Advanced Thermodynamics (Midterm Exam)

Advanced Thermodynamics (Midterm Exam)

M2794.007900-001 April 30, 2019

1. Equation of State (EOS): (P + ^a

v²) (v − b) = RT P = RT

v − b− a v²

The critical point is identical to inflection point of isothermal line in pressure-specific volume graph.

(∂P

∂v)

T

= − RT_c

(v_c − b)²+2a v_c³ = 0 (∂²P

∂v²)

T

= 2RT_c

(v_c− b)³−6a v_c⁴ = 0 Solving the simultaneous equation

2RT_c

(v_c− b)³ = RT_c

(v_c− b)²× 2

v_c− b =2a v_c³ × 2

v_c− b= 6a v_c⁴ v_c = 3b

T_c = 8a 27Rb Substituting the above critical values into EOS

P_c = a 27b² (Answer)

(1) Critical temperature: 𝐓_𝐜 = ^𝟖𝐚

𝟐𝟕𝐑𝐛

(2) Critical specific volume: 𝐯_𝐜 = 𝟑𝐛 (3) Critical pressure: 𝐏_𝐜= ^𝐚

𝟐𝟕𝐛^𝟐

2. (1) Joule-Thomson coefficient μ_JT = (^∂T

∂P)

h

Using cyclical relation

(∂T

∂P)

h

= − (∂T

∂h)

P

(∂h

∂P)

T

From a definition of isobaric heat capacity c_p ≡ (∂h

∂T)

P

Another derivative term is obtained from the first law of thermodynamics dh = Tds + vdP = T (∂s

∂T)

P

dT + [T (∂s

∂P)

T

+ v] dP = (∂h

∂T)

P

dT + (∂h

∂P)

T

dP

(∂h

∂P)

T

= T (∂s

∂P)

T

+ v

Using Maxwell relations

(∂s

∂P)

T

= − (∂v

∂T)

P

So Joule-Thomson coefficient can be expressed as follows.

(∂T

∂P)

h

= 1

c_p[T (∂v

∂T)

P

− v]

EOS of the problem

(P + a

Tv²) (v − b) = RT Pv − bP + a

Tv− ab

Tv² = RT

Differentiating all of the terms with temperature and specific volume in constant pressure P ∂v + a (1

v×−1

T² ∂T +1 T×−1

v² ∂v) − ab (1 v² ×−1

T² ∂T +1 T×−2

v³ ∂v) = R ∂T (P − a

Tv²+2ab Tv³) (∂v

∂T)

P

= R + a

vT² − ab v²T²

(∂v

∂T)

P

=

R + a

vT²− ab v²T² P − a

Tv²+2ab Tv³

= Rv³T²+ av³ − abv Pv³T²− avT + 2abT (Answer)

(𝛛𝐓

𝛛𝐏)

𝐡

= 𝟏

𝐜_𝐩[𝐑𝐯^𝟑𝐓^𝟐+ 𝐚𝐯^𝟑− 𝐚𝐛𝐯 𝐏𝐯^𝟑𝐓 − 𝐚𝐯 + 𝟐𝐚𝐛 − 𝐯]

Or

(𝛛𝐓

𝛛𝐏)

𝐡

= 𝟏 𝐜_𝐩[𝐓 ×

𝐑 + 𝐚

𝐯𝐓^𝟐− 𝐚𝐛 𝐯^𝟐𝐓^𝟐 (𝐏 − 𝐚

𝐓𝐯^𝟐+𝟐𝐚𝐛 𝐓𝐯^𝟑)

− 𝐯]

(2) Isentropic expansion coefficient μ_s = (^∂T

∂P)

s

(∂T

∂P)

s

= − (∂T

∂s)

P

(∂s

∂P)

T

From the above problem (1)

(∂s

∂P)

T

= − (∂v

∂T)

P

Using the first law of thermodynamics and differentiation (∂s

∂T)

P

= 1 T(∂h

∂T)

P

= c_P T

Thus the isentropic expansion coefficient can be expressed as follows.

