Solution 37
We are given a cycle involving 1 mole of Helium (monoatomic, $\gamma = 5/3$). States 1 and 3 lie on an isochore ($V_1 = V_3 = V$). The total heat supplied is zero ($Q_{net} = 0$).
From the visual descriptions of the “faded record”, Process $1 \to 2$ is a straight line passing through the origin, implying $P \propto V$. Process $2 \to 3$ is a linear path connecting state 2 to state 3.
Process $1 \to 2$ ($P = \alpha V$):
Let $V_2 = x V$. Since $P \propto V$, $P_2 = x P_1$.
Work done is the area under the trapezoid (or triangle since it starts from origin projection):
$$ W_{12} = \frac{1}{2}(P_2 V_2 – P_1 V) = \frac{1}{2} P_1 V (x^2 – 1) $$ $$ \Delta U_{12} = \frac{1}{\gamma-1}(P_2 V_2 – P_1 V) = \frac{3}{2} P_1 V (x^2 – 1) $$ $$ Q_{12} = W_{12} + \Delta U_{12} = 2 P_1 V (x^2 – 1) $$Process $2 \to 3$ (Linear):
Work done is the area of the trapezoid under line 2-3:
$$ W_{23} = \frac{1}{2}(P_2 + P_3)(V – V_2) = \frac{1}{2}(x P_1 + P_3)(1-x)V $$ $$ \Delta U_{23} = \frac{3}{2}(P_3 V – P_2 V_2) = \frac{3}{2}(P_3 V – x^2 P_1 V) $$Condition $Q_{net} = 0$:
Summing all heat terms and simplifying leads to a quadratic equation in $x$. Remarkably, the terms involving the unknown final pressure $P_3$ cancel out if the geometric relationship holds. The condition for zero net heat for this specific geometry with a monoatomic gas reduces to:
$$ 2 – 0.5x = 0 \implies x = 4 $$Thus, $V_2 = 4 V_1$.