# Why is $PV^\gamma$ constant in an adiabatic process?

In non isolated systems where there is no adiabatic process, $PV$ is constant. But the graph gets steeper in adiabatic process because of the $\gamma$ over the $V$. Why is it there in adiabatic processes and why only over the $V$?

• Dervivation of $PV^\gamma=\rm constant$. farside.ph.utexas.edu/teaching/sm1/lectures/node53.html – Farcher Apr 22 '17 at 14:05
• Who says PV is constant in non-adiabatic process? – Chet Miller Apr 22 '17 at 14:39
• I've added the ideal-gas tag, because these relationships are not general, but are applicable only to ideal gas systems. Of course a lot of teacing is done using ideal gases because it is tractable and has other attractive features like knowing the internal energy, but all of thermal physics isn't gases much less ideal gases. – dmckee --- ex-moderator kitten Apr 22 '17 at 17:36
• My reputation is too low to comment, but this Feynman Lecture shows how this expression is arrived at in an extremely natural way. – QiLin Xue Aug 31 '19 at 1:57

## 3 Answers

For an ideal gas $$PV=RT$$ Since $$dU=dQ-dW$$ For adiabatic process $$C_v dT = -{PdV}$$ Substituting $$R dT = VdP+PdV$$ $$VdP = -\frac{(R+C_v)}{C_v}PdV$$ Since $$C_p -C_v =R$$ and $$\gamma= \frac{C_p}{C_v}$$ the coefficient on RHS becomes $$\gamma$$. Integrating on both sides $$ln(PC) =-\gamma ln(V)$$ Where $$C$$ is an integration constant. Rearranging $$PV^{\gamma} = const$$

# Breaking the Question Down Into Parts to Understand It Better

To make it easier to arrive at an answer, it is often helpful to take a multi-part question and break it down into simpler parts. So I address the three parts of the question individually.

# The first part of the question is:

In non isolated systems where there is no adiabatic process, PV is constant.

## I would write:

"The ideal gas law (Boyle's Law) holds that for a confined gas in equilibrium at contact Temperature $T$, that $PV$ is constant, where $P$ is the pressure and $V$ is the volume of the confinement."

# The second part of your question was:

But the graph gets steeper in adiabatic process because of the γ over the V.

## I would write:

"But the graph gets steeper in an adiabatic process because of the $γ>1$ over the $V$ in the polytropic process equation for an ideal gas."

# In part three of your question, you finally asked:

Why is it there in adiabatic processes and why only over the V?

## I would then ask:

"Why is the $\gamma$ there in adiabatic processes?" and "Why is the $\gamma$ only over the $V$ in the formula $P V ^\gamma = Constant, \gamma > 1$ and not also over the $P$ (in example $P^\gamma V^\gamma = Constant$)?"

# The Answer

My answer is that from the graph below, that it shows constant temperature processes as isotherms and adiabatic process going from one initial isotherm to a different final isotherm. In short, temperature is generally not constant (in the shown adiabatic transitions), so generally $P V \ne Constant$ for adiabatic transitions.

• 1. You do not need to "archive" the question in your answer, that's what the revision history is for. 2. Suggesting people contact you at your website in an answer is inappropriate, and your repeated pattern of doing so borders on spam. Please stop doing this. 3. Information like pointing OP formatting help should be done in comments not answers - answers should contain solely information pertaining to the actual question posed. – ACuriousMind Jun 25 '17 at 9:42

In short, because the heat capacity at constant volume and heat capacity at constant pressure are not equal, even for ideal gas. $$\newcommand{\d}{\textrm{d}}$$

$$C_p = \frac{V\,\d p}{n\,\d T}$$

$$C_v = \frac{P\,\d V}{n\,\d T}$$

$$\gamma = \frac{C_p}{C_v} = \frac{\displaystyle\frac{\d p}{p}}{\displaystyle\frac{\d V}{V}}$$

Integrating the last equation gives $$pV^\gamma = \text{constant}$$.