( )
(
)
(
)
And hence it might be expected that
The difference between cp and cv is calculated as follows:
(
Hence
( )
)
(
( )
)
(
)
(
)
( )
But
(
)
(
)
And therefore
(
)
(
) (
)
(
) (
)
(
)
) [
(
) ]
Hence
(
The two expressions for cp-cv differ by the term (
(
⁄
) (
)
⁄
(
)
)
As the system was adiabatically contained and no work was performed, then the first law ,
And hence
( )
( )
Thus as dT=0 (experimentally determined )and dV=0 then the term(
⁄
)
Must be zero. For an ideal gas
But for real gases (
Nevertheless ,if (
⁄
⁄
)
(
)
(
)
.
)
Then from eq.(a)
And as, for one mole of ideal gas ,PV=RT, then
Reversible adiabatic processes:
During a reversible process during which the state of the gas is changed
,the state of the gas never leaves the equilibirium surface. Consequently,
during a reversible the gas passes through a continuum of equilibriuim
states , and the work w is given by the integral
w= ∫
only if the process is conducted reversibly.
In an adiabatic process q=0 and thus,from the first law ,
Consider a system comprising one mole of ideal gas dU=cvdt
And for reversible adiabatic process
Thus
As the system is one mole of ideal gas ,then P=RT/V and hence
Integrating between state 1 and 2 gives