Why joule Thomson coefficient is zero for ideal gases?

Answer 1

we all know determining the joule thomson coefficient is determined by performing irreversible adiabatic process, i.e. constant enthalpy process, it may be determined foe any gas .

Since enthalpy remains constant in this process

let h1 be the enthalpy before the irreversible expansion and h2 be the enthalpy after expansion, h1=h2 . for a real gas h enthalpy is a function of pressure and temperature so it may get cooled or get warm according to nature of gas and conditions

but for an ideal gas h=f(T) only since h= u +pv [they both are dependent on absolute temperature] so while going through irreversible expansion the gas does not depend on pressure .

mathematically,

let @ be joule thomson coefficient

$ be parial derivative

@= 1/ cp multiplied by [T multiplied by{$v/$t}at constant pressure -v]

for an ideal gas $v/$T at constant pressure becomes v/T also pv =RT {observe equations are written on per kg basis} so v/T= R/p so we get @=0

Answer 2

the value of joule Thomson coefficient for an ideal gas is 0 (zero)