(∂T

∂P)

s

= T c_P(∂v

∂T)

P

(Answer)

(𝛛𝐓

𝛛𝐏)

𝐬

= 𝐓(𝐑𝐯^𝟑𝐓^𝟐+ 𝐚𝐯^𝟑− 𝐚𝐛𝐯) 𝐜_𝐏(𝐏𝐯^𝟑𝐓^𝟐− 𝐚𝐯𝐓 + 𝟐𝐚𝐛𝐓)

(3) (Answer)

Isentropic process: T = 152 K (150 ~ 155 K) Isenthalpic process: T = 285 K (280 ~ 290 K)

3.

Tds = c_pdT − T (∂v

∂T)

p

dP

Using the first law of thermodynamics

Tds = dh − vdP From the derivative of enthalpy

Tds = [(∂h

∂T)_PdT + (∂h

∂P)_T dP] − vdP ds =1

T(∂h

∂T)_PdT +1 T[(∂h

∂P)_T − v]dP From the derivative of entropy

ds = (∂s

∂T)_PdT + (∂s

∂P)_TdP Since temperature and pressure are independent

(∂s

∂T)_P =1 T(∂h

∂T)_P & (∂s

∂T)_T = 1 T[(∂h

∂P)_T − v]

The differential entropy ds is exact. Therefore

∂²s

∂P ∂T= ∂²s

∂T ∂P (∂h

∂P)_T= −T (∂v

∂T)_P+ v Substituting all the term

∴ 𝐓𝐝𝐬 = 𝐜_𝐩𝐝𝐓 − 𝐓 (𝛛𝐯

𝛛𝐓)

𝐩

𝐝𝐏

4. (1) Temperature change

The container is adiabatic and no work is done

dU = 0 or U = Const Final temperature is

T₁ = T₀+ ∫ (∂T

∂v)

u

dv

v₁

v0

Using cyclical relation

(∂T

∂v)

u

= − (∂u

∂v)

T

(∂u

∂T)

Because internal energy of an ideal gas is independent of specific volume v

(∂u

∂v)

T

= 0 Substituting the relation

(Answer)

∆𝐓 = 𝐓_𝟏− 𝐓_𝟎 = 𝟎 (2) Pressure change

Since the air is treated as an ideal gas and temperature is constant regardless of free expansion

PV = Const

Volume that air takes increases as twice during expansion. Pressure of air at final state is P₁ =1

2𝑃₀ Thus pressure change during expansion is

(Answer)

∆𝐏 = −𝟏

𝟐𝑷_𝟎 = −(𝒏_𝒐𝟐𝟎+ 𝒏_𝑵𝟐𝟎)𝑹𝑻_𝟎 𝟐𝑽_𝟎

In view of the partial pressure of nitrogen and oxygen Nitrogen: ∆𝐏_𝑵𝟐𝟎= −𝟎. 𝟒𝑷_𝟎

Oxygen: ∆𝐏_𝒐𝟐𝟎= −𝟎. 𝟏𝑷_𝟎 (3) Specific volume change

Volume that air takes increases as twice

V₁ = 2𝑉₀

Considering mass conservation, change of specific volume is (Answer)

∆𝐯 = 𝟐𝑽_𝟎

𝒏_𝒐𝟐𝟎+ 𝒏_𝑵𝟐𝟎− 𝑽_𝟎

𝒏_𝒐𝟐𝟎+ 𝒏_𝑵𝟐𝟎 = 𝑽_𝟎 𝒏_𝒐𝟐𝟎+ 𝒏_𝑵𝟐𝟎 Or ∆𝐯 = 𝐯_𝟎

(4) Entropy change

From the first law of thermodynamics

TdS = dU + PdV

Because internal energy is constant and the air is treated as an ideal gas dS =𝑛𝑅

𝑉 ^𝑑𝑉 Integrating the equation, entropy change is

(Answer)

∆𝐒 = (𝒏_𝒐𝟐𝟎+ 𝒏_𝑵𝟐𝟎)𝐑𝐥𝐧𝟐

(5) Enthalpy change

From the definition of enthalpy

H = U + PV

Because internal energy and temperature is constant regardless of expansion (Answer)

∆𝐇 = 𝟎

(6) Helmholtz function change

From the definition of Helmholtz function

F = U − ST Since internal energy and temperature are constant (Answer)

∆𝐅 = −𝐓∆𝐒 = −(𝒏_𝒐𝟐𝟎+ 𝒏_𝑵𝟐𝟎)𝐑𝐓_𝟎𝐥𝐧𝟐

(7) Gibbs function change

From the definition of Gibbs function

G = U − ST + PV = F + PV

Since PV is constant during free expansion, the Gibbs function change is the same with the change of Helmholtz function

