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Showing posts with label THERMODYNAMICS:. Show all posts
Showing posts with label THERMODYNAMICS:. Show all posts
Thursday, October 1, 2020
A sample of gas is compressed by an average pressure of 0.50 atmosphere so as to decrease its volume from 400cm^3 to 200cm^3. During the process 8.00 J of heat flows out to surroundings. Calculate the change in internal energy of the system.
Sunday, October 21, 2018
SPONTANEOUS AND NON SPONTANEOUS PROCESS
SPONTANEOUS PROCESS: Those process which have natural tendency to take place ,they may or may not require initiation.
For example
(1) Flow of water from higher level to lower level .
(2) Flow of heat from high temperature to lower temperature.
(3) Radioactivity
(4) Cooling of cup of tea .
(5) Evaporation
(6) Condensation
(7) Sublimation
(8) Burning of candle
(9) Dissolution of salt and sugar in water
(10) Burning of fuel
(11) melting of ice at room temperature.
NON SPONTANEOUS PROCESS: All spontaneous processes are non spontaneous processes in reverse direction it requires spot of external energy for their progress.
CRITERIA FOR SPONTANEITY:
(1) If dS universe is greater than Zero then process is spontaneous (reversible)
(2) If dS is Zero then process is in reversible state of equilibrium
(3) If dS is lower than Zero then process is non spontaneous
For a spontaneous process entropy of univers is greater than Zero . In order to use entropy has a sole criteria , we need to have information about system as well as surrounding.
dS(system) + dS(surr) is greater than Zero
.
.
.
.
In order to explain the spontaneous behaviour by using the parameters of system we can established two criteria.
(1) Randomness
(2) Criteria of energy
A spontaneous process is one in which energy ( enthalpy) decreasing and Randomness of system increasing.
For spontaneous proceDH=-ve and dS=+ve
For example
(1) Flow of water from higher level to lower level .
(2) Flow of heat from high temperature to lower temperature.
(3) Radioactivity
(4) Cooling of cup of tea .
(5) Evaporation
(6) Condensation
(7) Sublimation
(8) Burning of candle
(9) Dissolution of salt and sugar in water
(10) Burning of fuel
(11) melting of ice at room temperature.
NON SPONTANEOUS PROCESS: All spontaneous processes are non spontaneous processes in reverse direction it requires spot of external energy for their progress.
CRITERIA FOR SPONTANEITY:
(1) If dS universe is greater than Zero then process is spontaneous (reversible)
(2) If dS is Zero then process is in reversible state of equilibrium
(3) If dS is lower than Zero then process is non spontaneous
For a spontaneous process entropy of univers is greater than Zero . In order to use entropy has a sole criteria , we need to have information about system as well as surrounding.
dS(system) + dS(surr) is greater than Zero
.
.
.
.
In order to explain the spontaneous behaviour by using the parameters of system we can established two criteria.
(1) Randomness
(2) Criteria of energy
A spontaneous process is one in which energy ( enthalpy) decreasing and Randomness of system increasing.
For spontaneous proceDH=-ve and dS=+ve
Saturday, September 29, 2018
RELATIONSHIP BETWEEN DH and DU
The difference between dH and dU becomes significant only when gases are involved (insignificant in solids and liquids)
DH = DU + D(PV)
If substance is not undergoing chemical reaction or phase change.
we know PV=nRT
PDV=DnRT
hence
DH = DE + DnRT
In case of chemical reaction
DH = DE + DngRT
Where Dng=number moles of product in gaseous state - number moles of reactant in gaseous state
Dng =(nP-nR)g
case-(1) If Dng=0 then DH = DE
case-(2) If Dng>0 then DH > DE
case-(3) If Dng<0 then DH < DE
SOLUTION: DH = DU + D(PV)
D(PV) = P2V2 – P1V1
= 4 × 30 – 2 × 40
= 40 liter -bar = 4 kJ
so DH = 35 + 4 = 24 kJ
EXAMPLE (2).: What is the relation between DH and DE in this reaction?
CH4(g) + 2O2(g) ---------> CO2(g) + 2H2O(l)
SOLUTION: DH = DE + DnRT
Dn = no. of mole of products - no. of moles of reactants = 1– 3 = –2
DH = DE – 2RT
EXAMPLE(3):
Consider the chemical
reaction at 300 K H2 (g)+Cl2
à2HCl(g) ΔH= -185KJ/mole calculate ΔU if
3 mole of H2 completely react
with 3 mole of Cl2(g) to form HCl.
