A heat engine performs the reversible cycle abca with 9.0 moles of an ideal gas,as shown in the figure.Path ca is an adiabatic process.The temperatures at points a and b are 300 K and 500 K,respectively.The volume at point c is 0.20 m³.The adiabatic constant of the gas is 1.60.The thermal efficiency of this engine is closest to A)0.070. B)0.10. C)0.13. D)0.16. E)0.19.

Question

Quizplus · Accepted Answer

The thermal efficiency of a reversible cycle is given by 
$\eta = 1-\frac{T_{	ext{cold}}}{T_{	ext{hot}}}$.
In this case, $T_{	ext{hot}}=500	ext{ K}$ and $T_{	ext{cold}}$ can be found from the adiabatic expansion process from point $c$ to point $a$.
Using $PV^{\gamma}$ = constant for an adiabatic process, and the fact that the process is reversible, we can write:

$\frac{P_cV_c^{\gamma}}{T_c}=\frac{P_aV_a^{\gamma}}{T_a}$

Since process $ca$ is adiabatic, we have:

$PV^{\gamma}=$ constant $\Rightarrow P_cV_c^{\gamma}=P_aV_a^{\gamma}$

Combining both equations gives:

$\frac{T_c}{T_a}=\left(\frac{V_a}{V_c}ight)^{\gamma}= 36$

Therefore, $T_{	ext{cold}}=300	ext{ K}/36 = 8.3	ext{ K}$.
The efficiency is therefore $\eta=1-\frac{8.3}{500}\approx0.13$, which is closest to choice (C).

A Heat Engine Performs the Reversible Cycle Abca with 9