20.9.6 Entropy and Life

The second law of thermodynamics is sometimes interpreted as saying that “things run down,” that order becomes disorder. But living systems seem to violate this rule:

Plants grow from simple seeds to complex, organized entities.
Single-cell fertilized eggs grow into complex, organized adults.
Over billions of years, life has evolved from simple unicellular organisms into the complex forms we see today.

Rather than running down, living systems seem to be ramping up and becoming more complex. How can this be?

The important qualification in the second law is that the inexorable increase in entropy applies to isolated systems, those that do not exchange energy or matter with their environment. Living organisms are neither isolated nor in equilibrium. All living organisms are maintained in a nonequilibrium state by a steady flow of energy passing through them, taking in low-entropy, high-quality energy in the form of food or sunlight and then producing thermal energy that increases the entropy of the environment. An organism can stay in a low-entropy state, or even decrease its entropy, because this flow of energy steadily increases the entropy of the environment. Life is possible without violating any laws of physics.

CHAPTER 20 CHALLENGE EXAMPLE Thermal energy in an adiabatic expansion

The pressure of a $1.0 \times 10^{4} {cm}^{3}$ cylinder of oxygen gas is 45 atm at a temperature of 20°C.

What is the thermal energy of the gas?
The gas expands adiabatically until the pressure is 1.0 atm. What is the change in the thermal energy of the gas?

MODEL Oxygen is a diatomic molecule. The temperature of a gas decreases during an adiabatic expansion, so we expect the thermal energy will decrease.

SOLVE

A diatomic molecule has both translational and rotational kinetic energy. Equation 20.37 for the thermal energy of a diatomic gas is

$E_{th} = \frac{5}{2} n R T$

We could use the ideal-gas law to calculate the number of moles of gas. It’s more straightforward, however, to use the ideal-gas law $p V = n R T$ to write the thermal energy as

$E_{th} = \frac{5}{2} p V$

We noted in Chapter 19 that the product $p V$ has units of J if p and V have SI units of Pa and $m^{3}$ . Thus the initial thermal energy of the gas is

$\begin{array}{l} {(E_{th})}_{i} & = & \frac{5}{2} (45 atm \times \frac{1.013 \times 10^{5} Pa}{1 atm}) (1.0 \times 10^{4} {cm}^{3} \times \frac{1 m^{3}}{10^{6} {cm}^{3}}) \\ = & 1.1 \times 10^{5} J \end{array}$
We found in Chapter 19 that an adiabatic process—a process in which no heat is transferred—obeys the equation $p_{f} V_{f}^{γ} = p_{i} V_{i}^{γ}$ , where $γ = C_{P} / C_{V}$ is the specific heat ratio. A diatomic gas has $γ = 1.40$ . We can use $p_{i} = 45 p_{f}$ to calculate that the gas volume expands to

$V_{f} = (\frac{p_{i}}{p_{f}})^{1 / γ} V_{i} = 45^{1 / 1.40} V_{i} = 15.2 V_{i}$

Consequently, the thermal energy after the expansion is

$\begin{array}{l} {(E_{th})}_{f} & = \frac{5}{2} p_{f} V_{f} = \frac{5}{2} (\frac{p_{i}}{45}) (15.2 V_{i}) = 0.338 (\frac{5}{2} p_{i} V_{i}) \\ = 0.338 {(E_{th})}_{i} \end{array}$

The thermal energy decreases because the gas temperature falls during an adiabatic expansion. The change in the thermal energy of the gas is

$Δ E_{th} = (E_{th})_{f} - (E_{th})_{i} = - 0.662 (E_{th})_{i} = - 7.3 \times 10^{4} J$

REVIEW The gas temperature can fall dramatically during an adiabatic process. Our analysis is valid only if the system remains a gas, so we should verify that this is true. We can use $p_{f} V_{f} = 0.338 p_{i} V_{i}$ and the ideal-gas law to infer that $T_{f} = 0.338 T_{i} = 99 K$ . This is a very low temperature, but it is above the 90 K boiling point of oxygen.