²

¿¯²

(

55²−1

)

¿ ln55

1 −3.492∗10⁻⁵ 2

1

¿ W=

8.314∗J∗373.15 K mol∗K ∗1 k J

1000 J ¿

W=3.102 kJ

mol (3.9545)

W=12.268 kJ

mol Answer

The answers for part (a) and (b) differ because the relations between the two sets of parameters are exact only for infinite series

Problem 3.32:

Calculate Z and V for ethylene at 25 ^oC and 12 bars by the following equations:

Calculate Z and V for ethylene at 25 ^oC and 12 bars by the following equations: