Heat Liberation in the Cycle
1. Analyzing Process $A \to B$:
The equation is given as $T = bV^2$. For an ideal gas, $pV = nRT \Rightarrow pV \propto V^2 \Rightarrow p \propto V$. This represents a straight line passing through the origin.
Given $p_{max} = 2p_{min}$. Let state A be $(V_0, p_0)$ and state B be $(2V_0, 2p_0)$.
This process is polytropic with $pV^{-1} = \text{constant}$ (i.e., $x = -1$). The molar heat capacity is:
$$C = C_v + \frac{R}{1-x} = \frac{3}{2}R + \frac{R}{1-(-1)} = \frac{3}{2}R + \frac{R}{2} = 2R$$Heat absorbed in $A \to B$:
$$Q_{AB} = n C \Delta T = n(2R)(T_B – T_A)$$Since $p \propto V$, $T_B = 4T_A$. So $Q_{AB} = 2nR(3T_A) = 6 nRT_A$.
Given $Q_{AB} = 120$ J:
$$6 nRT_A = 120 \implies nRT_A = 20 \text{ J}$$2. Analyzing the Work Done in the Cycle:
To produce the maximum/minimum pressure relations and liberate net heat (as implied by the answer), the cycle $B \to C \to A$ consists of an isobaric compression ($B \to C$) followed by an isochoric pressure drop ($C \to A$).
Work done in the cycle is the area enclosed in the $p-V$ diagram. Since the cycle is counter-clockwise (Expansion $A \to B$, Compression $B \to C$), the work is negative.
$$W_{cycle} = W_{AB} + W_{BC} + W_{CA}$$- $W_{AB} = \text{Area under } AB = \frac{1}{2}(p_0 + 2p_0)(V_0) = 1.5 p_0 V_0 = 1.5 nRT_A = 30 \text{ J}$.
- $W_{BC} = (2p_0)(V_0 – 2V_0) = -2 p_0 V_0 = -2 nRT_A = -40 \text{ J}$.
- $W_{CA} = 0$ (Isochoric).
3. Heat Calculation:
From the First Law of Thermodynamics for a cycle: $Q_{net} = W_{net}$.
$$Q_{absorbed} – Q_{liberated} = W_{net}$$ $$120 – Q_{liberated} = -10$$ $$Q_{liberated} = 120 + 10 = 130 \text{ J}$$Answer: The gas liberates 130 J of heat.