Some Special Cases of the First Law of Thermodynamics — Question 10

Question 10

A two-stage route separates isochoric and isothermal heating

Use $\Delta U=Q-W$, with work by the gas positive. The supplied gas model is $pV=BT$, with $B=10.0\,\mathrm{J/K}$, and $\Delta U=C_V\Delta T$, with total constant-volume heat capacity $C_V=15.0\,\mathrm{J/K}$. A gas starts at A with $T_A=300$ K and $V_A=.0100\,\mathrm{m^3}$. It first warms at fixed volume to B at 450 K, then expands quasistatically at 450 K to C with $V_C=.0200\,\mathrm{m^3}$. Only pressure-volume work occurs.

1. Find $Q$, $W$ and $\Delta U$ for each stage and for the full route.

2. Compare with the reversed order: isothermal expansion at 300 K to .0200 m$^3$, followed by fixed-volume warming to 450 K. Find the work and heat differences and explain why the final internal energy agrees.

Question 10 - Solution

A two-stage route separates isochoric and isothermal heating

(a) Method and calculation

$$W_{AB}=0,\quad\Delta U_{AB}=Q_{AB}=15(150)=2250\,\mathrm J.$$$$\Delta U_{BC}=0,\quad W_{BC}=Q_{BC}=4500\ln2\approx3119.162\,\mathrm J.$$$$\Delta U_{AC}=2250\,\mathrm J,\quad W_{AC}=3119.162\,\mathrm J,\quad Q_{AC}=5369.162\,\mathrm J.$$

(b) Interpretation and further calculation

$$W_{\mathrm{other}}=3000\ln2\approx2079.442\,\mathrm J,\quad Q_{\mathrm{other}}=2250+2079.442=4329.442\,\mathrm J.$$$$\Delta W=\Delta Q=1500\ln2\approx1039.721\,\mathrm J.$$The hotter isotherm gives more expansion work. Both routes end at the same $T,V$, so their common $\Delta U$ cancels when their first-law equations are subtracted.

Physical check

The additional heat for the first route equals its additional work because endpoint internal energy is unchanged between routes.