Thermodynamics, Chapter Notes, Class 11, Chemistry (IIT-JEE & AIPMT) PDF Download

Thermodynamics

Definition:

Thermodynamics deals with energy interaction b/w two bodies & its effect on the properties of matter.

Scope of thermodynamics:

→ Feasibility of a process

→ Extent of a process

→ Efficiency of a process

System Boundary Surrounding = Universe

System : The part of the universe under thermodynamical observation is called system.

surroundings : All the part of the universe excepting system is called surroundings.

Boundary : The part which separates system and surroundings is called boundary it may be rigid or flexible.

It may be diathermic (Heat can be exchanged) or adiabatic

Types of Systems

(Mass and energy both (Only energy can be (Neither mass nor energy

can be transfered) transfered) can be transfered)

State or condition of a system is described by certain measurable properties & these measurable properties are called state variables. e.g. mass, temperature, volume, pressure etc.

State function depends only on initial & final state of the system. If does not depend on the path or how process was carried out.

e.g. DU =  Where DU = u_f - u_i

= DT = T_f - T_i

Sol. The parameters which are required to completely define the state of the system are called state functions.

State functions are path independent function.

Path independent means the difference in state functions will be same for any path followed between two states.

Physical Properties

Path function depends on the initial as well as final state of a system & also depends on the path of the process.e.g. heat and work.

condition for a function to be state function

Euler's theorem f = f(T, V). If f is a state function then

= 

PV = nRT  ⇒  P = 

if pressure is a state function then

=     

 =   

Hence P is a state function.
 

e.g. f(x,y) = ye^x +  xy + xln y

Show that f(x,y) is a state function.

Sol.  = ye^x  y ln y  = e^x  x x. 

 =  = e^x  1  

 = 

= e^x  + 1 +  

Hence f (x, y) is a state function.

Two other important result from differential calculus will be used frequently.

Consider a function. z = f(x, y). which can be rearranged x = g(y, z) or y = h(x, z)

For example, PV = nRT, P = , T = 

in this case

The cyclic rule will also be used.

 = - 1

(5) Cyclic Process: System undergoes series of changes & ultimately comes back to initial state.

(1) Quasi-Static(Reversible)Process: If system & surrounding can restore their original state by reversing the direction of the process then process is called reversible process. In reversible process, there is no loss of energy.

These are slow process as it takes infinite time. System & surrounding are always in equilibrium. Reversible process is a theoretical process. Reversible process is most efficient with respect to work.

In reversible process P_ext = P_int

If all the above criteria are not fulfilled by any process, then it is known as irreversible process.

* irreversible process is a fast process. It takes definite time for completion

* In irreversible process P_ext  is not equal to P_int.

*It is an actual process. It is carried out in multiple stages and it tends towards reversible process.

Heat & work both are forms of energy. Both are boundary phenomena and take place at the interface of the system & surroundings.

Expansion  W  = - ve

Compression → W =+ ve

Heat given to the system +ve

Heat loss (it released) = - ve

Types of equilibrium

(1) Thermal equilibrium  Equality of temperature

(2) Mechanical equilibrium Equality of pressure

(3) Material equilibrium → no. of moles constant

When all the three equilibrium are established in a system, system is in true thermodynamics equilibrium

Internal energy : Internal energy of gaseous molecule present in a system or body is equal to sum of all possible kinds of energy.

Energy U = TE+ RE + VE + Chemical energy + nuclear energy + electron spin energy + PE

For ideal gas

C_p - C_v = R C_P =  = (f/2 1) R f --> degree of freedom

C_p/C_v = g C_v =  = f/2 R g --> Poisson's Ratio

For Isobaric Process :

Q = nC_pDT and Q = DH

For isochoric process

Q = nC_vDT and Q = DE

DE = n C_v DT = 

For liquid and solids

U = 

Total degree of freedom = 3n

where n = no of atoms

for vibrational u = fnRT

Total 

Based on thermal equilibrium if A & B, & B & C are in thermal equilibrium then A & C must be in thermal equilibrium.

