Thermodynamics B.Sc. 2nd Year

Thermodynamics B.Sc. 2nd Year Thermodynamics B.Sc. 2nd Year

Thermodynamics


Spontaneous Process and Nonspontaneous Process

The process which proceeds of its own accord without any outside assistance or help is called Spontaneous Process or Natural Process and the reverse process which does not proceed on its own accord is called nonspontaneous process.
Spontaneous process is unidirectional, for reverse change work has to be done. For spontaneous process time is no factor. It may takes palce rapidly or very slowly. If a system is not in equilibrium state, a spontaneous change is inevitable. The change will continue untill equilibrium exist. Once the system attains the equilibrium it does not undergo further spontaneous change. Spontaneous process or change is accompanied by decrease of internal energy of enthalpy.
The tendency of a process to occur naturally is called Spontaniety.

Examples
A rolling ball down the hill spontaneously but it will not rall uphill unless work is done on it.
When two metal ball one hot and one cold are connected to each other, the heat flow from hotter to colder ball spontaneously but can never from colder to hotter ball spontaneously.


Entropy (S)

It is a thermodynamic state function that is a measure of the randomness and disorderness of the molecules of the system. Greater the randomness, greater would be entropy. It is denoted by 'S'. As it is a state function so depends upon initial and final satate only. So change in entropy
ΔS = Sfinal − Sinitial
For a reversible process at constant temperature,the change in entropy is equal to energy absorbed or evolved divided by the temperature.
ΔS = q/T
If heat is absorbed, ΔS is positive and entropy increases while If heat is evolved, ΔS is negative and entropy decreases.
If heat change takes place at different temperatures, then-
dS = dq1/T1 + dq2/T2 + dq3/T3 + ... = Σ(dqrev/dT)
or, ∫ds = ∫dqrev/T
ΔS = ΔS + ΔS
at equilibrium, ΔS = 0


→→→→→→→
→→→→→→→
→→→→→→→
→→→→→→→
    well orderd
   Low Entropy
→↑←↓→←↓→
→←↓→↑←↓→
→↓←↓→↑←↓
←↑→↓←↓→↑
      Random
  High Entropy

Unit
Cal K-1
or, J K-1

Entropy Change in mixing of ideal gases

When two ideal gases are brought in contact , they mix with each other and incerase in their entropy.
Let n1 moles of gas A whose volume is V1 mixed with n2 moles of other gas B whose volume is V2 and form a new volume V1 + V2 of mixed gas. So the entropy change in gas A due to mixing is-
ΔS1 = n1 Rln(V1 + V2/V1)
Similarly, for gas B, entropy change is-
ΔS2 = n2 Rln(V1 + V2/V2)
So, the total entropy change-

ΔSmix = ΔS1 + ΔS2
or, ΔSmix = n1 Rln(V1 + V2/V1) + n2 Rln(V1 + V2/V2)
or, ΔSmix = −R(n1 ln V1/(V1 + V2) + n2 ln V2/(V1 + V2)
If X1 and X2 br the mole fraction of gas A and B respectively, then-
X1 = n1/(n1+n2) = V1/(V1+V2)
and X2 = n2/(n1+n2) = V2/(V1+V2) at NTP
So, ΔSmix = −R(n1 lnX1 + n2 lnX2)
or, ΔSmix = −RΣni lnXi
Again from this equation
ΔSmix = −R(n1 lnX1 + n2 lnX2)
ΔSmix = −R[n1/(n1+n2) lnX1 + n2/(n1+n2) lnX2](n1+n2)
Now entropy change for one mole-

n1+n2 = 1
ΔSmix = −R(X1 lnX1 + X2 lnX2)
or, ΔSmix = −RΣXi lnXi
Since all Xi is less than one so, entropy of mixing of ideal gas is always positive.

