Thermodynamic Cycle Analysis ($C$ vs $\theta$)
We are analyzing a cyclic process $A \to B \to C \to A$ for one mole of an ideal mono-atomic gas. The process is defined on a graph of Molar Heat Capacity ($C$) versus Temperature ($\theta$).
1. Calculation of Heat Received ($Q_{in}$)
The heat exchange in a process involving an ideal gas is given by $dQ = nCdT$. Since $n=1$, the total heat is the area under the curve on a $C$ vs $T$ diagram:
$$Q = \int C \, d\theta$$Heat is absorbed (received) when temperature increases ($d\theta > 0$) and $C$ is positive. This corresponds to the path $A \to B \to C$.
The total heat absorbed, $Q_{in}$, is the area of the trapezoid formed by the path $A \to B \to C$ and the temperature axis:
$$Q_{in} = \text{Area}(\text{Trapezoid under } ABC)$$ $$Q_{in} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times (\text{Height})$$It is simpler to view this as the area of the triangle $ABC$ plus the rectangle beneath it ($A-C-\theta_{150}-\theta_{50}$).
$$Q_{in} = \text{Area}(\triangle ABC) + \text{Area}(\text{Rectangle beneath } AC)$$Base width ($\Delta \theta$) = $150 – 50 = 100\, \text{K}$.
Height of triangle = $3.0R – 1.5R = 1.5R$.
Height of rectangle = $1.5R$.
In terms of specific points: $$Q_{in} = \frac{9R}{4}(\theta_C – \theta_A) \quad (\text{Since } \frac{9}{4} \times 100R = 225R)$$ Substituting $R = 8.31 \, \text{J/(mol}\cdot\text{K)}$:
$$Q_{in} = 225 \times 8.31 = 1869.75 \, \text{J} \approx \mathbf{1.9 \, \text{kJ}}$$2. Work Done in the Cycle ($W$)
From the First Law of Thermodynamics for a complete cycle, $\Delta U = 0$, therefore:
$$W_{cycle} = Q_{net} = Q_{in} – |Q_{out}|$$Heat is rejected in the process $C \to A$ (where temperature decreases from $150 \to 50$):
$$Q_{out} = \int_{150}^{50} C_{CA} \, d\theta$$For $C \to A$, $C$ is constant at $1.5R$.
$$Q_{CA} = 1.5R(50 – 150) = -150R$$ $$|Q_{out}| = 150R$$Thus, the net work done is:
$$W = Q_{in} – |Q_{out}| = 225R – 150R = 75R$$Generally expressed as: $W = \frac{3R}{4}(\theta_C – \theta_A)$.
$$W = 75 \times 8.31 = 623.25 \, \text{J} \approx \mathbf{0.62 \, \text{kJ}}$$3. Efficiency ($\eta$)
The efficiency of the cycle is defined as:
$$\eta = \frac{W}{Q_{in}} \times 100\%$$ $$\eta = \frac{75R}{225R} \times 100\% = \frac{1}{3} \times 100\%$$ $$\eta \approx \mathbf{33\%}$$