2.4: Adiabatic processes - energy change without heat transfer

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In establishing the first law of thermodynamics, we've stated that all of the changes in the internal energy of a system can come from one of two places - heat transfer across the system boundary, and work performed on or by the system. We've stated further that this internal energy change in the system is a state energy - that it is only dependent upon the initial and final conditions of the system.

Those statements mean that this statement (which we proposed in the previous section) is true in general:

$Δ U = n \bar{C_{V}} Δ T$

This represents, of course, the heat transfer at constant volume. But because the work done by the system is zero when the volume is held constant, it also represents the internal energy change - and, because the internal energy is a state energy, that means that this statement represents the internal energy change under any condition, whether the volume is held constant or not.

It can even represent the internal energy change when no heat is transferred across the system boundary at all.

Processes where no heat is transferred are called adiabatic processes, and these special cases have a host of interesting consequences.

In an adiabatic process, the entirety of the internal energy change in a system is a result of work performed. It's worthwhile to frame both the internal energy changes and the work performed in terms of differentials, rather than full deltas, in order to consider the full range of changes to the system. Let's first rewrite the internal energy change as a differential change:

$d U = n \bar{C_{V}} d T$

We can remind ourselves of the differential definition of work:

$δ w = - P d V$

The differential statement of the first law of thermodynamics is an equation of the inexact differentials of heat and work (because heat and work are process energies) to the exact differential of internal energy:

$d U = δ q + δ w$

Because δq is equal to zero in an adiabatic process (if no heat energy is transferred across the boundary, not even differential heat counts), dU equates to δw, and we have a new equation:

$d upper U equals delta w left-parenthesis adiabatic process right-parenthesis$

$n \bar{C_{V}} d T = - P d V (adiabatic process)$

Now this is a relationship that is describing the differential change in two state variables - temperature and volume. By placing this in terms of differentials, we will be able to integrate both sides of the relationship with appropriate rearrangement, and arrive at a relationship that will describe how volume changes with changing temperature in an adiabatic process. Let's assume our system is an ideal gas for simplicity, and let's make the rearrangements to make this equation integrable.

Because we're assuming an ideal gas, we can replace P on the right side:

$n \bar{C_{V}} d T = - \frac{n R T}{V} d V (adiabatic process)$

The moles of the gas cancel, and we can easily divide both sides by T to gather like terms on the same side of the equation:

$\frac{\bar{C_{V}}}{T} d T = - \frac{R}{V} d V (adiabatic process)$

We've acknowledged before that the antiderivative of 1/V is ln V, and T doesn't behave any differently. We can then perform an integration and easily rid ourselves of an annoying negative sign:

$\int_{T_{i}}^{T_{f}} \frac{\bar{C_{V}}}{T} d T = \int_{V_{i}}^{V_{f}} - \frac{R}{V} d V (adiabatic process)$

$\bar{C_{V}} {[\ln T]}_{T_{i}}^{T_{f}} = - R {[\ln V]}_{V_{i}}^{V_{f}} (adiabatic process)$

$\bar{C_{V}} \ln (\frac{T_{f}}{T_{i}}) = - R \ln (\frac{V_{f}}{V_{i}}) (adiabatic process)$

$\bar{C_{V}} \ln (\frac{T_{f}}{T_{i}}) = R \ln (\frac{V_{i}}{V_{f}}) (adiabatic process)$

This relationship is powerful enough, but we can simplify it further. Remember that R has the same units as a molar specific heat, and further that the difference between the specific heats $\bar{C_{P}}$ and $\bar{C_{V}}$ is exactly equal to R. Therefore we can set up a very interesting ratio on the right-hand side:

$\ln (\frac{T_{f}}{T_{i}}) = \frac{R}{\bar{C_{V}}} \ln (\frac{V_{i}}{V_{f}}) (adiabatic process)$

$\ln (\frac{T_{f}}{T_{i}}) = \frac{\bar{C_{P}} - \bar{C_{V}}}{\bar{C_{V}}} \ln (\frac{V_{i}}{V_{f}}) (adiabatic process)$

$\ln (\frac{T_{f}}{T_{i}}) = (\frac{\bar{C_{P}}}{\bar{C_{V}}} - 1) \ln (\frac{V_{i}}{V_{f}}) (adiabatic process)$

We make this final rearrangement here because it's common, in the study of ideal gas relationships, to define the ratio between constant-pressure and constant-volume specific heats with a symbol of its own, γ:

$γ = \frac{\bar{C_{P}}}{\bar{C_{V}}}$

That ratio makes the statement of this relationship between changing temperature and changing volume easier to express:

$\ln (\frac{T_{f}}{T_{i}}) = (γ - 1) \ln (\frac{V_{i}}{V_{f}}) (adiabatic process)$

There's one more simplification we can make to this expression. It's a basic property of a logarithm that if a constant c is multiplied by a logarithm ln x, that the argument x of that logarithm can be raised to the c power:

$c \ln x = \ln x^{c}$

We can apply this to the new constant (γ - 1) in our simplified expression:

$\ln (\frac{T_{f}}{T_{i}}) = \ln {(\frac{V_{i}}{V_{f}})}^{γ - 1} (adiabatic process)$

At this point, the arguments of the logarithms, are equivalent, and the logarithms themselves are surplus to requirements:

$\frac{T_{f}}{T_{i}} = {(\frac{V_{i}}{V_{f}})}^{γ - 1} (adiabatic process)$

This result is somewhat surprising if you've committed the gas laws that emerge from the ideal gas law to memory. We're very used to seeing this relationship, the rule known as Charles' Law:

$\frac{T_{f}}{T_{i}} = \frac{V_{f}}{V_{i}} (closed container at constant pressure)$

But, as the text to the right indicates, that statement isn't true without any context; we keep moles and pressure constant in the ideal gas law to arrive at such a relationship. We started this derivation indicating that pressure was not constant; it changed as a function of volume of the container. Under adiabatic processes, as a result, when the volume of the gas decreases, the temperature of the gas increases. This is a result of the temperature change being driven by work done on the gas, increasing the internal energy of the gas.

Derivations of this sort are important, and it is very easy to get lost in the mathematics and lose sight of the importance of the result. Both are important; the mathematics are the tools of our theoretical work, and it's important to use the mathematics well. But we're not blindly rearranging (much as it might feel like it sometimes!); we're rearranging to build simplified relationships that describe behavior under unique conditions.