Solution 20
1. Force Analysis:
Let the spring constant be $k$, the area of the piston be $A$, and the mass of the piston be $M$.
Initially (when pumped out), the piston touches the bottom without exerting force. This implies the weight of the piston is perfectly balanced by the spring extension $x_0$. $$ kx_0 = Mg $$ When gas is introduced and the piston rises to height $h$, the spring relaxes by $h$ (or compresses, depending on the setup, but linearly related). The force equation at equilibrium at height $h$ is: $$ P_1 A + F_{spring} = Mg $$ Since the net restoring force from the weight/spring system behaves effectively as $F_{net} = kh$, the pressure $P$ is directly proportional to height $h$: $$ P \propto h \quad \Rightarrow \quad P = C \cdot h $$ (Where $C$ is a constant involving $k$ and $A$).
Let the spring constant be $k$, the area of the piston be $A$, and the mass of the piston be $M$.
Initially (when pumped out), the piston touches the bottom without exerting force. This implies the weight of the piston is perfectly balanced by the spring extension $x_0$. $$ kx_0 = Mg $$ When gas is introduced and the piston rises to height $h$, the spring relaxes by $h$ (or compresses, depending on the setup, but linearly related). The force equation at equilibrium at height $h$ is: $$ P_1 A + F_{spring} = Mg $$ Since the net restoring force from the weight/spring system behaves effectively as $F_{net} = kh$, the pressure $P$ is directly proportional to height $h$: $$ P \propto h \quad \Rightarrow \quad P = C \cdot h $$ (Where $C$ is a constant involving $k$ and $A$).
2. Applying Ideal Gas Law:
State 1: Height $h$, Temperature $T$. $$ P_1 = C h $$ $$ V_1 = A h $$ $$ \frac{P_1 V_1}{T} = nR \Rightarrow \frac{(Ch)(Ah)}{T} = nR \Rightarrow \frac{C A h^2}{T} = nR $$
State 1: Height $h$, Temperature $T$. $$ P_1 = C h $$ $$ V_1 = A h $$ $$ \frac{P_1 V_1}{T} = nR \Rightarrow \frac{(Ch)(Ah)}{T} = nR \Rightarrow \frac{C A h^2}{T} = nR $$
3. Final State ($2T$):
State 2: New height $y$, Temperature $2T$. $$ P_2 = C y $$ $$ V_2 = A y $$ $$ \frac{P_2 V_2}{2T} = nR \Rightarrow \frac{(Cy)(Ay)}{2T} = nR \Rightarrow \frac{C A y^2}{2T} = nR $$
State 2: New height $y$, Temperature $2T$. $$ P_2 = C y $$ $$ V_2 = A y $$ $$ \frac{P_2 V_2}{2T} = nR \Rightarrow \frac{(Cy)(Ay)}{2T} = nR \Rightarrow \frac{C A y^2}{2T} = nR $$
4. Comparison:
Equating the $nR$ terms: $$ \frac{C A h^2}{T} = \frac{C A y^2}{2T} $$ $$ h^2 = \frac{y^2}{2} $$ $$ y^2 = 2h^2 $$ $$ y = \sqrt{2}h $$
Equating the $nR$ terms: $$ \frac{C A h^2}{T} = \frac{C A y^2}{2T} $$ $$ h^2 = \frac{y^2}{2} $$ $$ y^2 = 2h^2 $$ $$ y = \sqrt{2}h $$