CBE 231 - Thermodynamics of Fluids

Assignment #3 Solution

1.

1.a.

Use the cyclic rule

(∂P

∂T

)v

(∂v

∂P

)T

(∂T

∂v

)P

= −1

Apply the definitions of β and κ

(∂v

∂T

)P

= βv(∂v

∂P

)T

= −κv

Substitute in these expressions(∂P

∂T

)v

= −(∂P

∂v

)T

(∂v

∂T

)P

= −(βv

−κv

)=β

κ= 15.9 bar

◦C

We can use this to find ∆T

∆T =∆P(∂P∂T

)v

=(121 bar − 1 bar)

15.9 bar◦C

= 7.5 ◦C

1

1.b.

Rearrange the provided expression for CP

CP − Cv = T

(∂P

∂T

)v

(∂v

∂T

)P

= −T(∂P

∂v

)T

(∂v

∂T

)2

P

= Tvβ2

κ

Find the change in internal energy

∆u =

∫CvdT

=

∫ T2

T1

(CP − Tv

β2

κ

)dT

= CP (T2 − T1) −vβ2

2κ(T 2

2 − T 21 )

= 938 Jmol − 432 J

mol

= 506 Jmol

where v = MWρ = 7.35 × 10−5 m3

mol . Use this value to find ∆h

∆h = ∆u+ ∆(Pv)

= ∆u+ v∆P

= 1388 Jmol

Now, find the change in entropy

ds =

(∂s

∂T

)v

dT +

(∂s

∂v

)T

dv

=

(∂s

∂T

)v

dT =CvTdT

∆s =

∫ T2

T1

1

T

(CP − Tv

β2

κ

)dT

= CP ln

(T2T1

)− vβ2

κ∆T

= 1.70 Jmol K

2

1.c.

Perform an energy balance

∆u = Q = 506 Jmol

3

2.

Begin with the differential for internal energy

du = CvdT +

(∂u

∂v

)T

dv

Thus (∂u

∂T

)v

= Cv

(∂u

∂T

)P

= Cv +

[T

(∂P

∂T

)v

− P

](∂v

∂T

)P(

∂u

∂T

)P

−(∂u

∂T

)v

=

[T

(∂P

∂T

)v

− P

](∂v

∂T

)P

2.a.

For an ideal gas, P = RTv .

T

(∂P

∂T

)v

− P = T

(R

v

)− RT

v= 0

Therefore (∂u

∂T

)P

−(∂u

∂T

)v

= 0

2.b.

For a van der Waals gas

P =RT

v − b− a

v2

Use this to find that (∂P

T

)v

=R

v − b

So that

T

(∂P

∂T

)v

− P =a

v2

Also realize that

4

dP =R

v − bdT − RT

(v − b)2dv +

2a

v3dv

Rearrange this and solve for(∂v∂T

)P

(∂v

∂T

)P

=

R(v−b)

RT(v−b)2 − 2a

v3

=

[T

(v − b)− 2a(v − b)

Rv3

]−1Therefore

(∂u

∂T

)P

−(∂u

∂T

)v

=a

v2Tv−b −

2a(v−b)Rv

=aRv(v − b)

RTv3 − 2a(v − b)2

5

3.

3.a.

Recall the following definitions

β =1

v

(∂v

∂T

)P

κ = −1

v

(∂v

∂P

)T

Use them to find

β

κ= −

(∂v∂T

)P(

∂v∂P

)T

= −(∂v

∂T

)P

(∂P

∂v

)T

Apply the cyclic rule

−1 =

(∂v

∂T

)P

(∂P

∂v

)T

(∂T

∂P

)v

Thus,

β

κ=

(∂P

∂T

)v

3.b.

Write T = T (v, P ) and write

dT =

(∂T

∂v

)P

dv +

(∂T

∂P

)v

dP (3.1)

We can also write

ds =CvTdT +

(∂P

∂T

)v

dv =CPTdT −

(∂v

∂T

)P

dP

Rearrange this equation and solve for dT . Then, set it equal to Equation 3.1 and groupterms

(∂T

∂v

)P

dv +

(∂T

∂P

)v

dP =T

CP − Cv

(∂P

∂T

)v

dv +T

CP − Cv

(∂v

∂T

)P

dP

[(∂T

∂v

)P

− T

CP − Cv

(∂P

∂T

)v

]dv +

[(∂T

∂P

)v

− T

CP − Cv

(∂v

∂T

)P

]dP = 0

6

In order for this to be true, both terms in square brackets must be equal to zero. There-fore, we can write

(∂T

∂v

)P

=T

CP − Cv

(∂P

∂T

)v

CP − Cv = T

(∂P

∂T

)v

(∂v

∂T

)P

= Tβ

κ

(∂v

∂T

)P

=Tvβ2

κ

7

4.

Determine the ideal gas heat capacity from Table A.2.1

CPR

= 1.213 + 28.785 × 10−3T − 8.824 × 10−6T 2

This process is isentropic, so construct a solution so that the sum of the entropy changesfor each step is equal to zero.

4.a.

Choose two steps as follows

Write the differential of entropy

ds =

(∂s

∂T

)v

dT +

(∂s

∂v

)T

dv = 0

=CvTdT +

(∂P

∂T

)v

dv

For this system, use

P =RT

v − b− a

v2(∂P

∂T

)v

=R

v − b

to find the entropy change for the first step

8

∆s1 =

∫ds

=

∫ v2

v1

(∂P

∂T

)v

dv

=

∫ v2

v1

R

v − bdv

= R ln

[v2 − b

v1 − b

]Because P2 is low, we can assume that the gas at this state behaves as an ideal gas

∆s1 = R ln

[RT2P2

− b

v1 − b

]Next, calculate ∆s1

∆s2 =

∫ T2

T1

CvTdT = R

∫ T2

623.15 K

0.213 + 28.785 × 10−3T − 8.824 × 10−6T 2

TdT

Add both of the entropy changes and set them equal to zero

∆s = ∆s1 + ∆s2 = 0

Substitute in the expressions from above and solve for T2 = 448.3 K .

9

5.

Perform an energy balance to find that ∆u = 0. The gas is not ideal under the statedconditions, so we must create a path that connects the initial to the final state throughthree steps into a region where we can assume the gas behaves as an ideal gas, as shownbelow

First, find ∆u1

∆u1 =

∫ v=∞

vi

(∂u

∂v

)T

dv

=

∫ v=∞

vi

[T

(∂P

∂T

)v

− P

]dv

For the van der Waals equation of state(∂P

∂T

)v

=R

v − b+

a

T 2v2

Plug this into the expression for ∆u1

∆u1 =

∫ ∞vi=2.5×10−4

[RT

v − b+

a

Tv2− P

]dv =

∫ ∞vi=2.5×10−4

2a

Tiv2dv = 1120 J

mol

Perform a similar procedure for the third step

∆u3 =

∫ vf=5×10−4

∞

[RT

v − b+

a

Tv2− P

]dv =

∫ vf=5×10−4

∞

2a

Tfv2dv =

−168000

TfJ

mol K

The molar volume is infinite for step 2, so we can use the ideal gas heat capacity to find∆u2

10

∆u2 =3

2R(Tf − 300 K)

Sum the internal energies

∆u = ∆u1 + ∆u2 + ∆u3 = 1120 Jmol +

3

2

(8.314 J

mol K

)(Tf − 300 K) − 168000

Tf= 0

Solve for Tf = 262 K .

11