SOLUTION: H2
(g)+Cl2 à2HCl(g)
ΔH= -185KJ/mole
Δng=0
ΔH= ΔU+
ΔngRT
ΔH= ΔU
ΔHR= -185 KJ/mole ,ΔUR=
-185 KJ/mole
H2 (g)+Cl2
à2HCl(g) ΔH= -185KJ/mole
3 mole 3 mole
Hence ΔU= -185 X 3
KJ/Mole
EXAMPLE (4): The heat of combustion of naphthalene
(C10H8(s)) at constant volume was measured to be . 5133 kJ mol.1 at 298K.
Calculate the value of enthalpy change (Given R = 8.314 JK.1 mol.1).
SOLUTION: The combustion reaction of naphthalene.
C10H8(s) + 12O2(g) à10CO2(g) + 4H2O(l)
ΔE = -5133kJ
Δn = 10 -12 = -2 mol.
Now applying the relation.
ΔH = ΔE + (Δn) RT
= -5133 × 103 + (-2) (8. 314) (298)
= -5133000J - 4955.14J
= -5137955. 14 Joule
EXAMPLE(5) What is the true regarding complete
combustion of gaseous isobutene –
(A) ΔH = ΔE (B) ΔH > ΔE (C) ΔH = ΔE = O (D) ΔH < ΔE
SOLUTION: (D) C4H10(g) + 6.5O2
(g) à4CO2(g)
+ 5H2O(l)
Δn = [4 -7.5] = -3.5
ΔH = ΔE + ΔngRT
Δ H < ΔE
EXAMPLE (6): For a gaseous reaction: 2A2 (g) +
5B2(g) à2A2B5(g)
at 27ºC the heat change at constant pressure is found to be .50160J. Calculate
the value of internal energy change (ΔE). Given that R = 8.314 J/Kmol.
(A) -34689 J (B) -37689 J (C) -27689 J (D) -38689 J
SOLUTION : 2A2(g) + 5B2(g) à 2A2B5 (g); ΔH= -50160 J
Δ n = 2-(5 + 2) = -5 mol.
ΔH = ΔE + (Δn) RT
-50160 = ΔE + (Δn) RT
Δ E = -50160- (-5) (8.314) (300)
= -50160 + 12471 = -37689 J
The answer is (B)
Friday, September 28, 2018
ENTHALPY (H) INTRODUCTION
We known 1st
law of thermodynamics
dU=dq+dW
If Pext=constant
Then dW= -PdV
and dU=dq- PdV
dq=dU-PdV
dq=(U2-U1)+P(V2-V1)
dq=(U2+PV2)-(U1+PV1)
dQ=H2-H1 ( H2=U2+PV2 and H1=U1+PV1)
dq=dH
hence
The enthalpy of a system is defined as:
H
= U + PV
DH = DU + D(PV)
Where
H is the enthalpy of the system
U is the internal energy of the system
P is the
pressure at the boundary of the system and its environment.
(1) In thermodynamics the quantity U + PV is a new state function and known
as the enthalpy of the system and is denoted by H=U+PV. It represents the total energy stored in the system.
(2) It may be noted that change in
enthalpy is equal to heat exchange at constant pressure.(3) Enthalpy
is also an extensive property as
well as a state function. (4)The absolute value of enthalpy cannot
be determined, however the change in enthalpy can be experimentally determined.
DH = DU + D(PV)
(5) Change in enthalpy is a more useful quantity than its
absolute value.
(6) The unit of measurement for enthalpy (SI) is joule.
(7)The enthalpy is the preferred expression of system energy
changes in many chemical and physical measurements, because it simplifies
certain descriptions of energy transfer. This is because a change in enthalpy
takes account of energy transferred to the environment through the expansion of
the system under study.
(8)The change dH
is positive in endothermic reactions,
and negative in exothermic
processes. dH of
a system is equal to the sum of non-mechanical work done on it and the heat
supplied to it.
(9) For quasi static processes under constant pressure, dH is equal to the change in the
internal energy of the system, plus the work that the system has done on its
surroundings. This means that the change in enthalpy under such conditions is
the heat absorbed (or released) by a chemical reaction.
NOTE:
Transfer of heat at constant volume brings about a change
in the internal energy(DU) of the system whereas that at constant pressure
brings about a change in the
enthalpy (DH) of the system.
