GSO003 Problem 6

[State A] Given: $V_A = V_0$ , $P_A = P_0$ .

[State B] During process A $\to$ B, let $x$ be the upward displacement of the piston (compression of the spring) when volume is $V$ :

V - V_0 = Sx \implies x = \frac{V - V_0}{S}

Force balance on the massless piston (gas pressure upward; atmospheric pressure and spring force downward):

PS = P_0S + kx

Substituting $x$ :

P = P_0 + \frac{k}{S^2}(V - V_0)

At State B, $V_B = 4V_0$ . Using $k = \frac{2P_0S^2}{3V_0}$ :

P_B = P_0 + \frac{2P_0S^2}{3V_0 S^2}(4V_0 - V_0) = P_0 + 2P_0 = 3P_0

[State C] Process B $\to$ C is isochoric (constant volume), so $V_C = 4V_0$ . Cooling reduces pressure to $P_C = P_0$ .

Using the first law of thermodynamics $Q_{in} = \Delta U + W_{out}$ and $U = \frac{3}{2}PV$ for a monatomic ideal gas:

[Process A $\to$ B (heat absorption)] The $P$ – $V$ path is a straight line from $(V_0, P_0)$ to $(4V_0, 3P_0)$ . Work done by the gas:

W_{AB} = \frac{1}{2}(P_A + P_B)(V_B - V_A) = \frac{1}{2}(P_0 + 3P_0)(4V_0 - V_0) = 6P_0V_0

Change in internal energy:

\Delta U_{AB} = \frac{3}{2}(P_BV_B - P_AV_A) = \frac{3}{2}(12P_0V_0 - P_0V_0) = \frac{33}{2}P_0V_0

Heat absorbed:

Q_{AB} = \Delta U_{AB} + W_{AB} = \frac{33}{2}P_0V_0 + 6P_0V_0 = \frac{45}{2}P_0V_0

[Process B $\to$ C (heat release)] Isochoric, so $W_{BC} = 0$ .

\Delta U_{BC} = \frac{3}{2}(P_CV_C - P_BV_B) = \frac{3}{2}(4P_0V_0 - 12P_0V_0) = -12P_0V_0

$Q_{BC} = -12P_0V_0$ (heat released).

[Process C $\to$ A (heat release)] Isobaric at $P_0$ after the spring is removed.

W_{CA} = P_0(V_A - V_C) = P_0(V_0 - 4V_0) = -3P_0V_0

\Delta U_{CA} = \frac{3}{2}(P_AV_A - P_CV_C) = \frac{3}{2}(-3P_0V_0) = -\frac{9}{2}P_0V_0

$Q_{CA} = -\frac{9}{2}P_0V_0 - 3P_0V_0 = -\frac{15}{2}P_0V_0$ (heat released).

Net work done per cycle:

W_{net} = W_{AB} + W_{BC} + W_{CA} = 6P_0V_0 + 0 - 3P_0V_0 = 3P_0V_0

Total heat absorbed per cycle (only process A $\to$ B absorbs heat):

Q_{in} = Q_{AB} = \frac{45}{2}P_0V_0

Thermal efficiency:

e = \frac{W_{net}}{Q_{in}} = \frac{3P_0V_0}{\frac{45}{2}P_0V_0} = \frac{6}{45} = \frac{2}{15}

So $a = 2$ , $b = 15$ , and $a + b = 2 + 15 = 17$ .

Answer: 17

Thermodynamic Cycle with a Spring-Loaded and Stopper-Controlled Piston