First law of thermodynamics is based on energy conservation

E₂ = E_{1 +}  q +w (E₁ is the E_i)

E₂ - E₁ = q + w + (E₁₊ q + w is E_f)

(E_i = E_f)

or 

For an isolated system, q = 0, w = 0

DU = 0

or U = constant

For cyclic process.

Work done = -F_ext.dx

= - P. A dx

for expansion --> dW = - ve

compression --> dW = ve

dq = CdT dq_v = C_vdT

C_v = 

For an isochoric process dv = 0

dU = q_V  

dU = n C_VdT

If C_V is a function of temperature

DU = n 

We know that

U = f(T, V, P)

consider U = f(T, V)

du =    

For isochoric process dv = 0

for 1 mole of gas

C_VdT = dT

C_{V =}   

For an ideal gas U = f(T) only

 = 0

du = C_VdT

DU = nC_VDT = nC_V(T₂ - T₁)

- H = U + PV

dH = dU + d(PV)

 =   +  

From, I^st law of thermodynamics at constant pressure.

dU =  +  dW

dU =  - PdV

 = dU  + PdV

dH = dU + PdV

 ..........(1)

For an ideal gas expansion or compression

DH = nC_VDT + nRDT

= nDT[C_V + R]

We know that, H = f(T, P)

dH =    
 

At constant pressure

dP = 0

dH = 

**Calculation of Dn_g for any chemical reaction:

N₂  3H₂  2NH₃

Dn_g = - 2

N₂ + 3H₂ --> 2NH₃

10     30          0

0       0         20

Dn_g = 20 - (40) = -20

If reaction is 50% completed.

N₂ + 3H₂ --> 2NH₃

10     30          0

5      15         10

Dn_g = 30 - 40 = -10

(1) Isochoric process:

V = constant

dV = 0

dU = dq_V

DU = q_V = nC_VDT

(2) Isobaric process:

W = - P_ext (V₂-V₁)

Reversible & isobaric process

W = - P (V₂-V₁)

= - nR (T₂-T₁)

Irreversible & isobaric process

P₁ = P₂ = P_ext

For reversible & irreversible isobaric or isochoric process, workdone is same.

3. Isothermal process.

(a) Reversible Expansion or compression

W = 

= 

= 

In Expansion W = - ve

DE = 0

(b) Single stage irreversible expansion

W = - P_ext(V₂ - V₁)

|W_rev| > |W_irr| (in case of expansion)

W = - P_ext 

(c) Two Stage irreversible Expansion:

Stage I. P'_ext = 3 atm P_i = 5 atm

Stage II. P"_ext = 2 atm P_f = 2 atm

 Workdone in 2^nd stage > Workdone in I^st stage

(d) n- stage expansion

Compression - (One stage Compression )

| W_irr | = P_ext DV

P₁ = 1 atm , P₂ = 5 atm , P_ext = 5 atm

 | W_irr | > | W_rev | For compression

Two stage Comp. n stage Comp

Ex.1 2 moles of an ideal gas initially present in a piston fitted cylinder at 300 K, and 10 atm are allowed to expand against 1 atm but the piston was stopped before it stablished the mechanical equilibrium. If temperature were maintained constant through out the change and system delivers 748.26 J of work, determine the final gas pressure and describe the process on PV diagram.

Sol. W_irrv = - 748.26

W_irr = - P_ext [1/P₂- 1/P₁]nRT

P₂ = 4atm

Ex.2 1150 Kcal heat is released when following reaction is carried out at constant volume.

C₇H_16(l)  11O_2(g) --> 7CO_2(g)  8H₂O_(l)

Find the heat change at constant pressure.

