Koretsky Thermodynamic Solutions for Fugacity, VLE

CBE 231 - Thermodynamics of Fluids

Assignment #3 Solution

1. 1.a. Use the cyclic rule 

∂P ∂T

  v

∂v ∂P

  T

∂T ∂v

 = −1 P

Apply the definitions of β and κ 

 ∂v = βv ∂T P   ∂v = −κv ∂P T Substitute in these expressions         ∂P ∂P ∂v βv β =− =− = = 15.9 ∂T v ∂v T ∂T P −κv κ

bar ◦C

We can use this to find ∆T ∆T =

∆P (121 bar − 1 bar)
= = 7.5 ◦ C ∂P bar 15.9 ◦ C ∂T v

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1.b. Rearrange the provided expression for CP 

   ∂P ∂v ∂T v ∂T P     ∂P ∂v 2 = −T ∂v T ∂T P

CP − Cv = T

= Tv

β2 κ

Find the change in internal energy Z ∆u =

Cv dT  Z T2  β2 dT = CP − T v κ T1 = CP (T2 − T1 ) − = 938 = 506

where v =

MW ρ

= 7.35 × 10−5

m3 mol .

J mol

− 432

vβ 2 2 (T − T12 ) 2κ 2 J mol

J mol

Use this value to find ∆h

∆h = ∆u + ∆(P v) = ∆u + v∆P = 1388

J mol

Now, find the change in entropy 

   ∂s ∂s ds = dT + dv ∂T v ∂v T   ∂s Cv dT = dT = ∂T v T  Z T2  1 β2 ∆s = CP − T v dT κ T1 T   vβ 2 T2 = CP ln − ∆T T1 κ = 1.70

J mol K

2

1.c. Perform an energy balance ∆u = Q = 506

J mol

3

2. Begin with the differential for internal energy  du = Cv dT +



∂u ∂v

dv T

Thus 

∂u ∂T

 = Cv v



      ∂u ∂P ∂v = Cv + T −P ∂T P ∂T v ∂T P          ∂u ∂u ∂P ∂v − = T −P ∂T P ∂T v ∂T v ∂T P

2.a. For an ideal gas, P =

RT v .

 T

∂P ∂T

  R RT −P =T − =0 v v v



Therefore 

∂u ∂T



 − P

∂u ∂T

 =0 v

2.b. For a van der Waals gas P =

RT a − 2 v−b v



∂P T



∂P ∂T



Use this to find that = v

R v−b

So that  T

−P = v

a v2

Also realize that

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R 2a RT dv + 3 dv dT − v−b (v − b)2 v
∂v

dP = Rearrange this and solve for



∂T P

∂v ∂T

 = P

R (v−b) RT − 2a (v−b)2 v3



T 2a(v − b) = − (v − b) Rv 3

−1

Therefore 

∂u ∂T



 −

P

∂u ∂T

 = v

=

a v2 T v−b

−

2a(v−b) Rv

aRv(v − b) RT v 3 − 2a(v − b)2

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3. 3.a. Recall the following definitions  ∂v ∂T  P 1 ∂v κ=− v ∂P T

1 β= v



Use them to find ∂v ∂T P
∂v ∂P T




β =− κ

 =−

 

∂v ∂T

P



∂P ∂v

T

Apply the cyclic rule  −1 =

∂v ∂T

 

∂P ∂v

 

∂P ∂T



P

T

∂T ∂P

 v

Thus, β = κ



v

3.b. Write T = T (v, P ) and write ∂T ∂v



∂P ∂T



 dT =

 dv +

P

∂T ∂P

 dP

(3.1)

v

We can also write Cv ds = dT + T



CP dv = dT − T v



∂v ∂T

 dP P

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



∂T ∂v



∂T ∂v



 dv +

P

P

∂T ∂P



T − CP − Cv

T dP = C P − Cv v





∂T ∂P

∂P ∂T

 

 dv +

v

∂P ∂T





T dv + C P − Cv v

T − CP − Cv v





∂v ∂T

∂v ∂T

 dP P

  dP = 0 P

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In order for this to be true, both terms in square brackets must be equal to zero. Therefore, we can write   ∂P T = CP − Cv ∂T v P     ∂P ∂v CP − Cv = T ∂T ∂T P  v β ∂v =T κ ∂T P 

∂T ∂v



=

T vβ 2 κ

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4. Determine the ideal gas heat capacity from Table A.2.1 CP = 1.213 + 28.785 × 10−3 T − 8.824 × 10−6 T 2 R This process is isentropic, so construct a solution so that the sum of the entropy changes for each step is equal to zero.

4.a. Choose two steps as follows

Write the differential of entropy  ds =

∂s ∂T





∂s ∂v



dT +   ∂P Cv = dT + dv T ∂T v v

dv = 0 T

For this system, use RT a P = − 2 v−b v   ∂P R = ∂T v v−b to find the entropy change for the first step

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Z ∆s1 =

ds Z v2 

 ∂P dv ∂T v v1 Z v2 R dv = v1 v − b   v2 − b = R ln v1 − b =

Because P2 is low, we can assume that the gas at this state behaves as an ideal gas " RT # 2 P2 − b ∆s1 = R ln v1 − b Next, calculate ∆s1 Z

T2

∆s2 = T1

Cv dT = R T

Z

T2

623.15 K

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

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 .

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5. Perform an energy balance to find that ∆u = 0. The gas is not ideal under the stated conditions, so we must create a path that connects the initial to the final state through three steps into a region where we can assume the gas behaves as an ideal gas, as shown below

First, find ∆u1 v=∞ 

 ∂u dv ∂v T vi   Z v=∞   ∂P = T − P dv ∂T v vi Z

∆u1 =

For the van der Waals equation of state   R ∂P a = + 2 2 ∂T v v−b T v Plug this into the expression for ∆u1   Z ∞ Z ∞ RT a 2a ∆u1 = + − P dv = dv = 1120 2 2 Tv vi =2.5×10−4 v − b vi =2.5×10−4 Ti v

J mol

Perform a similar procedure for the third step Z

vf =5×10−4

∆u3 = ∞



 Z vf =5×10−4 RT a 2a −168000 + − P dv = dv = 2 2 v − b Tv Tf v Tf ∞

J mol K

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

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3 ∆u2 = R(Tf − 300 K) 2 Sum the internal energies

∆u = ∆u1 + ∆u2 + ∆u3 = 1120

J mol

+

3 8.314 2

J mol K




(Tf − 300 K) −

168000 =0 Tf

Solve for Tf = 262 K .

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