For Ideal
gas
H=U+PV and U=f(T)
PV=nRT
H=U+nRT and H=f(T) only for ideal gas
For other
substance and real gas
H=U+PV
U=f(P,V,T)
and H=f(,PV,T)
So H=f(P,T)/ f(V,T)/ f(P,V)
H=f(T,P)
dH=(dH/dT)p dT+(dH/dP)T
dP------------------------------------- (1)
H=f(V,T)
dH=(dH/dV)T dV+(dH/dT)V
dT------------------------------------- (2)
H=f(P,V)
dH=(dH/dV)P dV+(dH/dP)V
dP------------------------------------- (3)
Out of the
above three relation H as function
on of (T,P) Has a greater significance. The above differential equation simplified
for different substance for different condition.
For
isobaric process : dP = 0 (Molar Heat capacity at constant Pressure)
We known
QP=nCpmdT
(Molar Heat capacity at constant Pressure)
Cpm= (dQ/dT)P for 1 mole of gas
dQ=dH at dP=0
then Cpm= (dH/dT)P
For an ideal gas: change in
enthalpy at constant temperature with change in
pressure is zero. i.e.
Continue...........
THIRD LAW OF THERMODYNAMICS (TLOT)
We know that on increasing the entropy of a pure crystalline
substance increases because molecular motion increases with increase of
temperature and decreases on decreasing temperature. Or we can say that the entropy of a perfectly crystalline
solid approaches zero as the absolute zero of temperature is approached. Which
means that at absolute zero every crystalline solid is in a state of perfect
order and its entropy should be zero. This
is third law of thermodynamics.
We can calculate absolute value of entropy of any
substance in any state (solid ,liquid, gas) at any temperature by calculating
dS for the processes in going from the initial state to the state of the
substance for which entropy is to be calculated.
EXAMPLE: Find the entropy of change when 90 g
of H2O at 10C was converted into steam at 100C.(
Given Cp(H2O)=75.29 JK-1 and dHVap=43.932 k JK-1 mol-1)
SOLUTION:
EXAMPLE: Calculate dG for
(i) H2O
(l, 1 atm, 300 K ) ------à H2O (g, 1
atm, 300 K)
(ii) H2O
(l, 2 atm, 373 K ) -----à H2O
(g, 2 atm, 373 K)
Given: dH373 = 40 kJ ; CP(H2O,l)= 75 J / mol /K ; CP(H2O,g)
= 35 J / mol / K
SOLUTION: (1)
(2)
WORK DONE (PV-WORK ANALYSIS )
Energy
that is transmitted from one system to another in such a way that difference of
temperature is not directly involved. This definition is consistent with our
understanding of work as dw= Fdx. The force (F) can arise from electrical, magnetic,
gravitational & other sources. It is a path function.
(1) Work done in Isothermal Irreversible Process:
(2) Work done in Isothermal Reversible Process:
(3) Work done in Adiabatic Irreversible Process:
(4) Work done in Adiabatic Reversible Process:
PV-Work analysis: Consider a cylinder fitted with a friction less piston, which enclosed no more of an
ideal gas. Let an external force F pushes the piston inside producing displacement in piston.
Let
distance of piston from a fixed point is x and distance of bottom of piston at
the same fixed point is
l. This
means the volume of cylinder = (l – x) A where A is area of cross section of
piston.
For a small displacement dx due to force F, work done on
the system.
dw = F.dx
Also F = PA
dW = PA.dx
V = (l – x)A
dV
= –A . dx
dW = –Pext. dV
Note :
(1): Litre atmosphere term is unit of energy. It is useful to remember the conversion:
1 litre atm= 101.3 Joules = 24.206 Cal.
(2): During expansion dV is positive and
hence sign of w is negative since work is done by the system and negative sign representing
decease in energy content of system. During compression, the sign of dV is negative
which gives positive value of w
representing the increase in energy content of system during compression.
EXAMPLE.1 mole of ideal monatomic gas at 27°C expands adiabatically against a constant external
pressure of 1.5 atm from a volume of 4dm3 to 16 dm3. Calculate
(i) q (ii) w and (iii) DU
SOLUTION: (i)
Since process is adiabatic \ q = 0
(ii) As
the gas expands against the constant external pressure.
W = - PVd=-P(V2-V1)
W =-1.5(16-4)
W= - 18 dm3
(iii) DU = q + w = 0 + (-18) = -18 atm dm3
(1) Work done in Isothermal Irreversible Process:
(2) Work done in Isothermal Reversible Process:
(3) Work done in Adiabatic Irreversible Process:
(4) Work done in Adiabatic Reversible Process:
Wednesday, September 26, 2018
SECOND LAW OF THERMODYNAMICS (SLOT)
(1) FLOT is law of conservation of energy, and according to it all chemical and physical processes take place in a such way that energy remain constant
(2) In FLOT we introduce enthalpy and internal energy.