The pressure of liquid is a linear function of volume (P = a bV) and the internal energy of the liquid is U = 34 3PV find a, b, w, DE & DH for change in state from 100 Pa, 3m³ to 400 Pa, 6m³

Sol. 100 = a + bV

100 = a + 3b

Also, 400 = a + 6b

DU = 34 3(P₂V₂ - P₁V₁)

= 6300 J

DH = DU P₂V₂ - P₁V₁

= 6300 2100 = 8400 J

P is a linear function

P_ext =  = 250

W = - P_ext(dV)

= - 250(6 - 3) = - 750 J

Ex.3 4 moles of an ideal gas (C_v = 15 J) is subjected to the following process represented on P - T graph. From the given data find out whether the process is isochoric or not ? also calculate q, w, DU, DH,

Sol. PV = nRT

4V = 4R × 400

V = 400 R ........(1)

3V = 4R × 300

V = 400 R ........(2)

i.e., V is constant

w = 0

DU = nC_V DT 4 ×15 ×100 = 6000 J

DH = nC_P DT n (C_V  R ) DT

4 ×(15 8.3)×100

9320 J

Ex.4 2 mole of a gas at 1 bar and 300 K are compressed at constant temperature by use of a constant pressure of 5 bar. How much work is done on the gas ?

Sol. w = - nRT ×P_ext 

= 19953.6 J

Ex.5 2 moles of an ideal diatomic gas (C_V = 5/2 R) at 300 K, 5 atm expanded irreversibly and adiabatically to a final pressure of 2 atm against a constant pressure of 1 atm.

(1) Calculate final temperature q, w, DH & DU

(2) Calculate corresponding values if the above process is carried out reversibly.

Sol.. w = C_V (T₂ - T₁) = - P_ext R

Given 

P_ext = 1,P₂ = 2,P₁ = 5

q = 0

w = DU = nC_VDT

= -1247.1 J

DH = nC_PDT

= 1745.94 J

If process is reversible Pv^g = Constant

P^{1- g}T^g = Constant

dq = 0

dU = dW  

For an ideal gas C_P - C_V = R

nC_VdT = - P_ext dV

P_int = dP = P_ext

P_int = P_gas = 

C_V ln  = - Rln 

 ln  = - R ln    ln  = -ln 

ln  = -(g -1) ln   ln  = ln 

T₂(V₂)^{g - 1} = V₁^{g - 1}T₁

TV^{g - 1} = Constant
 

 = Constant

dU = dW

nC_V(T₂ - T₁) = - P_ext dV

nC_V(T₂ - T₁) = - P_ext [V₂ - V₁]

nC_V(T₂ - T₁) = - P_ext 

= - P_ext nR

(1) If final volumes are same.

Isothermal process.

P₁v₁ = P_iso V₂

Adiabatic process.

P₁v₁^g = P_adia V₂^g

> 1 >    > 

(2) If final pressures are same

Isothermal process.

P₁V₁ = P₂ V_iso

 =  .......(1)

P₁V1^γ = P₂V^γ_adia
 

= 

in ideal gas expansion, | W_iso | > | W_adia |

Hence  

Compression

(1) If final volumes are same

For isothermal process

P₁V₁ = P_isoV₂

 ..........(1)

Adiabatic process.

P₁V₁^γ = P_adiaV₂^γ

 ..........(2)

(2) If final pressures are same

P₁V₁ = P₂ V_iso ..........(1)

P₁V₁^γ = P₂ V^γ_adia ..........(2)

  < 

PV^x = Const

W = 

= - 

= - k  w = 

= 

dU = dq + dW

nC_VdT = nC_mdT (-PdV)

C_m = C_V    ..........(1)

PV = nRT

KV^-x V = nRT

kV^{-x 1} = nRT

k(-x 1)V^-x  = nR

C_m = C_v   

  x ¹ 1

1. First law of thermodynamics does not give information regarding the direction of propagation of a process

2. First law of thermodynamics does not tell us why an equilibrium is attained.

3. First law of thermodynamics does not tell us when an equilibrium will be attained.

4. First law of thermodynamics does not give information about why there can not be 100 percent conversion of heat into work

Statement(I) : Second law of thermodynamics states that heat can never be converted into work with 100% efficiency

Statement(II) : No engine in this world can be constructed which operates in cycles and converts all the heat from source to work.

