Spring and mass with varying force -- what is the change in temperature?

[TroyElliott]

Homework Statement

A system consists of a mass m moving in one dimension and attached to a rigid wall by a spring having stiffness constant $K$, as shown. The mass is subjected to a constant force $F$, and is in equilibrium with the surroundings at a temperature $T$. The partition function at constant $T$ and $F$ is given by

$$Z = \frac{e^{\frac{\beta F^{2}}{2K}}}{\hbar \beta \sqrt{(K/m)}}.$$

If the system is insulated from the surroundings, and the force is slowly and reversibly decreased from $F$ to zero, what will be the new temperature (in terms of the initial temperature $T$)? What will be the final temperature if the force is abruptly reduced to zero?

Homework Equations

$\Delta S = 0$
$\ln{(Z)} = \frac{\beta F^{2}}{2K}-\ln{(\beta)}-\ln{(\hbar \sqrt{(K/m)})}$
$U = -\frac{F^{2}}{2K}+k_{b}T$

The Attempt at a Solution

Using the equation $S = k_{b}ln{(Z)}+\frac{U}{T},$ where $U = -\frac{\partial \ln{(Z)}}{\partial \beta},$ I get that $S = k_{b}(1-\ln{(\beta \hbar \sqrt{(K/m)})}).$ This equation is independent of force. Since we are dealing with an adiabatic reversible situation we know $\Delta S = 0$, thus the temperature doesn't change? Does this sound correct?

As for the part about the final temperature if he force is abruptly reduced to zero, I am not sure where to begin. Any suggestions?

Thank you very much!

 

[mark726]

Your approach for the first part is correct. Since the system is insulated and undergoing a reversible process, the entropy remains constant and thus the temperature remains constant as well. This means that the final temperature, in terms of the initial temperature T, will be the same as the initial temperature T.

For the second part, if the force is abruptly reduced to zero, then the system is no longer in equilibrium and will start to oscillate back and forth due to the spring. In this case, the partition function will change and the temperature will also change. To find the final temperature, you can use the equation for internal energy: $U = -\frac{\partial \ln{(Z)}}{\partial \beta}$, and set it equal to the initial internal energy (since energy is conserved in this process). This will give you an equation with two unknowns, T and β. You can then use the fact that the force is abruptly reduced to zero to find a relation between T and β, and solve for the final temperature T.