The Second Law of Thermodynamics — Question 6

Question 6

The second law bounds cooling in an adiabatic gas expansion

One mole of monatomic ideal gas starts at 300 K and expands adiabatically to twice its initial volume. It exchanges work but no matter or heat. Use $\Delta S=\tfrac32R\ln(T_f/300)+R\ln2$ and $U=3RT/2$, with $R=8.314$ J/(mol K).

1. Find the lowest allowed final temperature and corresponding maximum work done by the gas.

2. Test a proposed final temperature of 100 K. Explain when equality in the temperature bound can be achieved.

Question 6 - Solution

The second law bounds cooling in an adiabatic gas expansion

(a) Method and calculation

$$\Delta S=S_{\mathrm{gen}}\ge0\ \Rightarrow\ln(T_f/300)\ge-\tfrac23\ln2.$$$$T_f\ge300\,2^{-2/3}\approx188.988\,\mathrm K,\quad W_{\max}=\tfrac32R(300-188.988)\approx1384.43\,\mathrm J.$$

(b) Interpretation and further calculation

$$\Delta S(100)=R[\tfrac32\ln(1/3)+\ln2]<0.$$$$T_f=T_{\min}\ \Leftrightarrow\ S_{\mathrm{gen}}=0.$$The proposed 100 K endpoint is impossible under the stated exchanges; the limiting endpoint belongs to a reversible adiabatic expansion.

Physical check

At fixed final volume, irreversibility leaves more energy in the gas and reduces extracted work.