Solution 27: Elastic Collision and Equilibrium
Work Done by the Gas:
The system is adiabatic ($Q=0$). The ball falls from height $H$ and eventually rests on the piston. The loss in potential energy of the ball increases the internal energy of the gas. $$ \Delta U = mgH $$ By First Law, $W_{\text{gas}} = -\Delta U = -mgH$.
The system is adiabatic ($Q=0$). The ball falls from height $H$ and eventually rests on the piston. The loss in potential energy of the ball increases the internal energy of the gas. $$ \Delta U = mgH $$ By First Law, $W_{\text{gas}} = -\Delta U = -mgH$.
Finding Height $H$:
Final state: Piston is at same height $h_i$ but now supports extra mass $m$. $$ P_f – P_i = \frac{mg}{A} $$ Since $V$ is constant ($A h_i$): $$ n R (T_f – T_i) = V(P_f – P_i) = (A h_i) \frac{mg}{A} = mgh_i $$ Energy conservation: $$ mgH = \Delta U = n C_v (T_f – T_i) $$ For Hydrogen (diatomic), $C_v = \frac{5}{2}R$. $$ mgH = \frac{5}{2} R (n(T_f – T_i)/R) = \frac{5}{2} (mgh_i) $$ $$ H = \frac{5h_i}{2} $$
Final state: Piston is at same height $h_i$ but now supports extra mass $m$. $$ P_f – P_i = \frac{mg}{A} $$ Since $V$ is constant ($A h_i$): $$ n R (T_f – T_i) = V(P_f – P_i) = (A h_i) \frac{mg}{A} = mgh_i $$ Energy conservation: $$ mgH = \Delta U = n C_v (T_f – T_i) $$ For Hydrogen (diatomic), $C_v = \frac{5}{2}R$. $$ mgH = \frac{5}{2} R (n(T_f – T_i)/R) = \frac{5}{2} (mgh_i) $$ $$ H = \frac{5h_i}{2} $$