Some Special Cases of the First Law of Thermodynamics — Question 4

Question 4

Adiabatic compression raises temperature through work alone

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 thermally insulated gas initially at 300 K is compressed. A measured 750 J of mechanical work is done on the gas; no other transfers occur.

1. Find $Q$, work by the gas, internal-energy change and final temperature.

2. If instead the same 750 J input is supplied by a paddle in a rigid insulated vessel, find the final temperature. Can either work measurement alone determine the final gas volume?

Question 4 - Solution

Adiabatic compression raises temperature through work alone

(a) Method and calculation

$$Q=0,\quad W=-750\,\mathrm J,\quad\Delta U=0-(-750)=750\,\mathrm J.$$$$T_f=300+750/15=350\,\mathrm K.$$

(b) Interpretation and further calculation

$$\Delta U_{\mathrm{paddle}}=750\,\mathrm J,\quad T_{f,\mathrm{paddle}}=350\,\mathrm K.$$$$W_{pV,\mathrm{paddle}}=0,\qquad W_{\mathrm{shaft,by}}=-750\,\mathrm J.$$The paddle vessel keeps its given initial volume. The compression endpoint volume requires an additional pressure or process relation; the measured work alone does not specify it.

Physical check

Adiabatic specifies zero heat transfer, not zero work or zero temperature change.