(3)When two body at different temperatures are kept closed toeach other there will be transfer of heat while the transfer of heat take place called not explain by FLOT.
(4) we can not explain spontaneous and irreversible behaviour of processes by using FLOT.
In 2nd law of thermodynamics we will introduce entropy as criteria of spontaneity.
STATEMENT OF 2ND LAW :. "It is impossible to construct the heat engine which can take heat from a source and completely convert into work without creating any disturbance in the surrounding.
FIRST LAW OF THERMODYNAMICS (FLOT)
(1) It is
also known as law of energy conservation.
(2)The
first law of thermodynamics states that energy can neither be created nor
destroyed, although it can be
transformed from one form to another.
(3)
Another Statement is “total energy of universe is remain constant”
(4) In FLOT we introduce the term ENTHALY and
INTERNAL ENERGY.
MATHEMATICAL EXPRESSION OF FIRST LAW
Let UA be the energy of a system in its state
A and UB be the energy in its state B. Suppose the system while
undergoing change from state A to state B absorbs heat q from the surroundings
and also performs some work (mechanical
or electrical), equal to w. The absorption of heat by the system tends to raise
the energy of the system. The performance of work by the system, on the other
hand, tends to lower the energy of the system because performance of work
requires expenditure of energy. Hence the change of internal energy, DU, accompanying the above process will
be given by
dU =UB -UA = q + w
In
general, if in a given process the quantity of heat transferred from the
surrounding to the system is q and work done in the process is w, then the
change in internal energy,
dU = q + w
This
is the mathematical statement of the first law of thermodynamics.
ACCORDINGTO IUPAC:
(1)
If work is done by the surroundings on
the system (compression of a gas), w is taken as positive so that dU = q + w.
(2)
If however work is done by the system
on the surroundings ( expansion of a gas), w is taken as negative so that dU = q – w.
(3) q+w it will be independent of the way the change is carried out, it only depend on initial and final state.
(4) If their is no transfer of energy as heat or work (Isolated System)
ie, if w=0 and q=0. then dU=0
EXAMPLE: A system gives out 25 J of heat and also does 35 J of work. What is the internal energy change ?
SOLUTION: According to FLOT dU = q + w
dU =-25 J+(-35 J)
dU = -60 J
Note:
(1) dU = q + w is invalid for open system.
(2) 1st
law of T.D. is applicable for closed system in which system is at rest or
moving with constant velocity and in
absence of external fields.
(3) The
macroscopic energy changes with velocity and elevation of the system are not
considered in internal energy change of system.
EXAMPLE: The pressure of a fluid is a linear function of volume P=a+bV and internal energy of fluid is U=34+3PV in SI unit. find a ,b, W ,dU for change in state (100 Pascal ,3m cube ) to (400 Pascal ,6m cube). (given Pascal =1J) .
SOLUTION: We know by FLOT DE=q+W
LIMITATIONS
OF FIRST LAW THERMODYNAMICS:
A major limitation of the first law of thermodynamics is
that it’s merely indicates that in any process there is an exact equivalence
between the various forms of energies involved, but it provides no information
concerning the spontaneity or feasibility of the process.
For example, the first law does not indicate whether heat
can flow from a cold end to a hot end or not.
The
answers to the above questions are provided by the second law of
thermodynamics.
PARAMETER INVOLVE IN FIRST LAW THERMODYNAMICS
Continue reading.........
(2) Work done (W)
(3) Enthalpy (H)
Tuesday, September 25, 2018
HEAT, HEAT EXCHANGE AND HEAT CAPACITY(Q)
Heat is defined as the energy
that flow into or out of the system because of a difference in temperature between
system and surrounding.
HEAT CAPACITY(Q)
Total degree of freedom = 3N (Where N is Number of atom in molecule)
Total degree of freedom:= ftrans + f rot + f vib and fvib= 3N- ftrans + f rot
Note; Work is more organised way
of energy transfer as compared to hear exchange.
IUPAC Sign convention of Heat: sign of heat will negative (-Ve) if
heat is released by the system given by system while sign of heat will be
positive (+Ve) if heat is given to the system.
