# 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 = S_{final} − S_{initial}

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 = dq_{1}/T_{1} + dq_{2}/T_{2} + dq_{3}/T_{3} + ... = Σ(dq_{rev}/dT)

or, ∫ds = ∫dq_{rev}/T

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 n_{1} moles of gas A whose volume is V_{1} mixed with n_{2} moles of other gas B whose volume is V_{2} and form a new volume V_{1} + V_{2} of mixed gas. So the entropy change in gas A due to mixing is-

ΔS_{1} = n_{1} Rln(V_{1} + V_{2}/V_{1})

Similarly, for gas B, entropy change is-

ΔS_{2} = n_{2} Rln(V_{1} + V_{2}/V_{2})

So, the total entropy change-

ΔS_{mix} = ΔS_{1} + ΔS_{2}

or, ΔS_{mix} = n_{1} Rln(V_{1} + V_{2}/V_{1}) + n_{2} Rln(V_{1} + V_{2}/V_{2})

or, ΔS_{mix} = −R(n_{1} ln V_{1}/(V_{1} + V_{2}) + n_{2} ln V_{2}/(V_{1} + V_{2})

If X_{1} and X_{2} br the mole fraction of gas A and B respectively, then-

X_{1} = n_{1}/(n_{1}+n_{2}) = V_{1}/(V_{1}+V_{2})

and X_{2} = n_{2}/(n_{1}+n_{2}) = V_{2}/(V_{1}+V_{2}) at NTP

So, ΔS_{mix} = −R(n_{1} lnX_{1} + n_{2} lnX_{2})

or, ΔS_{mix} = −RΣn_{i} lnX_{i}

Again from this equation

ΔS_{mix} = −R(n_{1} lnX_{1} + n_{2} lnX_{2})

ΔS_{mix} = −R[n_{1}/(n_{1}+n_{2}) lnX_{1} + n_{2}/(n_{1}+n_{2}) lnX_{2}](n_{1}+n_{2})

Now entropy change for one mole-

n_{1}+n_{2} = 1

ΔS_{mix} = −R(X_{1} lnX_{1} + X_{2} lnX_{2})

or, ΔS_{mix} = −RΣX_{i} lnX_{i}

Since all X_{i} 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 = −(S_{1} − S_{2})

or, −ΔS = (dF_{2}/dT)_{p} −(dF_{1}/dT)_{p}

or, −ΔS = [(F_{2} − F_{1})/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-

dQ_{rev} = dU + W_{max}

or, TdS = dU + W_{max}

So, dF = −Wmax − SdT

at constant temperature-

or, (−dF)_{T} = W_{max}

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 T^{o}K 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 = −W_{max} + PΔV (as W = −ΔA from 1st law of thermodynamics)

or, −ΔG = W_{max} − 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 dq_{rev}/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 lnP_{2}/P_{1}

or, ΔF = nRT lnV_{1}/V_{2} (as P_{1}V_{1} = P_{2}V_{2} )

## 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-

dG_{1} = V_{1}dP − S_{1}dT -----(equation-2)

and for Phase-2, change in free energy is-

dG_{2} = V_{2}dP − S_{2}dT -----(equation-3)

At equilibrium- dG_{1} = dG_{2} (i.e. ΔG = 0

so, V_{2}dP − S_{2}dT = 𝑉_{1}dP − S_{1}dT

V_{2}dP − V_{1}dP = S_{2}dT − S_{1}dT

(V_{2} − V_{1})dP = (S_{2} − S_{1})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 = Δ_{fus}H/T_{f}.ΔV -----(equation-7)

where Δ_{fus} is latent heat of fusion and T_{f} is melting point.

For vaporisation, equilibrium (i.e. liquid ⇌ vapour),

dP/dT = Δ_{vap}H/T.V_{v} -----(equation-8)

If the vapor act as an ideal gas-

then,V = RT/P

so, the above equation becomes-

dP/dT = Δ_{vap}H.P/RT^{2} -----(equation-9)

or, 1/P(dP/dT) = Δ_{vap}H/RT^{2} -----(equation-10)

or, dlnP/dT = Δ_{vap}H/RT^{2} -----(equation-11)

The equation-11 is known as the Clausius-Clapeyron equation.

