Some Special Cases of the First Law of Thermodynamics — Question 2

Question 2

Constant-pressure heating must also supply expansion work

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 $C_V=15.0\,\mathrm{J/K}$. The gas starts at 300 K and expands quasistatically at constant pressure 100 kPa until it reaches 360 K. Only pressure-volume work occurs.

1. Find initial and final volumes, work, internal-energy change and heat input.

2. Derive the total constant-pressure heat capacity for this gas. Compare heat needed for the same 60 K rise at fixed volume.

Question 2 - Solution

Constant-pressure heating must also supply expansion work

(a) Method and calculation

$$V_i=10(300)/100000=.0300\,\mathrm{m^3},\quad V_f=.0360\,\mathrm{m^3}.$$$$W=p\Delta V=600\,\mathrm J,\quad\Delta U=15(60)=900\,\mathrm J.$$$$Q=\Delta U+W=1500\,\mathrm J.$$

(b) Interpretation and further calculation

At constant pressure, $p\Delta V=B\Delta T$, so$$Q=(C_V+B)\Delta T\ \Rightarrow\ C_p=C_V+B=25.0\,\mathrm{J/K}.$$$$Q_V=C_V\Delta T=900\,\mathrm J,\quad Q_p-Q_V=600\,\mathrm J.$$The difference funds expansion work, rather than additional internal-energy gain.

Physical check

The capacities here describe this entire fixed gas sample, so their units are J/K rather than J/(mol K).