Some Special Cases of the First Law of Thermodynamics — Question 1

Question 1

Rigid-vessel heating converts all thermal input into internal energy

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 rigid vessel of volume $.0100\,\mathrm{m^3}$ initially holds the gas at 300 K. It receives 900 J of heat, and pressure-volume work is the only possible work mechanism.

1. Find work, final temperature and final pressure.

2. How much heat must subsequently leave to restore the initial pressure at the same volume? Explain why pressure changes despite zero displacement work.

Question 1 - Solution

Rigid-vessel heating converts all thermal input into internal energy

(a) Method and calculation

$$W=\int p\,dV=0,\quad\Delta U=Q=900\,\mathrm J.$$$$T_f=300+900/15=360\,\mathrm K,\quad p_f=\frac{10(360)}{.0100}=360\,\mathrm{kPa}.$$

(b) Interpretation and further calculation

$$p_i=\frac{10(300)}{.0100}=300\,\mathrm{kPa},\quad\frac pT=\frac BV\quad\text{constant}.$$$$Q_{\mathrm{return}}=15(300-360)=-900\,\mathrm J.$$Work depends on boundary displacement; a changing force on an unmoving wall does no displacement work.

Physical check

At fixed volume the ideal-gas pressure ratio equals the absolute-temperature ratio, $360/300=1.20$.