Another form of Clausius-Clapeyron equation-

from equation-11

dlnP = (Δ_{vap}H/RT^{2}).dT

If the temperature changes from T_{1} to T_{2} and pressure is varied from P_{1} to P_{2}, then -

∫dlnP = ∫(Δ_{vap}H/RT^{2}).dT

lnP_{2}/P_{1} = Δ_{vap}H/R ∫dT/T^{2}

** lnP _{2}/P_{1} = (Δ_{vap}H/R) [1/T_{1} − 1/T_{2}]** -----(equation-12)

The equation-12 is another form of Clausius-Clapeyron equation.

converting ln into log-

2.303 logP

_{2}/P

_{1}= (Δ

_{vap}H/R) [1/T

_{1}− 1/T

_{2}] -----(equation-13)

or, logP

_{2}/P

_{1}= (Δ

_{vap}H/ 2.303 R) [1/T

_{1}− 1/T

_{2}] -----(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(T_{2}) and (T_{1}). It takes heat from a source at higher temperature (T_{2}) and does work and gives out the unused heat into a sink at lower temperature (T_{1}).

**Stroke-1**

*Isothermal Expansion*

The gas is allowed to expand reversibly and isothermally at the temperature T_{2} so that the volume increases from V_{1} to V_{2}.

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 q_{2} be the heat absorbed by the system at temperature T_{2} and work w be the work done by the system on the surroundings.

then- q_{2} = w_{1}

= RT_{2} ln V_{2}/V_{1} -----(equation-1)

**Stroke-2**

*Adiabatic Expansion*

The gas is now allowed to expand reversibly and adiabatically from volume V_{2} to V_{3}.

Since work has been done by the system adiabatically, where heat is not absorbed. So, the temperature of the system falls from T_{2} to T_{1}. Therefore, q = 0. Thus, the 1st law of thermodynamics becomes-

ΔE = - w

or, -ΔE = w

we know that-

C_{v} = (δE/δT)_{v}

or, C_{v} . dT = dE

or, C_{v} . (T_{1} - T_{2}) = -w

or, C_{v} . (T_{2} - T_{1}) = w

Let work done by the system in this step is denoted by w_{2} then-

w_{2} = C_{v} . (T_{2} - T_{1}) -----(equation-2)

**Stroke-3**

*Isothermal Compression*

Here gas is subjected to a reversible and isothermal compression at the lower temperature T_{1} so the volume decreases from V_{3} to V_{4}.

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 q_{1} be the heat given out to the surrounding at temperature T_{1} and work w_{3} be the work done on the system.

so in this process-

q_{1} = -w_{3}

or, -w_{3} = RT_{1} ln V_{4}/V_{3} -----(equation-3)

**Stroke-4**

*Adiabatic Compression*

Finally by adiabatic and reversible compression, the gas is restored to the original volume V_{1} and temperature T_{1}. 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 = C_{v} (T_{2} - T_{1})

Let w_{4} be the work done in this stage-

then- w_{4} = C_{v} (T_{2} - T_{1})

or, -w_{4} = -C_{v} (T_{2} - T_{1}) -----(equation-4)

where, T_{2} - T_{1} 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 = q_{2} + (-q_{1})

q = RT_{2} ln V_{2}/V_{1} + RT_{1} ln V_{4}/V_{3}

or, q = RT_{2} ln V_{2}/V_{1} - RT_{1} ln V_{3}/V_{4} -----(equation-5)

According to the expression governing adiabatic changes-

(T_{2}/T_{1}) = (V_{3}/V_{2})^{γ-1}---for adiabatic expansion

(T_{1}/T_{2}) = (V_{1}/V_{4})^{γ-1}---for adiabatic compression

or, V_{3}/V_{2} = V_{4}/V_{1}

Therefore, substituting the value of V_{3}/V_{4} in (equation-5), the Net Heat may be give as-

q = RT_{2} ln V_{2}/V_{1} - RT_{1} ln V_{2}/V_{1}

or, q = R(T_{2} - T_{1})ln V_{2}/V_{1} -----(equation-6)

Net work done in one cycle-

W = w_{1} + w_{2} + (-w_{3}) + (-w_{4})

W = RT_{2} ln V_{2}/V_{1} + C_{v}(T_{2} - T_{1}) + (-RT_{1} ln V_{4}/V_{3}) + (-C_{v}(T_{2} - T_{1})

or, W = RT_{2} ln V_{2}/V_{1} + RT_{1} ln V_{3}/V_{4} -----(equation-7)

Equation-7 is work done in one cycle.

or, q = RT_{2} ln V_{2}/V_{1} - RT_{1} ln V_{2}/V_{1}

or, q = R(T_{2} - T_{1})ln V_{2}/V_{1} (as V_{2}/V_{1} = V_{3}/V_{4}) -----(equation-8)

Equation-8 is Net Heat absorbed in one cycle.