One molar of a perfect gas of $C_v = \pu{20.18 J/K}$ at $\pu{3.25atm}$ and $\pu{310 K}$ undergoes adiabatic expansion to reach a final state of $\pu{2.50atm}$. Calculate final volume and temperature.

Since $PV^{\gamma} = const$ and $\displaystyle T^{C_v/R}V = const$,

$$V_f = V_i\left(P_i\over P_f \right)^{1\over \gamma}$$

Which I got as $\pu{0.0113m^3}$ and $V_i = \pu{0.00782 m^3}$ .

To get final temperature I used $$T_f = T_i\left(V_i\over V_f \right)^{R\over C_v} = 310\left(0.00782\over 0.0113 \right)^{8.314\over 20.8} = \pu{267K}$$

But when I use equation of state to get final temperature I get $\pu{344K}$ which is the correct answer.

Why did not equation of state and $\displaystyle T^{C_v/R}V= const$ match in this case ?

  • $\begingroup$ How are you obtaining Gamma? Your equation for the volume and temperature also doesn't look right. Where are you getting $\frac {\text{C}_v}{\mathrm{R}}$? en.wikipedia.org/wiki/Adiabatic_process $\endgroup$
    – Tyberius
    Mar 31, 2017 at 2:21
  • $\begingroup$ @Tyberius Gamma is $C_p/C_v$ ? and $C_p - C_v = nR $ no ? $\endgroup$
    – user31607
    Mar 31, 2017 at 2:24
  • 2
    $\begingroup$ You raised it to the $\gamma$ instead of $1/\gamma$ $\endgroup$ Mar 31, 2017 at 3:08
  • 1
    $\begingroup$ With that new value for $V_f$, the equation of state should also give $T_f=288K$. The textbook has to be wrong, because an increase in temperature cannot happen for an adiabatic expansion. $\endgroup$
    – Tyberius
    Mar 31, 2017 at 3:19
  • 1
    $\begingroup$ 288 is correct. $\endgroup$ Mar 31, 2017 at 3:24

1 Answer 1


Your error stems from a miscalculation of $\mathrm{V_f}$ in the first calculation. As mentioned in the comments, it appears that you actually exponentiated by $\gamma$ rather than $\frac 1 \gamma$. When you differentiate by $\frac 1 \gamma$ , you should obtain $\mathrm{V_f}=0.0094 \mathrm{m^3}$.

Plugging this into your expression for $\mathrm{T_f}$, you should obtain:$$T_f = T_i\left(V_i\over V_f \right)^{R\over C_v} = 310\left(0.00782\over 0.0094 \right)^{8.314\over 20.8} = 288\mathrm{K}$$

Similarly, inserting this into the ideal gas equation, you should obtain: $$\mathrm{T_f= {P_fV_f \over R}}={(2.5)(0.0094) \over(8.206\times10^{-5})}=288\mathrm{K}$$ (The ideal equation will be slightly off as written due to rounding).

In regards to the textbook solution, the final temperature could not possibly be higher than the initial. An adiabatic expansion always reduces the temperature. Intuitively, we can understand this by noting that an expansion with no heat exchange should decrease the internal energy and thus decrease the temperature of the system.


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