Kvant Physics Problem 166

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Verdicts: PASS + PASS
Solve time: 15m47s
Source on kvant.digital

Problem

In a horizontally oriented cylinder, on one side of a fixed piston there is 1 mole of an ideal gas. The other part of the cylinder is a vacuum. A spring located between the piston and the cylinder wall is initially undeformed (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 may be neglected.

Fig. 2

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

Setup and Assumptions

A horizontal cylinder contains $n=1$ mole of an ideal gas. A movable piston separates the gas from a vacuum. A spring is positioned between the piston and the right wall of the cylinder. Initially the spring is undeformed and the piston is fixed. The initial state of the gas is described by pressure $p_1$, volume $V_1$, and temperature $T_1$. After releasing the piston, the gas expands and reaches equilibrium, occupying a volume $V_2 = 2V_1$. The final pressure and temperature are denoted by $p_2$ and $T_2$. The cylinder is thermally insulated, and the piston, spring, and cylinder have negligible heat capacities, so the total energy is contained solely in the gas and the spring. The gas obeys the ideal-gas equation $pV = nRT$. The piston is frictionless, and the system ultimately reaches a static equilibrium.

Physical Principles

The system is isolated thermally, so $Q=0$. The first law of thermodynamics for the gas reads

$\Delta U = Q - W,$

where $\Delta U$ is the change in internal energy and $W$ is the work done by the gas on the piston. With $Q=0$, this simplifies to

$\Delta U = - W.$

For an ideal gas, the internal energy is $U = n C_V T$, so the change in internal energy is

$\Delta U = n C_V (T_2 - T_1).$

The spring has elastic potential energy

$E_s = \frac{1}{2} k x^2,$

where $x$ is the spring deformation. In the final equilibrium state, the force exerted by the spring balances the gas pressure on the piston:

$p_2 S = k x,$

where $S$ is the piston area. The piston displacement corresponds to the volume change of the gas:

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

Consequently, the spring energy at equilibrium is

$E_s = \frac{1}{2} k x^2 = \frac{1}{2} k \left(\frac{V_1}{S}\right)^2.$

Substituting $k = \frac{p_2 S}{x} = \frac{p_2 S^2}{V_1}$ yields

$E_s = \frac{1}{2} \frac{p_2 S^2}{V_1} \frac{V_1^2}{S^2} = \frac{1}{2} p_2 V_1.$

This confirms that the work done by the gas is entirely stored in the spring:

$W = E_s = \frac{1}{2} p_2 V_1.$

The ideal-gas law applies to both initial and final states:

$p_1 V_1 = n R T_1, \qquad p_2 V_2 = n R T_2.$

Derivation

The total energy of the system is conserved. Initially, the energy is solely the internal energy of the gas:

$E_i = U_1 = n C_V T_1.$

Finally, the energy is divided between the gas and the spring:

$E_f = U_2 + E_s = n C_V T_2 + \frac{1}{2} p_2 V_1.$

Equating initial and final energies gives

$n C_V T_1 = n C_V T_2 + \frac{1}{2} p_2 V_1.$

The ideal-gas law for the final state provides $p_2 V_2 = n R T_2$. With $V_2 = 2 V_1$, it follows that

$p_2 \cdot 2 V_1 = n R T_2 \quad \Rightarrow \quad p_2 V_1 = \frac{1}{2} n R T_2.$

Substituting this expression into the energy conservation equation yields

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

Dividing both sides by $n$ results in a relation between temperatures:

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

Solving for the final temperature gives

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

This shows explicitly that the final temperature depends on the gas heat capacity and is not universal; it is lower than $T_1$ because part of the internal energy is stored in the spring.

The corresponding pressure ratio follows from the ideal-gas law:

$\frac{p_2}{p_1} = \frac{n R T_2 / V_2}{n R T_1 / V_1} = \frac{T_2}{T_1} \cdot \frac{V_1}{V_2} = \frac{T_2}{T_1} \cdot \frac{1}{2}.$

Therefore, the pressure decreases relative to the initial pressure by a factor of $\frac{1}{2} \cdot \frac{C_V}{C_V + R/4}$.

For a monatomic ideal gas with $C_V = \frac{3}{2} R$, the temperature and pressure ratios are

$\frac{T_2}{T_1} = \frac{3/2}{3/2 + 1/4} = \frac{6}{7}, \qquad \frac{p_2}{p_1} = \frac{1}{2} \cdot \frac{6}{7} = \frac{3}{7}.$

For a diatomic ideal gas with $C_V = \frac{5}{2} R$, the ratios are

$\frac{T_2}{T_1} = \frac{5/2}{5/2 + 1/4} = \frac{10}{11}, \qquad \frac{p_2}{p_1} = \frac{1}{2} \cdot \frac{10}{11} = \frac{5}{11}.$

These results confirm that the final temperature and pressure decrease due to the work transferred from the gas to the spring and are correctly determined by energy conservation and the ideal-gas law.

Result

The final temperature is

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

and the final pressure is

$p_2 = \frac{1}{2} \frac{C_V}{C_V + \frac{R}{4}} p_1.$

For specific gases, these reduce to

$\text{monatomic: } T_2 = \frac{6}{7} T_1, \quad p_2 = \frac{3}{7} p_1,$

$\text{diatomic: } T_2 = \frac{10}{11} T_1, \quad p_2 = \frac{5}{11} p_1.$

The gas temperature decreases because energy is stored in the spring, and the pressure decreases both due to expansion and the reduction in temperature.

Sanity Checks

The dimensions are consistent, as the ratios $T_2 / T_1$ and $p_2 / p_1$ are dimensionless. The qualitative behavior is physically reasonable: the gas expands into a larger volume, performing work on the spring, which reduces its internal energy and therefore its temperature. The final pressure is lower than the initial pressure, consistent with both the volume increase and the temperature drop. Energy conservation is explicitly satisfied for arbitrary ideal gases, as demonstrated by substituting the heat capacities of monatomic and diatomic gases into the equations. The triangular-work argument correctly accounts for the non-constant pressure during expansion.