Statement(III) : No refrigertator can be designed which operates in cycles and rejects heat from sink to source, perpetually (self - functioning).

Entropy : Entropy is the direct measurment of randomness or disorderness. Entropy is an extensive property & it is a state function

ds =  for reversible process. entropy is related with complexity of the molecule within the system.

EtOH > MeOH

C₂H₆ (g) > C₂H₅(g)

N₂O₄ > NO₂

O₂ > N₂ (molecular wt.)

Gas > Liq > Amorphous solid > crystalline solid

Entropy always increases in the following process

(1) s → ,  → g, s → g,

(2) Isothermal expansion of ideal gas.

(3) Mixing of two non reacting gases.

(4) In chemical reaction in which

Dn_g > 0

(5) Heating of any substance

Classification of process Based on spontaneity

Why Spontaneity

Points to ponder :

Why a system always moves towards disorderness ?

Answer : A system moves towards disorderness because the probability of moving towards disorderness is very high.

For Reversible process :

DS =  Q_revesible = constant

DS =  Q_revesible = Variable

Note : Irreversible process

DS_irreversible = 

The entropy change for an irreversible process can be calculated by substituting it with equivalent reversible process. Both will have same entropy change.
 

DS_system = 

DS_surrounding = 

DS_universe = DS_system + DS_surrounding

1. DS_universe > 0 Spontaneous

2. DS_universe = 0 Equilibrium

3. DS_universe < 0 Non-spontaneous.

(A) General heating or cooling

ds = 

DS =  = 

If C is temperature independent

If C is a function of Temperature

C = a + bT

DS = n 

DS = n

(B) In phase transformation

DS_fus = 

*A_(l)  A_(g).

*A_(s)  A_(g)

(C) Entropy change during chemical reaction.

aA + bB --> cC + dD

For any chemical reaction

(D) Calculation of entropy change during expansion/compression of ideal gas from P₁V₁T₁ to P₂V₂T₂

From I^st law of thermodynamics

dE = dq + dW

dq = - dW + dE

TdS = PdV + nC_VdT

dS = dV + nC_V

dS = dV  + 

DS =  = nRln + 

For ideal gas

DS = nC_Vln   + nRln . 

= nC_Vln   + nRln   + nRln 

= nRln   +(C_V  R)n ln

Conclusion

DS = n  nRln

DS = ndT nRln

DS = n nR ln 

Calculation of entropy change during isothermal Expansion;

because  = = -nRln

nRln = 

W_rev > W_irr

DS_total = + ve

Calculation of entropy change during isothermal compression.

q_irr + w_irr = 0

| w_irr | > | w_rev |

< 0 >0

| q_irr | > | q_rev |

Calculation of entropy change during for adiabatic expansion of ideal gas.

For reversible process

Where K is constant

For irreversible process

This means that the final temperature of irreversible process is greater than reversible process.

Calculation of entropy change in adiabatic compression.

dT = 110

For reversible

(T₂ - T₁) = 100K

T₂ = T₁  100 K

for irreversible  = T₁  110 K

DS_irr = nC_Vln   nRln

= nC_Vln - nC_Vln

= nC_vln  > 0

Carnot cycle is based on 4 reversible process.

(1) Reversible isothermal expansion from A to B.

DE_AB = 0 ,

W_AB = -nRT₁ln

(2) Reversible adiabatic expansion from B to C

DE_BC = nC_V(T₂ -T₁)

W_BC = DE_BC

(3) Isothermal compression from C to D

DE_CD = 0,

W_CD = -nRT₂ln

(4) Adiabatic compression from D to A.

