1
Physical Chemistry I for Biochemists
Chem340
Lecture 9 (1/31/11)
Yoshitaka Ishii Partial Derivatives Ch3.1-3.3 HW3 (Continued)
Announcement
• Homework 4 is uploaded at the web site (Monday)(Monday)
• exp(x) = ex in case that you do not know the definition of exp(x).
2
Derivatives)()(
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fxx 0
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Partial Derivatives
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Slope along the X axis
dy
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3
How to Obtain Partial Derivatives? (text p44)
dTT
PdV
V
PTVdP
VT
),(
Step 1: Rewrite P as a function of T and V.
TT V
VnRT
V
P
/
p
Step 2: Put the constants outside the derivative.
VnRT
P
/1 T is considered to be a
t t f thi ti l
?
VT
P
TT VnRT
V
constant for this partial derivative.
Step 3: Perform the derivative with respect to V
2
1
VnRT
V
P
T
Sample Question (Text p44)
Calculate for an ideal gas.VT
P
Step 1: Rewrite P as a function of T and V.
VV T
VnRT
T
P
/
p
Step 2: Put constants outside the derivatives.
TnRP
VV TVT
Step 3: Perform the derivative with respect to T
V
nR
T
P
V
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Example from A.6
=
• HW3. 5. Calculate Cv = (∂U/∂T)V and Cp = {∂(U + PV)/∂T}P for n mole of ideal gas assuming that U = 5nRT/2 and n is a constant. What kind of molecular properties do they depend on?
• In general, U can be given by U(T, V) or U(T, P). For an ideal gas, U(T) = 5nRT/2.
(∂U/∂T) = (∂(5nRT/2)/∂T)(∂U/∂T)V = (∂(5nRT/2)/∂T)V
= 5nR/2 (∂T/∂T)V
Cp = {∂(U + PV)/∂T}P = {∂(3nRT/2 + nRT)/∂T}P
= {∂(5nRT/2)/∂T}P = ? Change this into a function of T and P
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Ch3.1 Math Properties of State Function
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yxZdx
x
yxZyxZdyydxxZ
xy
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ZZ )()(dy
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yxZdx
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yxZyxdZ
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If Z is a state function,
yxZyxZ
),(),(
yxxy y
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),(),(
Total Differential, Exact Differential, & the Test for Exactness
yxZyxZ )()(
dZ is called total differential of Z
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yxZdx
x
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xy
),(),(),(),(
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dZ is called “an exact differential” if
yxxy y
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xx
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y
),(),(
yx
yxgx
yxfy
),(),(or
yxy
When dZ is exact, Z is a state function.Namely, the change in Z does not depend on a path of (x, y). Z =Z(xfin, yfin) - Z(xini, yini)
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Sample Question(Test of inexactness of Dw =-PdV)
We define work as Dw = -PdV. HW4 #1.
Show Dw is an inexact differential with respect to V and T for an ideal gas.
Dw = -PdV + 0 dT.
?)()( TVTVP
0 ?),(),(TV
TVV
TVPT
0
VnRVnRTT
TVPT VV
//),(
Partial Differentials for P, T, V• There are 6 possible partial differentials
P
P
mV
mV
T
T
TmV
VmT
P
m
T
T
m
P
VmP
PmV
Q. How many are independent functions?
In general, Z
xyy
x
1
TmV
P
T
m
P
V
=1/
Zx
1
YXZx
z
z
y
y
x
TmV
P
P
m
T
V
VmP
T
= -1
Any two of them having different colors are independent.
Cyclic Rule:
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Sample Question (HW4 P3.3)
• For ideal gas show the cyclic relationship i tis correct.
Use P =nRT/V, V =nRT/P, T=PV/nR
VPTVPT P
nRPV
T
PnRT
V
VnRT
P
T
T
V
V
P
///
?
21VnRT
V
VnRT
V
VnRT
TT
///
Some of the Partial Derivatives Have Special Meaning
PT
V
V
1 : Volumetric thermal expansion
coefficient
TP
V
V
1 : Isothermal Compressibility
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How to Use Them?
dTT
PdV
V
PTVdP
),(TV VT
TP
V
V
1
VV
P
T 1
/
TPV V
P
T
V
T
P
1dTdV
VTVdP
1),(
inifinini
fin
T
T
V
V
TTV
VdTdV
VP
fin
ini
fin
ini
)ln(
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Sample Question (p46 text)
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3.2 Dependence of U on V and T• U varies by changing V and T as
dTU
dVU
TVdU
)(
Using dU = q + w = Dq - PextdV
dTT
dVV
TVdUVT
),(
dTT
UdV
V
UdVPDq
VText
To be derived inSec. 5.3
VT When dV = 0 dT
T
UDq
VV
vV
V
CdT
Dq
T
U
P
T
PT
V
U
VT
w & q in various process for ideal gas Type of work
w q U T
Expansion for Pext = ext
const
isotherm -PextV -w 0 0adiabatic -PextV 0 -PextV U/CV
Reversible expansion/ compression
isotherm -nRT ln(Vfin/Vini) -w 0 0
adiabatic CvT 0 CvT Tini{(Vfin/Vini)a -1)}
a=1-CP/CV=1-
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HW 3
• P2.7 For 1.00 mol of an ideal gas, Pexternal
= P = 200.0 103 Pa. The temperature is changed from 100 0°C to 25 0°C andchanged from 100.0 C to 25.0 C, and CV,m = 3/2R. Calculate q, w, U, and H.
