2
$\begingroup$

Is $p\cdot V^γ= \mathrm{const.}$ valid for only adiabatic reversible process and not for adiabatic irreversible process if not why?
Today I was doing an objective question as mentioned above and chose option "any adiabatic process" but the ans was "only reversible adiabatic process" but i couldn't understand why. All i know is that path followed by both process will be different but how will it affect the adiabatic equation i.e $p\cdot V^γ= \mathrm{const}$.

$\endgroup$
2
$\begingroup$

An adiabatic and reversible process (i.e. isentropic) for ideal gases can be described as follows1:

$p \cdot V^\gamma = \mathrm{const.}$

Here $\gamma$ is the heat capacity ratio. Other processes can be described by changing the exponent. The equation for a polytropic process is:

$p \cdot V^n= \mathrm{const.}$

For $n=1$ this equation describes a isothermal process and for values greater $\gamma$ this equation can be used to describe irreversible processes.


[1]: Wikipedia: Isentropic process

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.