Some Special Cases of the First Law of Thermodynamics — Question 5

Question 5

Free expansion preserves internal energy but need not preserve temperature

Use $\Delta U=Q-W$, with work by the gas positive. A gas expands into vacuum inside a rigid insulated enclosure, from $V_i=.0100$ to $V_f=.0200\,\mathrm{m^3}$. The initial temperature is 300 K. Neglect partition energy. Compare two supplied caloric models: ideal $U=CT+U_0$ and nonideal $U=CT-a/V+U_0$, with $C=15.0\,\mathrm{J/K}$ and $a=6.00\,\mathrm{J\,m^3}$.

1. Find $Q$, $W$, $\Delta U$ and the ideal-model final temperature.

2. Find the nonideal-model final temperature. Explain the physical role of the volume-dependent energy term in the result.

Question 5 - Solution

Free expansion preserves internal energy but need not preserve temperature

(a) Method and calculation

$$Q=0,\quad W=0,\quad\Delta U=0.$$$$C(T_f-T_i)=0\ \Rightarrow\ T_{f,\mathrm{ideal}}=300\,\mathrm K.$$

(b) Interpretation and further calculation

$$0=C(T_f-T_i)-a\left(\frac1{V_f}-\frac1{V_i}\right).$$$$T_f=300+\frac6{15}(50-100)=280\,\mathrm K.$$The term $-a/V$ increases by 300 J as volume grows; the $CT$ term must fall by the same amount. Constant internal energy alone does not imply constant temperature for every substance.

Physical check

No quasistatic adiabatic pressure law should be imposed on free expansion.