(i) qV = nCvdT (for constant volume
process)
(ii) qp = nCpdT (for constant
pressure process)
(iii) Cp,m – Cv,m = R
(iv) Cv & Cp depends on
temperature even for an ideal gas.( C = a + bT + cT2 .....)
(v) It is a path function
Cv, Cp are heat capacity of system
and Cv,m, Cp,m are heat capacity of one mole system at
constant volume and pressure
respectively.
“Exchange of heat and work(P–V) in
between system and surrounding always
occur through boundry of system”
Note: heat exchange can be measured with the help of Heat
Capacity.
We know that usually on increasing in
temperature is proportional to the heat transfer
q=coefficient x ∆T
The coefficient depend on the size,
composition and nature of the system so we can also write it as
q=C∆T
where C is called Heat capacity
The heat capacity (C):
q=C∆T or
q=C∆T or
C= q/∆T unit- J/K
If ΔT=10=1K
Then C=q
It is equal to amount of heat needed
to raise the temperature of the sample of any substance by one degree Celsius
(or Kelvin).
heat
Capacity depend on quantity, nature as
well as physical state of the system. And the heat capacity is extensive. it may be made intensive as specific heat capacity
Specific Heat Capacity(Cs ):
q=Csm∆T or
q=Csm∆T or
Cs= q/m∆T
If ΔT=10=1K and m = 1g
Then
Cs=q
It is equal to amount of heat required to raise the temperature of 1gm
substance by one degree centigrade .it is intensive properties.
Molar Heat Capacity(Cm ):
q=Cmn∆T or
q=Cmn∆T or
Cm= q/n∆T
If ΔT=10=1K and n=1mole
Then Cm=q
It is equal
to amount of heat required to raise the temperature of 1 Mole substance by one
degree centigrade
EXAMPLE(1): The latent heat of fusion of ice at
0ºCis 80 cal/gm the amount of heat needed to convert 200 gm ice into water at
0ºC is ?
(A) 80 cal
(B) 16000 J (C) 16000 cal (D) 1600 cal
SOLUTION:
Ans (C) q = m.L = 200 × 80 = 16000 cal
EXAMPLE(2): Calculate the amount of heat required
to raise the temperature of 50 gm water from 25ºC to 55ºC.Specific heat
capacity of water = 4.2 J/ºC-gm.
(A) 126 J (B)
210 J (C) 6300 J (D) 1500 J
SOLUTION: Ans. (C)
q = m.s.ΔT = 50 × 4.2 × (55 – 25) = 6300 J
EXAMPLE (3): Five moles of a monatomic ideal gas is
heated from 300K to 400K at constant pressure. the amount of heat absorbed is :
(A) 500 cal (B) 1500 cal (C) 2500 cal (D) 2500 J
SOLUTION: Ans. (C)
qp = CpΔT = n.Cp,m ΔT
= 5 × 5/2× (400
– 300) = 2500 cal
EXAMPLE (4): 2
moles of an ideal gas absorbs 720 cal heat when heated from 27ºC to 87ºC, at
constant volume. 'ɤ' for the gas is :
(A) 1.5 (B) 1.4 (C) 1.6 (D) 1.33
SOLUTION :Ans. (D)
qv
= n.Cv,m. ΔT
Cv,m =qv/nΔT
=
720/2x(87 – 27)
= 6 cal/K-mol
Now, r = 1 +R/Cv,m
= 1 + 2/6
= 1.33
EXAMPLE (5): 500 gm ice at 0ºC is added in 2000 gm water
at tºC. If the final temperature of system is 0ºC, then the value of 't' is
(latent heat of fusion of ice = 80 cal/gm and specific heat capacity of water=cal/gm-ºC)
(A) 20 (B) 40
(C) 10 (D) 2
SOLUTION: Ans. (A)
Heat lost by water = heat gained by ice
or, (m.s. ΔT)water = (m.L)ice
or, 2000 × 1 × (t – 0) = 500 × 800
Δ t = 20ºC
EXAMPLE(6): What is the heat in Joules required to raise the temperature of 25 grams of water from 0 °C to 100 °C? What is the heat in calories?
(Given: specific heat of water = 4.18 J/g·°C)
SOLUTION: Use the formula q = mcΔT
where
q = heat energy
m = mass
c = specific heat
ΔT = change in temperature
q = 25gx4.18 J/g·°x(100 °C - 0 °C)
q = 25gx4.18 J/g·°Cx(100 °C)
q = 10450 J
where
q = heat energy
m = mass
c = specific heat
ΔT = change in temperature
q = 25gx4.18 J/g·°x(100 °C - 0 °C)
q = 25gx4.18 J/g·°Cx(100 °C)
q = 10450 J
We know 1 Calorie=4.18 J
So 10450 J in Calorie = 10450/4.8=2500 calorie
FOR LARGE HEAT CHANGE :
Q= nCm (T2-T1)
Case- (2) Cm = f(T)
Cm = a + bT+
cT2 +……..