(Answer)

∆𝐆 = −(𝒏_𝒐𝟐𝟎+ 𝒏_𝑵𝟐𝟎)𝐑𝐓_𝟎𝐥𝐧𝟐

5. (1) From mass conservation

n₀^v+ n₀^l = n₁^v+ n₁^l

n₀^v− n₁^v = −(n₀^l − n₁^l) Gibbs function of each state is

G₀ = n₀^vg^v+ n₀^lg^l

G₁ = n₁^vg^v+ n₁^lg^l Since the process is reversible

G₀ = G₁ or (∆G)_T,P = 0

g^v(n₀^v− n₁^v) + g^l(n₀^l − n₁^l) = 0 From the molecular relation of liquid and vapor state

g^v = g^l (2) From the above relation

dg^v = dg^l Using a differential expression of Gibbs function

−s^𝑙𝑑𝑇 + 𝑣^𝑙𝑑𝑃 = −s^𝑣𝑑𝑇 + 𝑣^𝑣𝑑𝑃 𝑑𝑃

𝑑𝑇 = 𝑠^𝑣− 𝑠^𝑙 𝑣^𝑣− 𝑣^𝑙 From the definition of entropy

s^𝑣− 𝑠^𝑙= 𝐿

Substituting the equation to entropy term and ignoring specific volume of saturated 𝑇 liquid

𝑑𝑃 𝑑𝑇 ≈ 𝐿

𝑇𝑣^𝑣 Treating the vapor as an ideal gas

𝑑𝑃 𝑑𝑇 = 𝐿𝑃

𝑅𝑇²

Integrating the equation by considering reference values lnP − ln P₀ = 𝐿

𝑅(⁻¹

𝑇 −−1 𝑇₀) Saturation pressure as a function of saturation temperature is (Answer)

P = P₀exp (−𝐿 𝑅(¹

𝑇− 1 𝑇₀))

6. (1) The equation of state (EOS)

Using a differential expression of Gibbs function dg = −sdT + vdP

v = (𝜕𝑔

𝜕𝑃)

𝑇

Differentiating specific Gibbs function with pressure in constant temperature v = ∂

∂P[𝑅𝑇𝑙𝑛 (𝑃

𝑃₀) − 𝐴(𝑇)𝑃]

𝑇

v = 𝑅𝑇

𝑃 − 𝐴(𝑇) (Answer)

∴ 𝐏(𝐯 + 𝐀(𝐓)) = 𝐑𝐓 (2) The specific entropy

Using the above differential expression of Gibbs function s = − (∂g

∂T)

𝑃

Differentiating specific Gibbs function with temperature in constant pressure s = − ∂

∂T[𝑅𝑇𝑙𝑛 (𝑃

𝑃₀) − 𝐴(𝑇)𝑃]

𝑃

(Answer)

𝐬 = −𝐑𝐥𝐧 (𝑷

𝑷_𝟎) + 𝐏 (𝝏𝑨

𝝏𝑻)

𝑷

(3) The specific Helmholtz function

From the definition of Helmholtz function f = g − Pv Substituting the above calculated equations

f = [𝑅𝑇𝑙𝑛 (𝑃

𝑃₀) − 𝐴(𝑇)𝑃] − 𝑃 (𝑅𝑇

𝑃 − 𝐴(𝑇)) (Answer)

𝐟 = 𝐑𝐓 (𝒍𝒏 (𝑷

𝑷_𝟎) − 𝟏)

7. Chemical potential can be expressed by Gibbs function and molecules.

μ =𝐺 𝑛 From the definition of Gibbs function

G = U − TS + PV = H − TS When we treat entropy as a function of temperature and volume

TdS = (∂U

∂T)

V

dT + (∂U

∂V)

T

dV + PdV

dS =𝑛c_v

T dT +nR V dV Substituting entropy by taking reference value into account

S = nc_vlnT + nRlnV + S₀ Differential expression of enthalpy is

dH = (𝜕𝐻

𝜕𝑇)

𝑃

𝑑𝑇 + (𝜕𝐻

𝜕𝑃)

𝑇

𝑑𝑃 When considering only temperature and volume

H = nc_𝑝𝑇 + 𝐻₀

Substituting enthalpy and entropy into the Gibbs function equation G = nc_𝑝𝑇 + 𝐻₀ − 𝑇(nc_vlnT + nRlnV + S₀) From the above relation

