Thermodynamics

Question. An aerobic fermentation runs in a bioreactor vented to hold a constant headspace pressure of 101.3 kPa (1 atm). Over one hour, the culture's respiration is exothermic enough that the cooling jacket has to remove 850 kJ of heat to hold temperature — so $q_p = -850$ kJ. Off-gas analysis shows CO$_2$ production exceeds O$_2$ consumption enough that the gas phase expands by 50 L ($5.0 \times 10^{-2}$ m³) over that hour. Find $\Delta U$ for the hour, and compare it with $\Delta H$.

Solution.

1. At constant pressure, $\Delta H = q_p = -850$ kJ.
2. Compute the pressure–volume work term: $P\Delta V = 101{,}300\ \text{Pa} \times 0.050\ \text{m}^3 = 5{,}065\ \text{J} \approx 5.1$ kJ.
3. Rearrange $\Delta H = \Delta U + P\Delta V$ to solve for $\Delta U$: $\Delta U = \Delta H - P\Delta V = -850 - 5.1 = -855.1$ kJ.

Answer: $\Delta U \approx -855.1$ kJ, compared with $\Delta H = -850$ kJ — a difference of about 5 kJ, or roughly 0.6% of the total.

That 5 kJ is real energy: it is what left the system as gas-phase expansion work against the vent, not as heat the jacket could remove. It is small here because the gas volume change is small relative to the heat released, but it is exactly the term you would need to close an energy balance around the headspace or size an off-gas system — and in a sealed, gas-generating process where $\Delta V$ is larger, it stops being a rounding error.

Question. Over one hour, a growing microbial culture assembles disordered nutrients (glucose, amino acids) into ordered biomass, decreasing the system's entropy by $\Delta S_{sys} = -2.5$ kJ/K. In that same hour, the culture's respiration releases 850 kJ of heat to the surroundings (the reactor's cooling water), held at a constant $T = 310$ K (37 °C). Find $\Delta S_{surr}$ and $\Delta S_{universe}$, and check whether the second law holds.

Solution.

1. For a large reservoir at constant temperature, heat released by the system raises the surroundings' entropy by $\Delta S_{surr} = q / T$.
2. $\Delta S_{surr} = \dfrac{850{,}000\ \text{J}}{310\ \text{K}} \approx 2{,}742\ \text{J/K} \approx 2.74$ kJ/K.
3. $\Delta S_{universe} = \Delta S_{sys} + \Delta S_{surr} = -2.5 + 2.74 = +0.24$ kJ/K.

Answer: $\Delta S_{universe} \approx +0.24$ kJ/K — positive, so the second law holds. The cell's own entropy fell, but the heat it released into the surroundings raised their entropy by more, so the total for system plus surroundings still increased.