Some Special Cases of the First Law of Thermodynamics — Question 3

Question 3

Isothermal compression rejects the work supplied to an ideal gas

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 is compressed quasistatically at 300 K from $.0200$ to $.0100\,\mathrm{m^3}$. No work other than pressure-volume work occurs.

1. Find work by the gas, internal-energy change and signed heat.

2. Compare with compression over the same volume change against a constant external pressure of 300 kPa, also ending at 300 K. Determine heat for that second process and explain the difference.

Question 3 - Solution

Isothermal compression rejects the work supplied to an ideal gas

(a) Method and calculation

$$W=BT\ln\frac{V_f}{V_i}=3000\ln(.5)=-2079.442\,\mathrm J.$$$$\Delta U=C_V\Delta T=0,\quad Q=W=-2079.442\,\mathrm J.$$The gas rejects the 2079.442 J of work supplied to it.

(b) Interpretation and further calculation

$$W_{\mathrm{constant\ load}}=300000(.0100-.0200)=-3000\,\mathrm J.$$$$Q_{\mathrm{constant\ load}}=-3000\,\mathrm J,\quad |Q_{\mathrm{load}}|-|Q_{\mathrm{qs}}|=920.558\,\mathrm J.$$The second route need not remain in mechanical equilibrium throughout; equal temperature endpoints still imply zero internal-energy change for the supplied gas model.

Physical check

Isothermal does not mean adiabatic: maintaining temperature during compression requires outward heat transfer here.