This preview shows page 1. Sign up to view the full content.
Unformatted text preview: BioE
BioE
Entropy calculation and
ideal work
ideal work
Entropy calculation for ideal gases in a
simple closed system
Ideal work done by simple closed systems
work done by simple closed systems
Helmholtz free energy
Entropy balance over a flow system
balance over flow system
Ideal work done by a flow system
Gibbs free energy
free energy
Molecular BioEngineering Laboratory 1. Calculation of ΔS for ideal gases under an
Calculation of
for ideal gases under an
isothermal process
dU = TdS − PdV , dH = TdS + VdP
dS = ( dU + PdV ) / T = ( dH − VdP ) / T
dS = nCV , m dT / T + nRdV / V = nC P , m dT / T − nRdP / P
1. Isothermal process, dS = nRdV / V = − nRdP / P
Isothermal
Δ S = nR ln V2
P
= − nR ln 2
V1
P1 BioE 2. Calculation of
2. Calculation of ΔS for ideal gases under a constant
for ideal gases under constant
volume process dU = TdS − PdV , dH = TdS + VdP
dS = (dU + PdV ) / T = (dH − VdP) / T
dS = nCV ,m dT / T + nRdV / V = nCP ,m dT / T − nRdP / P
2. Constant volume process, dS = nCV ,m dT / T
T2
ΔS = nCV ,m ln , if CV ,m is a constant.
if
T1 BioE
3. Calculation of ΔS for ideal gases under an isobar
Calculation of
for ideal gases under an isobar
process
dU = TdS − PdV
dS = ( dU + PdV ) / T = ( dH − VdP ) / T
dS = nC P , m dT / T + nRdP / P
3. Constant pressure process, dS = nC P , m dT / T
process
ΔS = nC P , m ln T2
, if C P , m is a constant.
if
is
T1 BioE Th thi
The third law of thermodynamics
th
Tf (
S (T ) − S (0) = ∫ C Ps ) dT +
0 ΔH f
Tf Tb ΔH vap Tf Tb (
+ ∫ C Pl ) dT + T (
+ ∫ C Pg ) dT
Tb CP/T S ( 0) = 0 BioE
4. Calculation of ΔS for ideal gases under an
Calculation of
for ideal gases under an
adiabatic process
dU = TdS − PdV
dS = (dU + PdV ) / T = (dH − VdP) / T
dS = nCP,m dT / T − nRdP / P
dS = nCV ,m dT / T + nRdV / V
4  1. Adiabatic reversible, dS = 0
Adiabatic
4  2. Adiabatic irreversible, dS = nCP ,m dT / T
V2
T2
ΔS = nCV ,m ln + nR ln , if CV ,m is a constant.
V1
T1
ΔS = nCP ,m ln T2
P
− nR ln 2 , if CP,m is a constant.
T1
P
1 BioE A system is divided into two chambers by a movable diathermic
piston and is placed in a constant temperature water bath at T1.
i
l
Chamber A contains gas A at a pressure of PA1 and chamber B
contains gas B at a pressure of PB1. Initially chambers A and B
ll
are kept at equal size by a mechanical lock and the initial
temperature of the system is T1. Estimate the changes after the
mechanical lock is released.
1. Gas A is an ideal gas.
PB1 = 0
The system is completely insulated.
system is completely insulated.
2. Gas A is a van der Waals gas.
PB1 = 0
The wall of the system is diathermic.
wall of the system is diathermic B A BioE
Two gas chambers are connected as the following diagram.
Chamber A is filled with ideal gas A and chamber is filled with
ideal gas B. The initial temperatures and pressures in both
chambers are the same, T0 and P0. Please calculate the entropy
change after the valve is opened. Assume the outside wall of the
system is (1) adiabatic, (2) diathermal. A B BioE Maximum obtainable work 1
海面溫度為TH = 330K, 深
海洋流溫度為TL=280K,
請問利用此洋流做功, 每
小時最多可做多少功? TH E W TL BioE Maximum obtainable work 2
Assuming that there is no heat
released to the environment, calculate
released to the environment, calculate
the maximum work can be done by a
1000g metal blocks. The initial
temperature of Metal A is 800K and
the surrounding temperature is 300K.
