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**THERMODYNAMICS:**. Show all posts## Thursday, October 1, 2020

## 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

## 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)

**D**

**H =**

**D**

**U +**

**D**

**(PV)**

If substance is not undergoing chemical reaction or phase change.

we know PV=nRT

PDV=DnRT

hence

**DH = DE + D**

**nRT**

In case of chemical reaction

**D**

**H =**

**D**

**E +**

**D**

**n**

_{g}RT
Where Dn

_{g}=number moles of product in gaseous state - number moles of reactant in gaseous state
Dn

_{g}=(n_{P}-n_{R})_{g }
case-(1) If

**D****n**_{g=0 then }**D****H =****D****E**
case-(2) If

**D****n**_{g>0 then }**D****H >****D****E**
case-(3) If

**D****n**_{g<0 then }**D****H <****D****E**

_{}

**EXAMPLE (1):**1 mole of a real gas is subjected to a process from (2 bar, 40 lit.,300K) to (4 bar, 30 lit., 400 K). If change in internal energy is 35 kJ then calculate enthalpy change for the process.

**SOLUTION:**

**DH = DU + D**

**(PV)**

D(PV) = P

_{2}V_{2}– P_{1}V_{1}_{ }= 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?

CH

_{4}(g) + 2O_{2}(g) ---------> CO_{2}(g) + 2H_{2}O(l)**SOLUTION:**

*DH = DE + D***nRT**

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 H

_{2 }(g)+Cl

_{2}à

_{2}HCl(g) ΔH= -185KJ/mole calculate ΔU if 3 mole of H

_{2 }completely react with 3 mole of Cl

_{2}(g) to form HCl.

**SOLUTION:**H

_{2 }(g)+Cl

_{2}à

_{2}HCl(g) ΔH= -185KJ/mole

**Δng=0**

**ΔH= ΔU+ ΔngRT**

ΔH= ΔU

ΔH

_{R}= -185 KJ/mole ,ΔU_{R}= -185 KJ/mole
H

_{2 }(g)+Cl_{2}à_{2}HCl(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)}+ 12O

_{2(g)}à10CO

_{2(g)}+ 4H

_{2}O

_{(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)**C

_{4}H

_{10(g)}+ 6.5O

_{2}

_{(g)}à4CO

_{2}(g) + 5H

_{2}O

_{(l)}

Δn = [4 -7.5] = -3.5

**ΔH = ΔE + ΔngRT**

Δ H < ΔE

**EXAMPLE (6):**For a gaseous reaction:

**2A**

_{2}(g) + 5B_{2}(g)**à**

**2A**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.

_{2}B_{5}(g)
(A) -34689 J (B) -37689 J (C) -27689 J (D) -38689 J

**SOLUTION :**

**2A**

_{2}(g) + 5B_{2}(g)**à**

**2A**ΔH= -50160 J

_{2}B_{5}(g);
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 1

(Molar Heat capacity at constant Pressure)

^{st }law of thermodynamics
dU=dq+dW

If P

_{ext}=constant
Then dW= -PdV

and dU=dq- PdV

dq=dU-PdV

dq=(U

_{2}-U_{1})+P(V_{2}-V_{1})
dq=(U2+PV2)-(U1+PV1)

dQ=H2-H1 ( H

_{2}=U_{2}+PV_{2}and H_{1}=U_{1}+PV_{1})**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

**d****H 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,

**d****H 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

We known

Q

_{P}=nC_{pm}dT (Molar Heat capacity at constant Pressure)
C

dQ=dH at dP=0_{pm}= (dQ/dT)_{P }for 1 mole of gas
then

**C**_{pm}= (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 H

_{2}O at 10

^{C }was converted into steam at 100

^{C}.( Given Cp(H

_{2}O)=75.29 JK

^{-1}and dH

_{Vap}=43.932 k JK

^{-1}mol

^{-1})

**SOLUTION:**

###
EXAMPLE: Calculate dG for
(i) H_{2}O
(*l*, 1 atm, 300 K ) ------à H_{2}O (g, 1
atm, 300 K)
(ii) H_{2}O
(*l*, 2 atm, 373 K ) -----à H_{2}O
(g, 2 atm, 373 K)
Given: dH_{373} = 40 kJ ; C_{P}(H_{2}O,*l*)= 75 J / mol /K ; C_{P}(H_{2}O,g)
= 35 J / mol / K

_{2}O (

*l*, 1 atm, 300 K ) ------à H

_{2}O (g, 1 atm, 300 K)

_{2}O (

*l*, 2 atm, 373 K ) -----à H

_{2}O (g, 2 atm, 373 K)

_{373}= 40 kJ ; C

_{P}(H

_{2}O,

*l*)= 75 J / mol /K ; C

_{P}(H

_{2}O,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.

