Question 16.7: A free expansion Let’s return to one of our examples of an i...
A free expansion
Let’s return to one of our examples of an irreversible process, a thermally insulated box that is divided by a partition into two compartments, each having volume V (Figure 16.12). Initially, one compartment contains n moles of an ideal gas at temperature T, and the other compartment is evacuated. We then break the partition, and the gas expands to fill both compartments. What is the entropy change in this free-expansion process?
SET UP For the given process, Q = 0, W = 0, ΔU = 0, and therefore (because the system is an ideal gas) ΔT = 0. We might think that the entropy change is zero because there is no heat exchange. But Equation 16.10 is valid only for reversible processes; the free expansion described here is not reversible, and there is an entropy change. To calculate ΔS, we use the fact that the entropy change depends only on the initial and final states. We can devise a reversible process having the same endpoints, use Equation 16.10 to calculate its entropy change, and thus determine the entropy change in the original process.
\mathrm{\Delta S=S_2 – S_1=\frac{Q}{T} } (16.10)
SOLVE The appropriate reversible process in this case is an isothermal expansion from V to 2V at temperature T, shown on the pV diagram of Figure 16.12c. We choose this process so that the final temperature of gas remains the same, as it does in the free expansion. The gas does work during this substitute expansion, so heat must be supplied to keep the internal energy constant. The total heat equals the total work, which is given by Equation 15.17:
\mathrm{W=Q=nRt \ln\frac{V_2}{V_1} .}
Thus, the entropy change is
\mathrm{\Delta S=\frac{Q}{T}=\frac{nRT \ln\frac{2V}{V} }{T}=nR \ln 2, }
and this is also the entropy change for the free expansion. For 1 mole,
\mathrm{\Delta S=( 1 mol)[8.314 J/(mol \cdot K)](0.693) }
= 5.76 J/K.
REFLECT The entropy change ΔS is proportional to the amount of substance (n moles) and is independent of the temperature T ( provided that it is constant). The increase in disorder is due to the increased randomness of position resulting from the increase in volume.
Practice Problem: In a free expansion, 2.00 mol of an ideal gas expands from an initial volume of 1.50 L to a final volume of 6.00 L. What is the increase in entropy? Answer: 23.1 J/K.