The constant volume heat capacity of
metal A is 3 J K1 g1. A W E Surrounding BioE dS ≥ δQ
Tsurr δQ ≤ Tsurr dS
dU = δQ + δW
− δW = δQ − dU ≤ Tsurr dS − dU
− Wmax = Tsurr ΔS − ΔU = −(ΔU − Tsurr ΔS )
ΔS = ∫ Tsurr TA 0 ΔU = ∫ dS = ∫ Tsurr TA 0 Tsurr TA 0 dU = ∫ CV dT
Tsurr
= CV ln
TA 0
T Tsurr TA 0 CV dT = CV (Tsurr − TA0 ) − Wmax = +CV (TA0 − Tsurr ) − CV Tsurr ln TA 0
Tsurr BioE Helmholtz free energy, A A ≡ U − TS
at constant temperature T
dA = dU − TdS
− δWmax = TdS − dU = −dA BioE Entropy balance for an open system d
&
&
&
(mS )sys + dS surr = (mS )in − (mS )out + SG
dt
dt
d
&
&
(mS )sys + dS surr + Δ(mS ) = SG ≥ 0
dt
dt
&
d
Q
&
&
(mS )sys −
+ Δ(mS ) = SG ≥ 0
Tsurr
dt BioE Entropy balance for a steady state flow
py
system
&
d
Q
&
&
(mS )sys −
+ Δ (mS ) = S G ≥ 0
dt
Tsurr
d
(mS )sys = 0
dt
&
Q
&
&
+ Δ (mS ) = S G ≥ 0
∴−
Tsurr
&
Q
&
Δ (mS ) ≥
Tsurr
Q BioE Id
Ideal work
&
Q
&
Δ (mS ) ≥
Tsurr
if Tsurr &
Q
&
≈ Tsys , Δ (mS ) =
Tsys 1
⎡
⎤&&
&
Δ ⎢ m ( H + u 2 + gz ) ⎥ = Q + W s
2
⎣
⎦
1
⎡
⎤
&
&
&
&
Ws ( rev ) = Wideal = Δ ⎢ m ( H + u 2 + gz ) ⎥ − Tsys Δ (mS )
2
⎣
⎦
&
&
&
&
W
= Δ ( mH ) − T Δ (mS ) = m (ΔH − T ΔS )
ideal sys Wideal = ΔH − Tsys ΔS sys BioE Gibbs free energy
G ≡ H − TS
at constant temperature T
dG = dH − TdS
− δWideal = TdS − dH = −dG
dG BioE Maximum obtainable work 3
Assuming that there is no heat
released to the environment, calculate
released to the environment, calculate
the maximum work can be done by
two 1000g metal blocks and the final
temperature of these two blocks. Metal
A is at 800K and metal B is at 200K.
The constant volume heat capacity of
metal A is 3 J K1 g1 and that of metal
B is 5 J K1 g1. A E W B BioE Maximum obtainable work 4
E
W
TA0
PA0
A0 T0, P0 Calculate the maximum work
the maximum work
can be done by a N2 gas
cylinder. The initial
cylinder. The initial
temperature and pressure of
the N2 gas are 400 K and 10
bar. The work is done by
connecting an ideal engine to
the cylinder and exhaust the
N2 gas into the outside
environment at 1 bar and
300K. BioE HW#10
HW#101
1. Estimate the maximum
work that can be done by
the big bird. The bird has m
grams ether in its body. Its
head is always wet by
keeping on dipping in the
water The atmospheric
water. The atmospheric
temperature is Ta and the
wet bulb temperature is Tw.
The heat of evaporation per
gram of ether is ΔHvap. BioE Answer#101
m ΔHvap x (1 Tw/ Ta) BioE HW#10
HW#102
2. Without contacting with
environment please calculate the
maximum work that can be
obtained from the three metal
blocks A, B, and C. The initial
temperatures of metal blocks, A,
B, and C are TA0, TB0, and TC0,
respectively. The masses of
,
blocks A, B, and C are mA, mB,
and mC, respectively. The specific
heat capacities of the three
blocks are CPgA, CPgB, and CPgC,
respectively. A C
B BioE Answer #102
#10
Tf A
CP, g TA 0 T ΔS = ∫ m A Tf B
CP, g TB 0 T dT + ∫ mB Tf C
CP, g TC 0 T dT + ∫ mC dT Solve for Tf − W ≤ TΔS − ΔU = −ΔU
Tf = ∫ m AC
TA 0 A
P, g Tf dT + ∫ mB C
TB 0 B
P,g Tf C
dT + ∫ mC C P , g dT
TC 0 BioE ...
View
Full
Document
This note was uploaded on 11/11/2010 for the course CHE CH2005 taught by Professor 曹恒光 during the Spring '10 term at National Central University.
 Spring '10
 曹恒光
 Physical chemistry, pH

Click to edit the document details