F.Y. B. Sc., Semester -I Physical Chemistry Savitribai Phule Pune University, Pune

1. Arts, Commerce & Science College, Tryambakeshwar (Nashik)
Email ID : Manohar@mvptryambakcollege.ac.in
Chapter – 1 : Chemical Energetics

2. Third Law of Thermodynamics
•Statements :
•The entropy of all perfectly crystalline solids may be
taken as zero at the absolute zero temperature.
The law is also stated as follows:
• Every substance has finite positive entropy but at the
absolute zero temperature, the entropy may become zero
and in fact it does become zero in case of perfectly
crystalline solids.
Mathematical Equation :
dS =
𝜹𝒒
𝑻
dS = Entropy change, T = Temperature, 𝜹𝒒 = heat
Chapter – 1 : Chemical Energetics

3. Importance of the Third Law :
• Calculation of absolute entropies :
• It helps in the calculation of the absolute entropies of
chemical compounds at any desired temperature. This is
explained as under. The infinitesimal entropy change is given
by equation
dS =
𝜹𝒒
𝑻
…….. (1)
• But CP =
𝒅𝒒
𝒅𝑻
so that, dq = CP x dT ………. (2)
Substituting this value in equation (1), we have
• dS =
𝑪𝑷 𝒅𝑻
𝑻
………….. (3)
Entropy change of substances when its temperature can be
Changes from absolute zero to temperature (T) can be
calculated using the equation 3 as
Chapter – 1 : Chemical Energetics

4. 𝑺=𝑺𝟎
𝑺=𝑺
𝒅𝑺 = 𝑻=𝟎
𝑻=𝑻 𝑪𝑷 𝒅𝑻
𝑻
or
S- S0 = 𝟎
𝑻 𝑪𝑷
𝒅𝑻
𝑻
………….. ( 4 )
Where S0 = entropy of the substances at absolute zero
S = entropy of the substances at temperature T
According to third law of thermodynamics, S0 = 0
ST = 0
𝑇 𝑪𝑷
𝒅𝑻
𝑻
………… (5)
ST = 0
𝑇
𝑪𝑷 𝒅 𝒍𝒏 𝑻 ………….. (6)
Thus entropy (S) of the substances at temperature (T) can be
calculated by using heat capacities (Cp)
Chapter – 1 : Chemical Energetics

5. • Heat Capacity at different temperature is calculated by using
equation,
1. ST = 0
𝑇
𝑪𝑷 𝒅 𝒍𝒏 𝑻 in temperature between 0 to T K
2. CP can be calculated by using plot between Cp vs ln T i.e. 2.303 log T and
than measuring area under curve which shown in
3. Fig plot between Cp vs ln T
Chapter – 1 : Chemical Energetics
CP
Log T (0K)
C
A
B
Fig. 1 Fig. 2

6. • Equation 6 can be simplify as,
S= Cp ln T = 2.303 Cp log T
From above equation we can directly calculated S from Cp and T.
But from above equation we can not be acutely calculated CP in
temperature range of 0 to 15 K,
Heat capacity is obtained by Debye T-cubed law.
𝑪𝑷 = aT3 ………. (7)
where, a = constant
• Let T1 be the temperature above which heat capacity can be
measured.
Chapter – 1 : Chemical Energetics

7. • This correspond to the point B in Fig. 1. C Corresponds to the
temperature T at which the entropy of solid is to be determined. In
the Fig 1 point A corresponds to 0 K. The integral in eq. 6 become,
• can be split as under
ST = 0
𝑇1 𝑪𝑷 𝒅𝑻
𝑻
+ 𝑇1
𝑇 𝑪𝑷 𝒅𝑻
𝑻
……………( 8 )
Put equation 7 in 8 we get,
0
𝑇1 𝑪𝑷
𝒅𝑻
𝑻
= 0
𝑇1
𝒂𝑻𝟑 𝒅𝑻
𝑻
= 0
𝑇1
𝒂𝑻𝟐dT =
𝟏
𝟑
aT1
3
Or ∆S1 =
𝟏
𝟑
CP (at T1) …………. (9)
This equation are use to calculate heat capacities of substances.
Chapter – 1 : Chemical Energetics

8. • Entropy of solid between temperature 0K and TK is given by
equation,
ST =
𝟏
𝟑
CP (at T1) + 𝑇1
𝑇
𝑪𝑷 𝒅 𝒍𝒏 𝑻 ….. (10)
If there is some allotropic change between the temperature range 0K
and TK, than the entropy of transition given by equation,
ST =
𝟏
𝟑
CP (at T1) + 𝑇1
𝑇
𝑪𝑷 𝒅 𝒍𝒏 𝑻 +
∆𝑯𝒕
𝑻𝒕
• The entropy of heating the gas from Tb to 25 0C (298 K) given by
equation,
• S = 0
𝑇𝑚
𝑪𝑷 𝒔 𝒅 𝒍𝒏 𝑻 +
∆𝑯𝒎
𝑻𝒎
+ 𝑇𝑚
𝑇𝑏
𝑪𝑷 (𝟏) 𝒅 𝒍𝒏 𝑻 +
∆𝑯𝒗
𝑻𝒃
+ 𝑇𝑏
𝑇
𝑪𝑷 (𝒈) 𝒅 𝒍𝒏 𝑻
I + II + III + IV + V
Chapter – 1 : Chemical Energetics