Some Special Cases of the First Law of Thermodynamics — Question 8

Question 8

Isothermal does not universally mean zero internal-energy change

Use $\Delta U=Q-W$, with work by the gas positive. A hypothetical substance obeys the supplied caloric relation $U(T,V)=CT+\lambda V+U_0$, with $C=50.0\,\mathrm{J/K}$ and $\lambda=1.00\times10^5\,\mathrm{J/m^3}$. During an isothermal process its volume increases by $.00200\,\mathrm{m^3}$ and it does 100 J of total work.

1. Find the internal-energy change and required heat.

2. For a separate isochoric process in which its temperature rises 4.00 K and only pressure-volume work is possible, find the heat. Explain which extra material assumption would justify $Q=W$ on an isotherm.

Question 8 - Solution

Isothermal does not universally mean zero internal-energy change

(a) Method and calculation

$$\Delta T=0,\quad\Delta U=C\Delta T+\lambda\Delta V=200\,\mathrm J.$$$$Q=\Delta U+W=200+100=300\,\mathrm J.$$

(b) Interpretation and further calculation

$$\Delta V=0\ \Rightarrow\ W=0,\quad Q=\Delta U=C(4)=200\,\mathrm J.$$$$Q=W\ \text{on an isotherm if}\ \Delta U=0.$$A caloric equation with $U$ depending only on temperature, as for the supplied ideal-gas model, provides that additional assumption; isothermal by itself does not.

Physical check

The units $\lambda\Delta V$ are joules, making the volume-dependent stored-energy contribution explicit.