Case- (3) The theoretical value of Cvm
and Cpm for Ideal gas can determined by using degree of freedom.
CHARACTERISTIC OF HEAT CAPACITY:
(1): The heat
capacity Of any system should depend upon temperature because by increasing temperature
of system different degree of freedom get excited.
(2): When
temperature approaches zero then heat absorbed by the solid mainly converted into vibration potential energy of molecule
resulting in very small increase in temperature, hence ‘C’ increases sharply
with increase in temperature.
Normally C Directly proportional T3
(3):
When the temperature at Melting point of solid ,then heat capacity becomes
nearly constant for solid elements.
Molar
heat capacity= 6.4 Cal/K mole
Or specific heat capacity x atomic
weight=6.4 (Dulong and petite’s law)
(4): Exactly
at melting point, the heat capacity become infinite as ΔT=0
(5):
the heat capacity of liquid is greater than that of solid because of rotational
degree of freedom also excited.
(6):
In liquid heat capacity also depend
upon temperature and also infinite at boiling point.
(7):
the heat capacity of gas become less than liquid because all vibrational and
rotational degree of freedom converted into translational degree of freedom.
(8):
the heat capacity of gases depend upon
their atomicity.
(9):
If the heat capacity depends upon temperature.
(10);
As heat (q) is path function, any substance may have infinite heat capacity.
Example
for any substance.
Isothermal process = infinite
Adiabatic
process = 0
Isobaric
process = Cp
Isochoric process = Cv
Normally ,we use Cp and Cv value as characteristic of substance.
DEDREE OF FREEDOM :
It is equal to number of modes of energy transfer when a gaseous molecule undergoes collision. OR
It represent the number of independent modes to describe the molecular motion.Total degree of freedom = 3N (Where N is Number of atom in molecule)
1-
Translational degree of freedom is 3 (three)
always for mono,di and tri atomic molecule.
2-
Rotational degree of freedom is zero
for mono atomic
gas,2 (two) for diatomic molecules and 3 (three)
for triatomic
molecule
3-Vibrational
degree of freedom is also zero for mono atomic gas and 1(one) diatomic gas molecule
and for polyatomic gases VDOF is
calculated individually.( fvib= 3N- ftrans+ frot)
Total degree of freedom:= ftrans + f rot + f vib and fvib= 3N- ftrans + f rot
Molecules N TDF
He 1 3
O2 2 6
CO2 3 9
NH3 4 12
PCl5 6
18
Case-1
Monoatomic Diatomic Triatomic
(linear) Triatomic (Non linear)
f total
=3 ftotal =6 ftotal =9 f total=9
f trans=3 ftrans =3 ftrans=3 ftrans=3
f
rot =0 frot =2
frot = 2 frot =3
f vib =0 fvib =1 fvib = 4 fvib =3
Q =n CmdT
QV=n CvmdT
Cvm=(dQ/dT)v
By FLOT dq=dU+dW and at constant volume
dW=0 so dQv=dU
Hence Cvm= (dU/dT)v
LAW OF EQUIPARTIAL OF ENERGY :
Average energy associate with each molecule per degree of freedom is U= 1/2KT (where K is Boltz’s man constant.
Average energy associate with each molecule per degree of freedom is U= 1/2KT (where K is Boltz’s man constant.
Let degree of freedom is =
f then U is U=1/2fkT
And U=1/2fkTNA per molecule we know kNA=R
U=1/2fRT
and dU/dT=1/2fR
And dU/dT=Cv hence
Cv=1/2fR
Cv=1/2ftransR +1/2frotR (Where
Vib degree inactive in chemistry)
For ideal gas Cpm-Cvm=R
and Gama= Cpm/Cvm
Adiabatic exponent :Adiabatic exponent (Gama) for a mixture
of gas with different heat capacity is defined as :
where n1, n2 ........................ are moles of
different gases
Example:Calculate change in internal energy of
10 gm of H2 ,when it's state is changed from(300K, 1Atm) to (500 K,
2Atm) ?
Solution: For ideal gas
Cv for H2 (diatomic) in low temperature
range will be 5R as vibrational part is not included.
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