I'm going through the exercises in a Thermodynamics book, just to revise and build my intuition. Right now, I'm working on:

Show that for a quasistatic adiabatic process in a perfect gas, with constant specific heats:

$$PV^\gamma = \left[\text{constant}\right]$$

with $\gamma = \frac{C_P}{C_V}$

where $P$ is pressure, $V$ is volume, and $C_V$ is the constant-volume heat capacity.

I'm not looking for the answer, just for a hint (I'm stuck and want to find the solution myself).

So those are my thoughts:

  • perfect gas means: $PV = RT$, ($R$ is the universal gas constant)
  • adiabatic means: $\mathrm{d}Q = 0$, ($Q$ for heat)
  • since there is no heat exchange, the process is reversible
  • reversible means: $\mathrm{d}W = -P \, \mathrm{d}V$, ($W$ is for work)
  • heat capacity is defined as $C_\text{V} = \left( \frac{\mathrm{d}Q}{\mathrm{d}T} \right)_V$, respectively $C_\text{P} = \left( \frac{\mathrm{d}Q}{\mathrm{d}T} \right)_\text{P}$

If I draw a $PV$ diagram for this situation, it looks like this:

Now I want to show that $PV^\gamma=\left[\text{const}\right]$ by going from $\text{State 1}$ to $\text{State 2}$ in the $PV$ diagram.

I've started like this: $$ W ~=~ -\int_{V_1}^{V_2}P \, \mathrm{d}V ~=~ -\int_{V_1}^{V_2} \frac{RT}{V} \mathrm{d}V ~=~ RT \ln{\left(\frac{V_2}{V_2}\right)} $$

This leads me into the wrong direction though. I thought about using $R = C_P - C_V$ here, but it doesn't seem to work. Any suggestions?

Please just give me a hint, not the solution.

  • 2
    $\begingroup$ You have to use $dU=c_V dT$. The answer can be found in any book. $\endgroup$
    – jinawee
    Feb 10 '14 at 16:07
  • 6
    $\begingroup$ NB: "since there is no heat exchange, the process is reversible". This is not true as far as I'm aware. The mixing of different substances is irreversible, but does not correspond to heat exchange. (On the flip side, there do exist plenty of reversible processes that involve heat exchange). When we talk about adiabatic expansions/compressions, we typically mean an expansion which is both a) reversible and b) involves no heat exchange. The key point is that the latter on its own does not imply the former. $\endgroup$
    – gj255
    Feb 10 '14 at 17:30
  • $\begingroup$ Try applying the first law of thermodynamics to get some expression with $C_p$ and/or $C_v$ and then use the definition of gamma & the ideal gas equation. $\endgroup$
    – xasthor
    Feb 18 '17 at 13:31

You have to equal the expression for $\mathrm{d}E$ given by the first law of thermodynamics with the expression that you get if you isolate $\mathrm{d}E$ from $C_V=\left(\frac{\partial E}{\partial T}\right)_V$. Then integrate both sides. With this you should be able to reach the final answer.

A more detailed explanation can be found here. Note that the solutions in the link are also done by me, and Studydrive is a free-access website.

  • $\begingroup$ Is study drive a free-access site, or is it a subscription service? (It might be useful to mention that for those who are not knowledgeable on it) $\endgroup$
    – Kyle Kanos
    Nov 19 '16 at 19:00

From the first law of thermodynamics, you have:

$$\Delta U = q + w$$

As $q$ is zero, $$\mathrm{d}U = -P \, \mathrm{d}V$$

Now put $\mathrm{d}U = C_v \, \mathrm{d}T$ (molar heat capacity at constant volume).

Therefore, $$C_v \, \mathrm{d}T = -P \, \mathrm{d}V$$

Using the ideal gas equation, substitute $P$ by $\frac{RT}{V}$.

$$-\frac{RT}{V} \mathrm{d}V = C_v \, \mathrm{d}T$$

Integrating the equation, you get $$ \begin{align} \int_{V_1}^{V_2} -\frac{RT}{V} \mathrm{d}V & = \int_{T_1}^{T_2} C_v \, \mathrm{d}T \\[10px] \frac{RT}{V} \ln \left(\frac{V_1}{V_2}\right) & = C_v \left(T_2 - T_1 \right) \end{align} $$

  • $\begingroup$ You can integrate differential equations only. $\Delta$ signifies a finite change whereas $d$ signifies an infinitely small change (differential change). An equation containing $\Delta$ cannot be integrated. $\endgroup$
    – Yashas
    Feb 18 '17 at 14:28

The mathematical condition you're missing is work = -change in internal energy (which you mentioned as dQ = 0).

dU is ideally written as n*Cv(dT).

take n=1 (for simplicity), equate PdV to -dU, bring in R = Cv (gamma - 1) (same thing as you mentioned), and integrate it. It will be of a logarithmic form with a ton of constants.

Hope this helps. Good Luck!


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