)( inifinVV TTCqU
VP CCnR VPUH
Note U < 0
)( inifinP TTCH
qwU
VP CCnR
Hqp (Reversible)
TnRU
VPUH ext
qUw
• P2.19) 3.50 moles of an ideal gas are expanded from 450. K and an initial pressure of 5.00 bar to a final pressure of 1.00 bar, and CP,m = 5/2R.Calculate w for the following two cases: a. The expansion is isothermal and reversible b Theexpansion is isothermal and reversible.b. The expansion is adiabatic and reversible.
• b. q = 0 w = U = CV(Tfin-Tini)
•)(
1
finfin VT VPVP
Cv = CP –nR So the question is how to get Tfin?
iniini VTfinfininiini VPVP
fin
ini
ini
fin
V
V
P
P/1
/)1()1(/1
ini
fin
ini
fin
ini
fin
P
P
P
P
T
T
(Xa)b =X(ab)Derive this
[Q1]
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• P2.26) One mole of an ideal gas, for which CV,m = 3/2R, initially at 298 K and 1.00 105 Pa undergoes a reversible adiabatic compression. At the end of the ad abat c co p ess o t t e e d o t eprocess, the pressure is 1.00 106 Pa. Calculate the final temperature of the gas. Calculate q, w, U, and H for this process.
Adi b ti • Adiabatic process
• q = 0 & U = w = Cv T
1
f
i
1
i
f
1
i
f
i
f
p
p
T
T
V
V
T
T
[Q1] [Q2] [Q3]
i
f
1
i
f
i
f
1
f
i
T
T
T
T
T
T
p
p
• P2.29) A cylindrical vessel with rigid adiabatic walls is separated into two parts by a frictionless adiabatic piston. Each part contains 50.0 L of an ideal monatomic gas with CV,m = 3/2R. Initially, Ti = 298 K and Pi = 1.00 bar in each part. Heat is slowly introduced into the left part using an electrical heater until the piston has moved sufficiently to the right to result in auntil the piston has moved sufficiently to the right to result in a final pressure Pf = 7.50 bar in the right part. Consider the compression of the gas in the right part to be a reversible process. (a) Calculate the work done on the right part in this process and the final temperature in the right part. (b) Calculate the final temperature in the left part and the amount of heat that flowed into this part.of heat that flowed into this part.
(a) Right part is subject to adiabatic reversible compression. qR =0 Obtain Tfin; wR = U = CVT
(b) Obtain the volume of the right part. VL = Vtotal – VR.
U = qL + wL; Note: wL = -wR
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Maxwell-Boltzmann Distribution of Speed P(C)
C = {vx2+vY
2+vZ2}1/22
22/3
2exp
24)( C
RT
MC
RT
MCP
O2 at 300K
12/1
24)(
e
RT
MCP mp
2
212
/
M
RTCmp
• HW3 Q3. (a) Calculate dP(C)/dC. (b) Show there two solutions for dP(C)/dC = 0 (C 0) and one solution gives the most probable speed, C = (2RT/M)1/2, where m is the weight of a gas molecule, M is its molar mass, and k is the Boltzmann constant.
RT
MCACC
RT
MC
RT
MCP
2exp
2exp
24)(
222
22/3
RT
MCC
dC
d
RT
MC
dC
dCA
dC
CdP
2exp}
2{exp
)( 22
22
dC
duu
du
d
RT
MC
dC
dexp
2exp
2
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P2.20) An ideal gas described by Ti = 300. K, Pi = 1.00 bar, and Vi = 10.0 L is heated at constant volume until P = 10.0 bar. It then undergoes a reversible isothermal expansion until P = 1.00 bar. It is then restored to its original state by the extraction of heat at constant pressure. Depict this closed-extraction of heat at constant pressure. Depict this closedcycle process in a P–V diagram. Calculate w for each step and for the total process. What values for w would you calculate if the cycle were traversed in the opposite direction?
(P, V) V = const T = const P = const (1 bar, 10 L) (10 bar, 10L) (1bar, 100L) (1 bar, 10L)
Which process involves work? How much work is involved? For the reverse process: Wreverse = -W for each step For the reverse cycle: Wreverse cycle = -Wcycle
• P2.43) One mole of N2 in a state defined by Ti = 300. K and Vi = 2.50 L undergoes an isothermal reversible expansion until Vf = 23.0 L. Calculate w assuming (b) that the gas is described by the van der Waals equation of state.der Waals equation of state.
• Isothermal process T = 0 but for a vdW gas, you cannot use U = CV T!
For a reversible process Pext = Pint
finfin VV
ext dVPdVPw intRT
P a
What is the percent error in using the ideal gas law instead of the van der Waals equation?
ViniVini
ext intbV
Pm
int 2mV
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• P2.23) A pellet of Zn of mass 10.0 g is d d i t fl k t i i dil tdropped into a flask containing dilute H2SO4 at a pressure of P = 1.00 bar and temperature of T = 298 K. What is the reaction that occurs? Calculate w for the process.
• The chemical equation is given by
(g) H (aq) SO (aq) Zn (aq) SOH (s)Zn 2-2
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42
W = -Pext ∆VH2
