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for the ideal gas
If we take into account internal energy of mol of ideal gas is equal
to U  C V RT and dA  pdV , then we have:

pdV  C V dT  0 (2.7.2)

The mentioned above equation is a differentiating equation for ideal gas
law, so pV  RT and we obtain

pdV  Vdp
pdV  Vdp  RdT  dT 
R (2.7.3)

Now, if we substitute equation (2.7.3) into equation (2.7.2), we will obtain:

p  dV V  dp
C V   p  dV  0 (2.7.4)
R

Since R  C  C , then
P
V

p dV V dp
C V   p dV 
C P  C V

C  p dV  C Vdp  C  p dV  C  p dV 
 V V P V  0 (2.7.5)
C P  C V

Whence, it follows that

C
C P  p dV  C V V  dp  0  P  p dV V dp  0. (2.77.6)
C
V

or

 p dV V dp  0, (2.7.8)

C
where P   is heat capacity ratio and by dividing both sides by PV we
C V
obtain differential equation:

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