∴ 𝛍 = 𝐜_𝐩𝐓 − 𝐜_𝐯𝐓𝐥𝐧𝐓 − 𝐑𝐓𝐥𝐧𝐕 − 𝐬_𝟎𝐓 + 𝐡_𝟎 where 𝐬_𝟎, 𝐡_𝟎 are reference values

8. (1) Molecular flux, Φ (Number of collisions per unit time and unit length) Local curve of velocity circle of which radii is v

𝑑𝐿

𝐿 = 𝑣𝑑𝜃 2𝜋𝑣 The number of molecules in the space is

d²𝑛_𝜃𝑣𝑑𝐴 = 𝑑𝑛_𝑣×𝑣𝑑𝜃

2𝜋𝑣× (𝑣𝑑𝑡𝑠𝑖𝑛𝜃𝑑𝑙) Where 𝑑𝑛_𝑣 means density in the range of v to v+dv.

The number of collisions per unit length and time 𝑑Φ =d²𝑛_𝜃𝑣𝑑𝐴

𝑑𝑙𝑑𝑡 = 1

2𝜋𝑣𝑑𝑛_𝑣𝑠𝑖𝑛𝜃𝑑𝜃 Integrating the equation in the given range

∫d²𝑛_𝜃𝑣𝑑𝐴 𝑑𝑙𝑑𝑡 = 1

2𝜋∫ 𝑣𝑑𝑛_𝑣

∞ 0

∫ 𝑠𝑖𝑛𝜃𝑑𝜃

𝜋

0

Since 𝑣̅ =¹

𝑛∫ 𝑣𝑑𝑛𝑣

∞ 0

(Answer)

𝚽 =𝒏𝒗̅ (2) Pressure of the gas, P 𝝅

Momentum change of one molecule

mvsinθ − (−mvsinθ) = 2mvsinθ Change in momentum due to collisions

dp = ∬ 2mvsinθ × 1

2𝜋𝑣𝑑𝑛_𝑣𝑠𝑖𝑛𝜃𝑑𝜃 Integrating in the given range

dp = 𝑚

𝜋∫ 𝑣²𝑑𝑛_𝑣

∞ 0

∫ sin²𝜃 𝑑𝜃

𝜋 0

𝑑𝑙𝑑𝑡 Since 𝑣̅̅̅ =^{2 1}

𝑛∫ 𝑣₀^{∞ 2}𝑑𝑛_𝑣

dp =1

2𝑚𝑛𝑣̅̅̅𝑑𝑙𝑑𝑡 ² From the definition of force

dF =𝑑𝑝 𝑑𝑡 Pressure acting on the given unit length (Answer)

𝐝𝐏 =𝒅𝑭 𝒅𝒍 =𝟏

𝟐𝒎𝒏𝒗̅̅̅^𝟐

9. Fraction of the total molecules lying in the range dN_v

N = f(v)dv Number of molecules lies in the range of v to v + dv

dN_v = Nf(v)dv Density of the molecules

ρ =dN_v

dv = Nf(v)

Because of the homogeneous distribution and elastic collisions, the form of f(v) is f(v) = αe^−βv2

Where α, β are constants

Number of molecules in one dimension is

dN_v = Nαe^−βv2dv

Considering two constraints, total number of molecules and its energy Total number of molecules is

N = ∫ dN_v

∞ 0

= Nα ∫ e^−βv2dv

∞

0

Because ∫ e₀^{∞ −ax}

2

dx =¹

2√^π_a

1 =α 2√π

β

Total energy of the molecules in one dimensional system is 1

2NkT =1

2m ∫ v²dN_v

∞ 0

=1

2m × Nα ∫ v²e^−βv2dv

∞

0

Because ∫ x₀^{∞ 2}e^−ax

2

dx = ¹

4√_a^π₃

kT =mα 4 √π

β³ Solving the simultaneous equations, two constants are

β = m 2kT α = √2m

πkT Substituting two constants into density function (Answer)

𝛒 =𝐝𝐍_𝐯

𝐝𝐯 = 𝐍√𝟐𝐦

𝛑𝐤𝐓𝐞^{−𝟐𝐤𝐓𝐦 𝐯𝟐}