Kvant Physics Problem 166

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Problem

In a horizontally oriented cylinder, on one side of a fixed piston there is 1 mole of an ideal gas. In the other part of the cylinder there is a vacuum. A spring located between the piston and the cylinder wall is in its undeformed state (Fig. 2). The cylinder is thermally insulated from the surroundings. The piston is released, and after equilibrium is established, the volume occupied by the gas doubles. How do the temperature and pressure of the gas change? The heat capacities of the cylinder, piston, and spring are negligible.

Fig. 2

E. I. Butikov, A. A. Bykov, A. S. Kondratyev

Setup and Assumptions

A horizontal cylinder contains a frictionless piston of mass negligible for energy accounting. On the left side of the piston there is $n=1$ mol of an ideal gas. On the right side there is vacuum, so no external pressure acts on that side. A spring connects the piston to the fixed cylinder wall on the right; initially the spring is undeformed.

The system is thermally insulated, so no heat is exchanged with the environment. The piston is released from rest and reaches a final static equilibrium. The gas expands until its volume doubles, so $V_2 = 2V_1$. The piston displacement is $x$, and the gas cross-sectional area is $A$, so $x = (V_2 - V_1)/A = V_1/A$.

The gas is ideal, with internal energy depending only on temperature: $U = n C_V T$. Heat capacities of the piston, cylinder, and spring are neglected, so all work done by the gas is stored in the gas internal energy and the spring elastic energy.

Unknowns are the final temperature $T_2$ and pressure $P_2$ of the gas.

Physical Principles

The first law of thermodynamics for the gas reads $ \Delta U = Q - W$, with $Q=0$ due to insulation, so $\Delta U = -W$.

The internal energy of an ideal gas satisfies $\Delta U = n C_V (T_2 - T_1)$.

The work done by the gas is stored entirely in the spring at the final state, since the system ends at rest. The spring potential energy is $E_s = \tfrac{1}{2} k x^2$, so $W = \tfrac{1}{2} k x^2$.

Mechanical equilibrium at the final state gives force balance on the piston: $P_2 A = kx$, hence $P_2 = kx/A$.

The ideal gas law holds at each equilibrium state: $P V = n R T$.

Derivation

The piston displacement is determined from the volume change,

$x = \frac{V_2 - V_1}{A} = \frac{V_1}{A}.$

The spring energy at the final state becomes

$E_s = \frac{1}{2} k x^2 = \frac{1}{2} k \frac{V_1^2}{A^2}.$

The thermodynamic work done by the gas equals this energy,

$W = \frac{1}{2} k \frac{V_1^2}{A^2}.$

The first law gives

$n C_V (T_2 - T_1) = -\frac{1}{2} k \frac{V_1^2}{A^2}.$

From force equilibrium,

$P_2 = \frac{kx}{A} = \frac{k V_1}{A^2}.$

This eliminates the mechanical constants by expressing $k/A^2$ in terms of $P_2$,

$\frac{k}{A^2} = \frac{P_2}{V_1}.$

Substituting into the energy relation,

$n C_V (T_2 - T_1) = -\frac{1}{2} \frac{P_2}{V_1} V_1^2 = -\frac{1}{2} P_2 V_1.$

The ideal gas law in the final state with $V_2 = 2V_1$ gives

$P_2 (2V_1) = n R T_2,$

$P_2 V_1 = \frac{n R T_2}{2}.$

Substituting this into the energy equation yields

$n C_V (T_2 - T_1) = -\frac{1}{2} \cdot \frac{n R T_2}{2} = -\frac{n R T_2}{4}.$

Dividing by $n$,

$C_V (T_2 - T_1) = -\frac{R}{4} T_2.$

Rearranging terms,

$C_V T_2 - C_V T_1 = -\frac{R}{4} T_2,$

$T_2 \left(C_V + \frac{R}{4}\right) = C_V T_1.$

Hence,

$T_2 = \frac{C_V}{C_V + \frac{R}{4}} T_1.$

The pressure follows from the ideal gas law. Initially,

$P_1 V_1 = n R T_1,$

so $P_1 = RT_1/V_1$. In the final state,

$P_2 V_1 = \frac{n R T_2}{2},$

which gives

$P_2 = \frac{R T_2}{2V_1}.$

Taking the ratio,

$\frac{P_2}{P_1} = \frac{T_2}{2T_1},$

$P_2 = \frac{T_2}{2T_1} P_1.$

Result

The final temperature is

$T_2 = \frac{C_V}{C_V + \frac{R}{4}} T_1.$

For one mole of gas, substituting $n=1$ implicitly in $C_V$ (molar heat capacity),

$\boxed{T_2 = \frac{C_V}{C_V + \frac{R}{4}}, T_1}.$

The final pressure is

$P_2 = \frac{T_2}{2T_1} P_1,$

$\boxed{P_2 = \frac{C_V}{2\left(C_V + \frac{R}{4}\right)}, P_1}.$

Sanity Checks

The combination $C_V + R/4$ has units of $R$, so the expression for $T_2$ is dimensionally consistent as a ratio of energies per mole. The pressure expression reduces correctly to $P_2 \propto P_1$, ensuring correct scaling with the initial state.

If the spring were infinitely stiff, $C_V \ll R$ in comparison would force $T_2 \ll T_1$, corresponding to strong cooling due to work stored in the spring. If the spring were absent, the model breaks because the final volume would not be bounded, showing that the spring is essential for equilibrium. The substitution $P_2 V_1 = nRT_2/2$ is the step where a missing factor of two would incorrectly propagate into both final temperature and pressure.