Solution: Temperature Change in Spring-Piston System
Step 1: Initial State Analysis
Relaxed length of spring is $l_0 = l/2$.
Current length of the left chamber is $l_A = l/2 + x$.
Length of the cylinder is $2l$, so length of right chamber is $l_B = 2l – (l/2 + x) = 3l/2 – x$.
Force Balance on Piston:
$$ P_A S – P_B S = k x $$
$$ \frac{nRT}{l_A} – \frac{nRT}{l_B} = kx $$
$$ nRT \left( \frac{1}{l_A} – \frac{1}{l_B} \right) = kx \implies k = \frac{nRT}{x} \left( \frac{l_B – l_A}{l_A l_B} \right) $$
Step 2: Energy Conservation
When the hole is made, the gas mixes and pressures equalize. The spring returns to its relaxed state (stored potential energy is released). The system is insulated, so total energy is conserved. $$ \Delta U_{gas} + \Delta U_{spring} = 0 $$ $$ 2n C_v (T_f – T) + (0 – \frac{1}{2} k x^2) = 0 $$ For monoatomic gas, $C_v = \frac{3}{2}R$. $$ 2n \left(\frac{3}{2}R\right) \Delta T = \frac{1}{2} k x^2 $$ $$ 3 n R \Delta T = \frac{1}{2} k x^2 $$
Step 3: Calculating Temperature Change
Substitute the expression for $k$ derived in Step 1:
$$ 3 n R \Delta T = \frac{1}{2} \left[ \frac{nRT}{x} \left( \frac{l_B – l_A}{l_A l_B} \right) \right] x^2 $$
$$ 3 \Delta T = \frac{1}{2} T x \left( \frac{l_B – l_A}{l_A l_B} \right) $$
Now substitute lengths $l_A = \frac{1}{2}(l+2x)$ and $l_B = \frac{1}{2}(3l-2x)$:
Numerator: $l_B – l_A = (3l/2 – x) – (l/2 + x) = l – 2x$.
Denominator: $l_A l_B = \frac{1}{4}(l+2x)(3l-2x)$.
$$ \Delta T = \frac{Tx}{6} \frac{l-2x}{\frac{1}{4}(l+2x)(3l-2x)} $$
$$ \Delta T = \frac{2Tx(l-2x)}{3(l+2x)(3l-2x)} $$