Gibbs-Helmholtz Equation

When a system undergoes a reversible change, the change in free energy with temperature and pressure is given by-
dF = VdP − SdT -----(equation-1)
at constant pressure, dP = 0
so, (equation-1) becomes-
[dF/dT]p = −S
∴ −ΔS = −(S1 − S2)
or, −ΔS = (dF2/dT)p −(dF1/dT)p
or, −ΔS = [(F2 − F1)/dT]p

or, −ΔS = [d(ΔF)/dT]p -----(equation-2)
we know that ΔF = ΔH − TΔS -----(equation-3)
so putting the value of ΔS from equation-2 in equation-3 we get-
ΔF = ΔH + T[d(ΔF)/dT]p -----(equation-4)
equation-4 is called Gibbs-Helmholtz Equation. and is applicable to all process which takes place at constant pressure.

Helmholtz Free Energy(F)

Mathematically, Helmholtz free energy is expressed as-
F = U - TS
on differentiation we get-
dF = dU − TdS − SdT
From 1st law of thermodynamics-
dQrev = dU + Wmax

or, TdS = dU + Wmax
So, dF = −Wmax − SdT
at constant temperature-
or, (−dF)T = Wmax
So, Helmholtz free energy is the function decrease of which measures maximum amount of work which is obtainable from the system isothermally and reversibly.It is a state function and extensive property.

Gibbs free energy(G or F)

Gibbs free energy is the property of the system whose decrease is the measure of the maximum external work available during the transformation of system reversibly at constant pressure and temperature and is state function. So depend only on initial and final state of the system.
If S is the entropy of a system at ToK and H is its enthalpy then, Gibbs free energy is mathematically expressed as-
G = H - TS
On differentiation we get,
ΔG = ΔH − TΔS − SΔT

we know that- ΔH = ΔE + PΔV + VΔP
so, ΔG = ΔE + PΔV + VΔP − TΔS − SΔT
or, ΔG = ΔA + PΔV + VΔP − SΔT (as ΔE − TΔS = ΔA (work function))
at constant pressure and temperature-
or, ΔG = ΔA + PΔV
or, ΔG = −Wmax + PΔV (as W = −ΔA from 1st law of thermodynamics)
or, −ΔG = Wmax − PΔV


PΔV is the work done due to expansion against a constant pressure. So, decrease in free energy accompanying a process which occurs at constant pressure and temperature is the maximum work obtained from the system other than PΔV. Hence it is also called Net Work.
So, Net Work = −ΔG

Variation of Free Energy with Temperature and Pressure

We know that-
F = H − TS -----(equation-1)
or, F = E + PV − TS (as H = E + PV)
differentiating this equation we get-
dF = dE + PdV + VdP − TdS − SdT
or, dF = dq + VdP − TdS − SdT (as dq = dE + PdV)
or, dF = TdS + VdP − TdS − SdT (as dqrev/T = dS)
or, dF = VdP − SdT -----(equation-2)
at constant Temperature-
or, dF = VdP


or, (dF/dP)T = V -----(equation-3)
and at constant Pressure-
or, dF = − SdT
or, (dF/dT)P = − S -----(equation-4)

Free Energy Change of Expansion of an Ideal Gas

When a system undergo a reversible change, the change in free energy with temperature and pressure is given as-
dF = VdP − SdT
at constant temperature-
dF = VdP
or, dF = nRT.(dP/p) (as PV = nRT)
on integrating the above equation we get-
∫dF = nRT∫dP/P
or, ΔF = nRT lnP2/P1
or, ΔF = nRT lnV1/V2 (as P1V1 = P2V2 )


Clausius-Clapeyron Equation

The Clausius-Clapeyron equation was initially proposed by a German physics Rudolf Clausius in 1834 and further developed by French physicist Benoît Clapeyron in 1850. This equation is extremely useful in characterizing a discontinuous phase transition between two phases of a single constituent.
We know that-

