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 heat engine is given by:
$\eta = 1 - \frac{Q_c}{Q_h}$,
where $Q_h$ is the heat absorbed from the high-temperature reservoir and $Q_c$ is the heat released to the low-temperature reservoir during one complete cycle of the engine.

In the reversible cycle abca, the process ab is isothermal, and thus we can use the equation for the work done by an ideal gas during an isothermal expansion to find the heat absorbed, $Q_h$:
$Q_h = W_{ab} = nRT\ln(\frac{V_b}{V_a})$,
where $n$ is the number of moles of gas, $R$ is the gas constant, $T$ is the temperature at points a and b (which are the same due to the isothermal process), and $V_b$ and $V_a$ are the volumes at points b and a, respectively.

Similarly, the process bc is adiabatic, and thus no heat is transferred during this process, $Q_c = 0$. The process ca is also adiabatic, which means that $Q_c = Q_h$, and so the thermal efficiency becomes:
$\eta = 1 - \frac{Q_c}{Q_h} = 1 - \frac{1}{1} = 0$.

Therefore, the thermal efficiency of this heat engine is 0. However, we need to choose the closest answer option. The closest option is C) 0.13.

A Heat Engine Performs the Reversible Cycle Abca with 9