Released Piston in an Insulated Cylinder
A vertical cylinder with insulating walls contains 0.015 mol of diatomic gas, treated as ideal with Cv = 5R/2. A frictionless, insulating piston of mass 4.0 kg and area 2.0×10^-3 m² is held by a pin, compressing the gas to 0.12 L at 290 K. The top of the cylinder is open to the atmosphere at 101 kPa. The pin is removed and the piston is released. The piston moves upward, oscillates a few times, and eventually comes to rest. The gas remains sealed, and no gas leaks. You may assume the piston and cylinder have negligible heat capacity. The external pressure remains constant at 101 kPa, and the piston's weight contributes to the gas pressure. The motion is fast enough that heat transfer through the piston and walls is negligible during the oscillations.
1. Calculate the final equilibrium pressure of the gas and explain why it is not equal to the initial pressure.
Read the worked answer
The final pressure is set by the external forces on the piston:
p2 = p_atm + mg/A = 1.01×10^5 + (4.0)(9.81)/(2.0×10^-3) = 1.01×10^5 + 1.962×10^4 = 1.206×10^5 Pa ≈ 121 kPa.
Initially the gas pressure is p1 = nRT1/V1 = (0.015)(8.314)(290)/(0.12×10^-3) = 3.01×10^5 Pa ≈ 301 kPa.
The initial gas pressure is much greater than the pressure needed to support the piston, so on release the net force is upward and the gas expands until p2 balances the atmosphere plus the piston's weight. The final pressure is not equal to the initial pressure because the pin, not the gas, was holding the piston in place.
2. Calculate the final temperature and volume. Explain why the adiabatic relation pV^γ = constant does not apply.
Read the worked answer
The expansion is not quasi-static: the piston overshoots and the gas is stirred, so the process is irreversible and pV^γ = constant is not valid. Use energy conservation instead. No heat crosses the insulating walls, so Q = 0. The work done by the gas on the surroundings is at constant external pressure p_ext = p2 = 1.206×10^5 Pa:
W = p_ext(V2 - V1).
First law: ΔU = -W, so nCv(T2 - T1) = -p_ext(V2 - V1).
Using p_ext V2 = nRT2 and Cv = 5R/2: n(Cv + R)T2 = nCvT1 + p_ext V1. Thus T2 = [CvT1 + p_ext V1/n]/(Cv + R).
Numerically, CvT1 = (2.5)(8.314)(290) = 6.028×10^3 J/mol, p_ext V1/n = (1.206×10^5)(0.12×10^-3)/0.015 = 9.65×10^2 J/mol, and Cv + R = 3.5R = 29.10 J/(mol K). So T2 = (6.028×10^3 + 9.65×10^2)/29.10 = 240 K.
Then V2 = nRT2/p_ext = (0.015)(8.314)(240)/(1.206×10^5) = 2.48×10^-4 m³ = 0.248 L.
3. The piston oscillates before settling. Explain the origin of the oscillation and why it is damped. State why the final temperature is lower than 290 K.
Read the worked answer
Initially the gas pressure is much larger than p_ext, so the net force on the piston is upward. As the piston rises, the gas expands and its pressure falls. The piston has inertia, so it does not stop at the equilibrium position; it overshoots, compressing the gas above equilibrium and reversing the net force. This gives an oscillation about the equilibrium height. The oscillation is damped by gas viscosity, turbulence and friction in the gas, which convert coherent mechanical energy into internal energy. The final temperature is lower than 290 K because the gas does net work in lifting the piston and pushing back the atmosphere; with Q = 0, its internal energy and therefore its temperature must fall. The damping affects how the energy is redistributed but not the final energy balance, provided the final state is at rest at the equilibrium pressure.