dG = VdP − SdT -----(equation-1)
Let us consider a single-constituent equilibria-
𝑃ℎ𝑎𝑠𝑒-1 ⇌ 𝑃ℎ𝑎𝑠𝑒-2
Where phase-1 may be solid, liquid, or gas; whereas phase-2 may be liquid or vapor depending upon the nature of the transition whether it is melting, vaporization or sublimation, respectively.
For phase-1, change in free energy is-
dG1 = V1dP − S1dT -----(equation-2)
and for Phase-2, change in free energy is-
dG2 = V2dP − S2dT -----(equation-3)
At equilibrium- dG1 = dG2 (i.e. ΔG = 0
so, V2dP − S2dT = 𝑉1dP − S1dT
V2dP − V1dP = S2dT − S1dT
(V2 − V1)dP = (S2 − S1)dT
ΔV. ΔP = ΔS. dT
dP/dT = ΔS/ΔV -----(equation-4)
Now, if the ΔH is the latent heat of phase transformation takes place at temperature (𝑇), then the entropy change is-
ΔS = ΔH/T -----(equation-5)

Now, putting the value of 𝛥𝑆 from (equation-5) into (equation-4), we get-
dP/dT = ΔH/T.ΔV -----(equation-6)
The (equation-6) is known as Calpeyron equation.
Now if phase-1 is solid while phase-2 is vapor (i.e. solid ⇌ melt), then the equation-6 becomes-
dP/dT = ΔfusH/Tf.ΔV -----(equation-7)
where Δfus is latent heat of fusion and Tf is melting point.
For vaporisation, equilibrium (i.e. liquid ⇌ vapour),
dP/dT = ΔvapH/T.Vv -----(equation-8)
If the vapor act as an ideal gas-
then,V = RT/P

so, the above equation becomes-
dP/dT = ΔvapH.P/RT2 -----(equation-9)
or, 1/P(dP/dT) = ΔvapH/RT2 -----(equation-10)
or, dlnP/dT = ΔvapH/RT2 -----(equation-11)
The equation-11 is known as the Clausius-Clapeyron equation.
Another form of Clausius-Clapeyron equation-
from equation-11
dlnP = (ΔvapH/RT2).dT
If the temperature changes from T1 to T2 and pressure is varied from P1 to P2, then -
∫dlnP = ∫(ΔvapH/RT2).dT
lnP2/P1 = ΔvapH/R ∫dT/T2
lnP2/P1 = (ΔvapH/R) [1/T1 − 1/T2] -----(equation-12)
The equation-12 is another form of Clausius-Clapeyron equation.
converting ln into log-
2.303 logP2/P1 = (ΔvapH/R) [1/T1 − 1/T2] -----(equation-13)
or, logP2/P1 = (ΔvapH/ 2.303 R) [1/T1 − 1/T2] -----(equation-14)


Necessity of the Second Law of Thermodynamics

According to first law of thermodynamics, the universe is a closed system and the amount of matter and energy in the universe is constant. It also tells that different forms of energies are interconvertible and establishes an exact relationship between heat and work.
First law also tells that heat and mechanical work both are interconvertible but does not give any idea about the conditions under which this conversion is possible. Thus, the first law of thermodynamics does not tell us how much heat energy is converted into work and also the direction and extent of energy transformation. Practically, it is impossible to convert the heat energy into an equivalent amount of work. So, another law is needed to explain this called second law of thermodynamics. The second law of thermodynamics predict whether the reaction is feasible or not and also tell the direction of the flow of heat.


Second Law of Thermodynamics

Second law of thermodynamics can be explained by the following statements which was given by various scientist-

1. Kelvin and Plank's Statement

It is impossible to obtainedwork by a colder body below it's lower temperature.

2. R Clausious Statement

Heat can not be itself passes from a colder body to hotter body without expenditure of external energy.
The energy of the universe remains constant but the entropy of the universe approaches maximum inreversible process.

3. Carnot's Statement

It is impossible to take heat from a hotter body and convert it completely into work without transferring a part of heat to the colder body.

4. All the natural and spontaneous process takes place only in one direction but can not be reverse.
5. For any process, the change in entropy of the system plus change in entropy of the surrounding is zero.
6. The gasses diffuses from the region of higher pressure to the region of lower pressure until the pressure becomes uniform.


