I have a formula for finding the final temperature in an irreversible adiabatic expansion: $$T_2=\left[\frac{C_V+\frac{P_2}{P_1}}{C_p}\right]T_1,$$ where $T_1$ is the initial temperature. $C_V$ and $C_p$ are the molar heat capacities at constant volume and constant pressure, respectively.

I am interested in how this formula is derived.

  • 3
    $\begingroup$ What have you yourself tried to derive this equation? This site requires you to show some effort otherwise your question might be in danger of being closed. As for the topic at hand: Where have you gotten this equation from? The equation, as it stands, cannot be correct, because if you make a dimensional analysis you see that $C_v$ and $\frac{p_2}{p_1}$ don't have the same unit. However, I've seen a similar equation that contains a term $\frac{p_2}{p_1}R$ instead of $\frac{p_2}{p_1}$. $\endgroup$
    – Philipp
    Commented Sep 13, 2014 at 21:23
  • $\begingroup$ I have not the slightest idea from where has this equation come. It has been given in one of the rather incomplete solutions. Maybe it is wrong. However, even if it were $\frac{p_2}{p_1}R$ instead of just $\frac{p_2}{p_1}$, how is it arrived at? Or is it necessary to have one more parameter to determine the final temperature? $\endgroup$
    – Tejas
    Commented Sep 14, 2014 at 18:39

2 Answers 2


When you first posted this question I was very busy, but now I've found the time to look at it again and can give you an answer:

As I already mentioned in one of my comments your equation is slightly incorrect. It should actually be

\begin{align} T_2=\left[\frac{c_{V} + R\frac{p_2}{p_1}}{c_{p}}\right]T_1 \end{align}

Now, I will come to the derivation: You should always start out by identifying what you already know about the system. You know the expansion will be performed adiabatically, so there can be not heat exchange with the environment, i.e.

\begin{align} \delta Q = 0 \ . \end{align}

This means that from the first law of thermodynamics you get:

\begin{align} \mathrm{d} U = \delta Q + \delta W = \delta W \end{align}

where $U$ is the internal energy and $W$ is the work. Furthermore, you know that the expansion process is irreversible. If the only work performed is the pressure-volume work, where the system has to expand against a constant external pressure, which is $p_2$ in your equation, then this work is given by:

\begin{align} \delta W = - p_{2} \mathrm{d} V \qquad \text{where:} \qquad p_{2} = \text{const.} \neq f(V) \end{align}

So, together with the first law of thermodynamics this yields:

\begin{align} \mathrm{d} U = - p_{2} \mathrm{d} V \ . \tag{1} \end{align}

Now, you need to assume that the expanding gas is an ideal gas. For ideal gases you know that

\begin{align} \mathrm{d} U &= \underbrace{\frac{\partial U}{\partial T}}_{= \, C_{V}} \mathrm{d} T + \underbrace{\frac{\partial U}{\partial V}}_{= \, \Pi \, = \, 0} \mathrm{d} V \\ \mathrm{d} U &= C_{V} \mathrm{d} T \tag{2} \end{align}

where $T$ is the temperature, $C_{V}$ is the heat capacity at constant volume, and $\Pi$ is the internal pressure, which is equal to zero for an ideal gas. Other relations that hold for ideal gases are the ideal gas law

\begin{align} p V = n R T \end{align}

where $n$ is the amount of the gas and $R$ is the universal gas constant, and

\begin{align} c_{p} - c_{V} = R \tag{3} \end{align}

where $c_{p} = \frac{C_{p}}{n}$ and $c_{V} = \frac{C_{V}}{n}$ are the molar heat capacities at constant pressure and constant volume, respectively.

