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**Entropy: Let**

then entropy change for a finite change of state of a system at constant temperature is given by

Unit of entropy is JK^{–1}.

Entropy is an extensive property. Its value depends upon the amount of the substance involved.

Entropy change for an ideal gas: According to first law, we have dqrev = dV + PdV

Dividing T; we have

or

This is the expression when both the volume and temperature or pressure and temperature of an ideal gas are changed.

For an isothermal process, the change is given by the relation

∵ T_{2} = T_{1}

_{and }

For the change of temperature at constant volume

For the change of temperature at constant pressure

**Entropy change in a few typical cases.**

Entropy change in a reversible phase transformat ion.

Since the reversible phase transformation takes place at constant equilibrium temperature then entropy.

where q_{rev} is the heat invo lved in the phase transformat ion.**Examples **

Entropy change in an irreversible phase transition.

A → B

then

**Problem. The phase transformation of one mole of liquid water at –10°C (T _{1}) to solid water –10°C can be calculated following the paths given below: **

(i) ΔC

(ii) ΔHf = 6088.2 J mol

(iii) C_{Ps} = 36.401 JK^{-1} mol^{-1}

**Then changes of entropy can be calculated Sol. **(i)

ΔS = -1.358 JK mol ^{-1}

then, the change of the entropy of the process

is given by

**Entropy change when two solid at different temperatures are brought together: **Let one at higher temperature Th and other at lower temperature Tc. When two bodies are brought together then

Heat lost = heat gain

(by conservat ion law)

i.e. Heat lost by hot body = Heat gained by cold body

Cp(T_{h} – T) = C_{p} (T – T_{c})

or

**Entropy changes of two bodies are bodies are:**

The total change of entropy is ΔS_{total} = ΔS_{h} + ΔS_{c }

**Problem. 5.0 gm ice at 273 K is added to 30 gm of water at 323 K in a thermally insulated container. What is the final temperature? **

Δ

C

Δ

**Sol. **Heat require to convert 5 gm of ice at 273 K to 5 gm water at 273 K

= (5 g) × (334.72 J gm^{-1}) = 1673.6 J

Final temperature after mixing

Heat gained by ice = Heat lost by hot water

(5 gm) (334.72 J gm^{–1}) + (5 g) (4.184 J gm^{–1} K^{–1}) (T – 273 K)

= (30 g) (4.184 JK^{–1} gm^{–1}) (323 K – T)

Solving for T, we get

T = 304.43 K

**Entropy of mixing: **Entropy of mixing is defined as the difference between the entropy of the mixture of gas and the sum of the entropy of the separate gases, each at a pressure P.

For a mole of an ideal gas,

S = C_{V} ln T + R ln V + S_{0}

where S_{0} is the integration constant.

We know that,

, we get S = C_{P} ln T – R ln P + R ln R + S_{0}

= C_{P} ln T – R ln P + S_{0}’ …(1)

Where S_{0} = (R ln R + S_{0}) is another constant.

If n_{1}, n_{2}, …. etc. are the number of moles of the various gases present in the mixture and p_{1}, p_{2}, … etc. are their partial pressures, then the entropy of the mixture is given by

S = n_{1 }(Cp ln T –R ln P_{1} + S0’) + n_{2}(Cp ln T – R ln P_{2} + S_{0}’) + ……

...............(2)

The partial pressure (p) of an ideal gas is given by P = xp where x is the mole fract ion and P is the total pressure.

Putting this value we get

...............(3)

Equation (3) gives the entropy of a mixture of ideal gases. “Entropy of mixing is difference between the entropy of the mixture of gases and the sum of the entropies of the separate gases”

Thus

where n_{i} and n_{x} are number of moles and mole fraction of each constituent of the mixture.

If n is the total number of moles then

n = n_{1} + n_{2} ……..

then

then molar entropy of gases are

**Problem. Calculate the entropy of mixing of one mole of nitrogen gas and three moles of hydrogen gas, assuming that no chemical reaction occurs. i.e. mixing is ideal. Sol.**

n

The mole fract ion are:

form equation

**Physical Significance of Entropy: **

**(1) Entropy as a measure of the disorder of the system: **Spontaneous process are accompanied by increase in entropy as well as increase in the disorder of the system.

Thus, entropy is a measure of the disorder of a system.

**(2) Entropy as a measure of probability:** All spontaneous process lead to increase in entropy and also to increase in disorder. It appears that there is a close relation between ‘entropy S and the probability (W).

This relationship was expressed by Boltzmann as

S = k ln (Probabilit y) = kln (W)

S = k ln W

where k is Boltzmann constant

S = k ln W = k ln (arrangement)N_{A}

= k N_{A} ln (arrangement)

S = R ln (arrangement)

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