Carnot Cycle

Carnot used a reversible cycle to show the maximum convertibility of heat into work. It is a cyclic process carried out in four reversible steps alternatively isothermal and adiabatic. The first two steps being of expansion and the last two of compression. The engine that works by the above process is known as Carnot engine or ideal gas engine. The working substance of the engine is one mole of an ideal gas.
The engine works between two temperature(T2) and (T1). It takes heat from a source at higher temperature (T2) and does work and gives out the unused heat into a sink at lower temperature (T1).

Stroke-1
Isothermal Expansion
The gas is allowed to expand reversibly and isothermally at the temperature T2 so that the volume increases from V1 to V2.
We know that in isothermal expansion of an ideal gas ΔE = 0. So, from first law of thermodynamics-
ΔE = q - w
or, q = w

Let q2 be the heat absorbed by the system at temperature T2 and work w be the work done by the system on the surroundings.
then- q2 = w1
= RT2 ln V2/V1 -----(equation-1)
Stroke-2
Adiabatic Expansion
The gas is now allowed to expand reversibly and adiabatically from volume V2 to V3.
Since work has been done by the system adiabatically, where heat is not absorbed. So, the temperature of the system falls from T2 to T1. Therefore, q = 0. Thus, the 1st law of thermodynamics becomes-
ΔE = - w
or, -ΔE = w
we know that-
Cv = (δE/δT)v
or, Cv . dT = dE
or, Cv . (T1 - T2) = -w
or, Cv . (T2 - T1) = w
Let work done by the system in this step is denoted by w2 then-
w2 = Cv . (T2 - T1) -----(equation-2)


Stroke-3
Isothermal Compression
Here gas is subjected to a reversible and isothermal compression at the lower temperature T1 so the volume decreases from V3 to V4.
In this case work is done on the system and hence heat will be produced and given up to the surroundings. Since compression takes place isothermally and reversibly,
so, ΔE = 0
and if q1 be the heat given out to the surrounding at temperature T1 and work w3 be the work done on the system.
so in this process-
q1 = -w3
or, -w3 = RT1 ln V4/V3 -----(equation-3)


Stroke-4
Adiabatic Compression
Finally by adiabatic and reversible compression, the gas is restored to the original volume V1 and temperature T1. In this case, work is done on the system. Hence, w is negative. Then, first law of thermodynamics becomes-
ΔE = q - (-w) = q + w
In adiabatic process, q = 0
Hence- ΔE = w = Cv (T2 - T1)
Let w4 be the work done in this stage-
then- w4 = Cv (T2 - T1)
or, -w4 = -Cv (T2 - T1) -----(equation-4)
where, T2 - T1 is increase in temperature produced by the adiabatic compression.
The net heat absorbed q by an ideal gas in the whole cyclic process is-
q = q2 + (-q1)
q = RT2 ln V2/V1 + RT1 ln V4/V3
or, q = RT2 ln V2/V1 - RT1 ln V3/V4 -----(equation-5)
According to the expression governing adiabatic changes-
(T2/T1) = (V3/V2)γ-1---for adiabatic expansion
(T1/T2) = (V1/V4)γ-1---for adiabatic compression
or, V3/V2 = V4/V1


Therefore, substituting the value of V3/V4 in (equation-5), the Net Heat may be give as-
q = RT2 ln V2/V1 - RT1 ln V2/V1
or, q = R(T2 - T1)ln V2/V1 -----(equation-6)
Net work done in one cycle-
W = w1 + w2 + (-w3) + (-w4)
W = RT2 ln V2/V1 + Cv(T2 - T1) + (-RT1 ln V4/V3) + (-Cv(T2 - T1)
or, W = RT2 ln V2/V1 + RT1 ln V3/V4 -----(equation-7)
Equation-7 is work done in one cycle.
or, q = RT2 ln V2/V1 - RT1 ln V2/V1
or, q = R(T2 - T1)ln V2/V1 (as V2/V1 = V3/V4) -----(equation-8)
Equation-8 is Net Heat absorbed in one cycle.


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