Now, you just have to combine equations $(1)$ and $(2)$:

\begin{align} - p_{2} \mathrm{d} V = C_{V} \mathrm{d} T \ , \end{align}

integrate the result (under the assumption that $C_{V}$ is temperature independent)

\begin{align} - p_{2} \int^{V_2}_{V_1} \mathrm{d} V &= C_{V} \int^{T_2}_{T_1} \mathrm{d} T \\ - p_{2} \left( V_2 - V_1 \right) &= C_{V} \left( T_2 - T_1 \right) \ , \end{align}

use the ideal gas law $V = \frac{n R T}{p}$ to substitute $p$ for $V$

\begin{align} - n R p_{2} \left( \frac{T_2}{p_2} - \frac{T_1}{p_1} \right) &= C_{V} \left( T_2 - T_1 \right) \\ - n R \left( T_2 - \frac{p_2}{p_1} T_1 \right) &= C_{V} \left( T_2 - T_1 \right) \ , \end{align}

introduce the molar heat capacity via $C_{V} = n c_{V}$

\begin{align} - R \left( T_2 - \frac{p_2}{p_1} T_1 \right) &= c_{V} \left( T_2 - T_1 \right) \ , \end{align}

and separate $T_1$ and $T_2$

\begin{align} T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &= T_2 \left( c_{V} + R \right) \ . \end{align}

Finally, you can use equation $(3)$ to get the desired result:

\begin{align} T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &= T_2 \bigl(\underbrace{c_{V} + R}_{= \, c_{p}} \bigr) \\ T_1 \left( c_{V} + \frac{p_2}{p_1} R \right) &= T_2 c_{p} \\ \Rightarrow \ T_2 &=\left[\frac{c_{V} + R\frac{p_2}{p_1}}{c_p}\right]T_1 \ . \end{align}

  • $\begingroup$ Hi there, you can use \tag{...} at the end of a line to make enumerations of equations. It's a bit easier, saves some coding and does not mess up the alignment. I have edited that in, hope you don't mind ;) Excellent answer! $\endgroup$ Commented Nov 20, 2014 at 2:30
  • $\begingroup$ @Martin Thanks for making me aware of \tag{...}. That's indeed much better than my manual approach. It never crossed my mind that there is a proper mathjax function for this kind of thing, but it's about time for me to be dragged kicking and screaming into the century of the fruit bat :) I will look over my old questions and replace my old label constructs with \tag. Up until now I usually avoided labelling equations because I thought it had to be done manually. Again, thank you very much. $\endgroup$
    – Philipp
    Commented Nov 20, 2014 at 2:46
  • $\begingroup$ No problem at all. Not long ago, only one tag was available per environment, so I had to do it also manually. But apparently they fixed that. Have a look at my sandbox on meta $\endgroup$ Commented Nov 20, 2014 at 3:02
  • $\begingroup$ why $p_2=const.$ $\endgroup$
    – RE60K
    Commented Nov 20, 2014 at 11:04
  • $\begingroup$ @eaxdpiotnyeantial In a reversible expansion you change the volume in infinitesimal steps from the initial to the final value and after each step you give the system time to equilibrate, i.e. the external pressure is equal to the inner pressure of the gas for all intermediate volume steps. So the external pressure is a function of the volume $p_{2} = f(V)$. In an irreversible expansion you change the volume abruptly from the initial to the final value, so the inner pressure of the gas is not equal to the external pressure during the process (only at the end after equilibrium is reached)... $\endgroup$
    – Philipp
    Commented Nov 20, 2014 at 12:18

$$\color{purple}{X-axis: Pressure},\color{green}{Y-axis: Volume}$$ enter image description here

Use conditions for reversible adiabatic process: $$P_1^{1-\gamma}T_1^\gamma=P_2^{1-\gamma}T_2^\gamma|\quad P_1V_1^\gamma=P_2V_2^\gamma|\quad T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}|\quad\gamma=\frac {C_p}{C_v}$$ and the combined gas law: $$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}=\frac{P_2V_3}{T_3}$$

  • 1
    $\begingroup$ The problem with this approach is that you can't solve for the final temperature because you don't know the final volume $V_{3}$. $\endgroup$
    – Philipp
    Commented Sep 14, 2014 at 11:45
  • 1
    $\begingroup$ @Philipp so there exists no definite formula for temperature as tere can ennumerous adiabatic cycles $\endgroup$
    – RE60K
    Commented Sep 14, 2014 at 13:34
  • 1
    $\begingroup$ Well, there is a way to treat the irreversible process but that entails some different conditions and equations compared to the chain of reversible processes you are suggesting. But you are right in so far that a treatment of the chain of reversible reactions would always be ambiguous as long as you don't know $V_3$ beforehand. $\endgroup$
    – Philipp
    Commented Sep 14, 2014 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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