# Final temperature of a adiabatic process

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 ?

• 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 – Tyberius Mar 31 '17 at 2:21
• @Tyberius Gamma is $C_p/C_v$ ? and $C_p - C_v = nR$ no ? – A---B Mar 31 '17 at 2:24
• You raised it to the $\gamma$ instead of $1/\gamma$ – Chet Miller Mar 31 '17 at 3:08
• 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. – Tyberius Mar 31 '17 at 3:19
• 288 is correct. – Chet Miller Mar 31 '17 at 3:24

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).