DE_DA = nC_V(T₁ - T₂)

W_DA = DE_DA

DE_Cycle = 0

W_cycle = -nRT₁ln  nC_V(T₂- T₁) - nRT₂ln  nC_V(T₁- T₂)

= 

For BC, 

For DA, 

W_cycle = -nR(T₁ - T₂)ln V₂/V₁

Efficiency of any engine may be given as

 =                                   

DE_cycle = q_cyc  w_cycle

w_cycle = - q_cycle = 

1  

  = 

This means DS is a state function

Gibb's Free Energy (G)

G_system = H_system - TS_system

W = W_expansion  W_{non-expansion}

W_{non - expansion} = w_useful (useful work)

DG = W_{non expansion} = W_useful

All those energy which is available with the system which is utilized in doing useful work is called Gibb's free energy :

R_x :

1. DG_system = DH_system - TDS_system = - TDS_universe = W_{non expansion} = W_useful

2. DG_system = - TDS_universe

3. (a) DS_universe > 0 or DG_system < 0 Spontaneous

(b) DS_universe = 0 or DG_system = 0 Equilirbium

(c) DS_universe < 0 or DG_system > 0 Non-Spontaneous

R_x :

(1) aA bB  cC dD

DG° = Standard Gibb's free energy change (P = 1 atm, 298 K)

DG = Gibb's free energy change at any condition.

DG = DG° 2.303 RT log Q ; Q = Reaction Quotient

At equilibrium, DG = 0 and Q = K_eq.                                       

0 = DG° 2.303 RT log k_eq

DG° = -2.303 RT log k_eq

å G°(product) - å G° (Reactant) = - 2.303 RT log k_eq

DH° - TDS° = -2.303 RT log k_eq

(2) W_cell = q × E

DG = - W_cell

DG = - q × E_cell

Now, one mole e^- have charge 96500 coulomb = 1 Faraday (F)

n mole of e^- will have charge = n × F or q = n × F

DG = - nFE_cell

DG° = -nFE°_cell

Third law of thermodynamics states that as the temperature approaches absolute zero, the entropy of perfectly crystalline substance also approaches zero.

Taking T₂ = T and T₁ = 0°k.

S_T - 0 =   S_T =  For perfectly crystalline substance

The entropy of perfectly crystalline substance can be determined using third law of thermodynamics.

or

With the help of third law of thermodynamics we can calculate the exact value of entropy.

(DG)_T,P is a measure of useful work a non PV work (non expansion work) that can be produced by a chemical transformation.e.g.electrical work .

For reversible reaction at constant T & P

dU = dq +dW_total

dU = dq + dW_{P, V}  + dW_{non P,V}

dU = T.dS - P. dV  + dW_{non P,V}

dU P.dV = T. dS  + dW_{non P,V}

dH = T.dS + dW_{non P,V}

(dG_sys)_T,P = dW_{non P,V}

Useful work done on the system = increase in Gibb's energy of system at constant T & P.

- (DG_sys)_T,P= - W_{non P,V}

- (DG_sys)_T,P= - W_{by, non P,V}

Useful work done by the system = decrease in Gibb's energy of system at constant T & P .

If (DG_sys)_T,P = 0, then system is unable to deliver useful work.

For reversible process in which non expansion work is not possible

dU = dq dW

H = U + PV

dH = dU + P.dV + VdP

dH = T.dS - PdV + PdV + V.dP

dH = T.dS +V.dP

G = H -TS

dG = dH - T.dS -S.dT

dG = T.dS V.dP - T.dS - S.dT

 For a particular system (s//g)

(1) At constant temperature dG = V.dP or  = V

(A) For a system is s/ phase

(B) For an ideal gas, expansion/compression :-

=

(2) At constant pressure : dG =-S.dT or  = - S

* For phase transformation/chemical reaction

d(DG) = DV.dP - DS dT

H₂O(s) --> H₂O(l)

DV = V_M(H₂O,l) - V_m(H₂O,s)

DS = S_M(H₂O,l) - S_m(H₂O,s)

A(s) --> B(g) 2C(g)

D_rS = S_M(B,g ) 2S_m(C,g) - S_m(A,s)

C (s, graphite) --> C(s, diamond)

DV = V_m(C, diamond) - V_m(C-graphite)

At constant temperature:-

d(DG) = DV. dP

At constant pressure

D_rG_T2 - D_rG_T1 = - D_rS(T₂ - T₁)