thermodynamics notes

ME 205 Thermodynamics Module 6



Thermodynamic Relations
Mathematical Theorems

Theorem 1
Some relation exists between 𝑥, 𝑦, 𝑧
𝜕𝑧 𝜕𝑧
𝑑𝑧 = ( ) 𝑑𝑦 + ( ) 𝑑𝑥
𝜕𝑦 𝑥 𝜕𝑥 𝑦
Now, let
𝜕𝑧 𝜕𝑧
𝑀 = (𝜕𝑦) and 𝑁 = (𝜕𝑥)
𝑥 𝑦

∴ 𝑑𝑧 = 𝑀𝑑𝑦 + 𝑁𝑑𝑥
Hence,
𝜕𝑀 𝜕 2𝑧
( ) =
𝜕𝑥 𝑦 𝜕𝑥𝜕𝑦
And
𝜕𝑁 𝜕 2𝑧
( ) =
𝜕𝑦 𝑥 𝜕𝑥𝜕𝑦

𝜕𝑀 𝜕𝑁
∴( ) =( )
𝜕𝑥 𝑦 𝜕𝑦 𝑥

This is the condition for exact or perfect differential.

Theorem 2
If a quantity 𝑓 is a function of 𝑥, 𝑦 and 𝑧 and a relation exists among 𝑥, 𝑦 and 𝑧, then 𝑓 is a function
of any two of 𝑥, 𝑦 and 𝑧.
Similarly, any one of 𝑥, 𝑦 and 𝑧 may be regarded to be a function of 𝑓 and any one of 𝑥, 𝑦 and 𝑧.
Thus if
𝑥 = 𝑥(𝑓, 𝑦)
𝜕𝑥 𝜕𝑥
𝑑𝑥 = ( ) 𝑑𝑓 + ( ) 𝑑𝑦
𝜕𝑓 𝑦 𝜕𝑦 𝑓

And if
𝑦 = 𝑦(𝑓, 𝑧)


Page | 3 Department of Mechanical Engineering, SSET

, ME 205 Thermodynamics Module 6

𝜕𝑦 𝜕𝑦
𝑑𝑦 = ( ) 𝑑𝑓 + ( ) 𝑑𝑧
𝜕𝑓 𝑧 𝜕𝑧 𝑓
Substituting for 𝑑𝑦 is the previous relation, we get
𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦
𝑑𝑥 = ( ) 𝑑𝑓 + ( ) [( ) 𝑑𝑓 + ( ) 𝑑𝑧]
𝜕𝑓 𝑦 𝜕𝑦 𝑓 𝜕𝑓 𝑧 𝜕𝑧 𝑓

𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦
∴ 𝑑𝑥 = [( ) + ( ) ( ) ] 𝑑𝑓 + ( ) ( ) 𝑑𝑧
𝜕𝑓 𝑦 𝜕𝑦 𝑓 𝜕𝑓 𝑧 𝜕𝑦 𝑓 𝜕𝑧 𝑓

Also,
𝑥 = 𝑥(𝑓, 𝑧)
𝜕𝑥 𝜕𝑥
⇒ 𝑑𝑥 = ( ) 𝑑𝑓 + ( ) 𝑑𝑧
𝜕𝑓 𝑧 𝜕𝑧 𝑓
Comparing the coefficients of 𝑑𝑧, we get
𝜕𝑥 𝜕𝑥 𝜕𝑦
( ) =( ) ( )
𝜕𝑧 𝑓 𝜕𝑦 𝑓 𝜕𝑧 𝑓

𝜕𝑥 𝜕𝑦 𝜕𝑧
∴( ) ( ) ( ) =1
𝜕𝑦 𝑓 𝜕𝑧 𝑓 𝜕𝑥 𝑓


Theorem 3
Among the variables 𝑥, 𝑦 and 𝑧, any one variable may be considered as a function of the other
two.
𝑥 = 𝑥(𝑦, 𝑧)
𝜕𝑥 𝜕𝑥
𝑑𝑥 = ( ) 𝑑𝑦 + ( ) 𝑑𝑧
𝜕𝑦 𝑧 𝜕𝑧 𝑦
Similarly,
𝑧 = 𝑧(𝑥, 𝑦)
𝜕𝑧 𝜕𝑧
𝑑𝑧 = ( ) 𝑑𝑥 + ( ) 𝑑𝑦
𝜕𝑥 𝑦 𝜕𝑦 𝑥
Substituting the value of 𝑑𝑧, we get
𝜕𝑥 𝜕𝑥 𝜕𝑧 𝜕𝑧
𝑑𝑥 = ( ) 𝑑𝑦 + ( ) [( ) 𝑑𝑥 + ( ) 𝑑𝑦]
𝜕𝑦 𝑧 𝜕𝑧 𝑦 𝜕𝑥 𝑦 𝜕𝑦 𝑥
𝜕𝑥 𝜕𝑥 𝜕𝑧 𝜕𝑥 𝜕𝑧
𝑑𝑥 = [( ) + ( ) ( ) ] 𝑑𝑦 + ( ) ( ) 𝑑𝑥
𝜕𝑦 𝑧 𝜕𝑧 𝑦 𝜕𝑦 𝑥 𝜕𝑧 𝑦 𝜕𝑥 𝑦


Page | 4 Department of Mechanical Engineering, SSET