Solution 21: Dynamics of Two Pistons
Reasoning:
Since the pistons move without friction and the system reaches a state where the distance between them is constant, both pistons must move with the same acceleration $a$. The internal gas pressure $P$ becomes uniform.
Since the pistons move without friction and the system reaches a state where the distance between them is constant, both pistons must move with the same acceleration $a$. The internal gas pressure $P$ becomes uniform.
Applying Newton’s Second Law (assuming inward forces compress the gas):
For Piston A ($m_1$):
$$ F_1 + P_0 S – P S = m_1 a \quad \dots(1) $$
For Piston B ($m_2$):
$$ P S – P_0 S – F_2 = m_2 a \quad \dots(2) $$
Eliminating acceleration $a$ by equating $\frac{F_{net,1}}{m_1} = \frac{F_{net,2}}{m_2}$:
$$ \frac{F_1 + P_0 S – P S}{m_1} = \frac{P S – P_0 S – F_2}{m_2} $$
Solving for Pressure $P$:
$$ m_2(F_1 + P_0 S – P S) = m_1(P S – P_0 S – F_2) $$ $$ m_2 F_1 + m_2 P_0 S – m_2 P S = m_1 P S – m_1 P_0 S – m_1 F_2 $$ $$ P S(m_1 + m_2) = m_2 F_1 + m_1 F_2 + P_0 S(m_1 + m_2) $$ $$ P = \frac{m_2 F_1 + m_1 F_2 + P_0 S(m_1 + m_2)}{S(m_1 + m_2)} $$
$$ m_2(F_1 + P_0 S – P S) = m_1(P S – P_0 S – F_2) $$ $$ m_2 F_1 + m_2 P_0 S – m_2 P S = m_1 P S – m_1 P_0 S – m_1 F_2 $$ $$ P S(m_1 + m_2) = m_2 F_1 + m_1 F_2 + P_0 S(m_1 + m_2) $$ $$ P = \frac{m_2 F_1 + m_1 F_2 + P_0 S(m_1 + m_2)}{S(m_1 + m_2)} $$
Final Calculation:
Using Ideal Gas Law $PV = nRT_0 \implies (Sx)P = nRT_0$: $$ x = \frac{nRT_0}{S \cdot P} $$ $$ x = \frac{nRT_0 (m_1 + m_2)}{m_1 F_2 + m_2 F_1 + P_0 S(m_1 + m_2)} $$
Using Ideal Gas Law $PV = nRT_0 \implies (Sx)P = nRT_0$: $$ x = \frac{nRT_0}{S \cdot P} $$ $$ x = \frac{nRT_0 (m_1 + m_2)}{m_1 F_2 + m_2 F_1 + P_0 S(m_1 + m_2)} $$