**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 4dm

^{3}to 16 dm

^{3}. 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 dm

^{3}^{ }(iii) DU = q + w = 0 + (-18) = -18 atm dm

^{3}

__(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 U

_{A}be the energy of a system in its state A and U_{B}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**d**

**U =U**

_{B -}U_{A =}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) d

**U = 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.

It is equal to amount of heat required to raise the temperature of 1gm
substance by one degree centigrade .it is intensive properties.

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

###

###

Average energy associate with each molecule per degree of freedom is U= 1/2KT (where K is Boltz’s man constant.

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) q

_{V}= nC_{v}dT (for constant volume process)
(ii) q

_{p}= nC_{p}dT (for constant pressure process)
(iii) C

_{p,m}– C_{v,m }= R
(iv) C

_{v}& C_{p}depends on temperature even for an ideal gas.( C = a + bT + cT^{2}.....)
(v) It is a path function

C

_{v}, C_{p}are heat capacity of system and C_{v,m,}C_{p,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.

**HEAT CAPACITY(Q)**

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**

**C= q/∆T unit- J/K**

**If ΔT=1**

^{0}=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(C**

_{s }):

**q=C**

_{s}m∆T or**C**

_{s}= q/m∆T

**If Δ**

**T=1**

^{0}=1K and m = 1g**Then C**

_{s}=q**Molar Heat Capacity(C**

_{m }):

**q=C**

_{m}n∆T or**C**

_{m}= q/n∆T

**If Δ**

**T=1**

^{0}=1K and n=1mole**Then C**

_{m}=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)

q

_{p}= C_{p}ΔT = n.C_{p,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**

C

_{v,m}=qv/nΔT
=
720/2x(87 – 27)

= 6 cal/K-mol

Now, r = 1 +R/C

_{v,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

We know 1 Calorie=4.18 J

So 10450 J in Calorie = 10450/4.8=2500 calorie

**FOR LARGE HEAT CHANGE :**

Q= nC

_{m}(T_{2}-T_{1})**Case- (2)**C

_{m }= f(T)

Cm = a + bT+
cT

^{2 +……..}**Case- (3)**The theoretical value of C

_{vm}and C

_{pm }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 T**

^{3}**(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 = C

_{p}
Isochoric process = C

_{v}
Normally ,we use C

_{p}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

**.(****f**_{vib}**=**_{ }**3N- f**_{trans}**+ f**_{rot}**)****Total degree of freedom:= f**

_{trans}+ f_{rot}+ f_{vib and }f_{vib= }3N- f_{trans}+ f_{rot}**Molecules**

**N**

**TDF**

He 1 3

O

_{2 }2 6
CO

_{2}3 9
NH

_{3 }4 12
PCl

_{5 }6 18**Case-1**

Monoatomic Diatomic Triatomic
(linear) Triatomic (Non linear)

f

_{total}=3 f_{total }=6 f_{total }=9 f_{total=}9_{}
f

_{ trans}=3_{ }f_{trans }=3_{ }f_{trans}=3 f_{trans}=3
f

_{ rot }=0_{ }f_{rot }=2_{ }f_{rot }= 2 f_{rot }=3
f

_{ vib }=0_{ }f_{vib }=1_{ }f_{vib }= 4_{ }f_{vib }=3_{ }

**Q =n C**

_{m}dT**Q**

_{V}=n C_{vm}dT

**C**

_{vm}=(dQ/dT)_{v}

**By FLOT dq=dU+dW and a**

**t constant volume dW=0 so dQv=dU**

**Hence**

**C**

_{vm}= (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.

Let degree of freedom is =
f then U is U=1/2fkT

And U=1/2fkTN

_{A }per molecule we know kN_{A=}R
U=1/2fRT
and dU/dT=1/2fR

And dU/dT=C

_{v }hence C_{v=}1/2fR**Cv=1/2f**

_{trans}R_{ }+1/2f_{rot}R (Where Vib degree inactive in chemistry)
For ideal gas C

_{pm}-C_{vm}=R and Gama= C_{pm}/C_{vm}
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 H

_{2},when it's state is changed from(300K, 1Atm) to (500 K, 2Atm) ?

**Solution:**For ideal gas

C

_{v}for H_{2}(diatomic) in low temperature range will be 5R as vibrational part is not included.
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