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WORK, HEAT, AND ENERGY: THE FIRST LAW OF THERMODYNAMICS

The Joule Experiment


The first attempt to determine whether (∂U/∂V)T, n vanishes for real gases was made by Joule around 1843. His apparatus is schematically depicted in Figure 5. A sample of a gas (the system) was placed in one side of the apparatus and the other side of the apparatus was evacuated. Mechanical vacuum pumps were not yet invented, so a water aspirator was used. The system was allowed to equilibrate and the initial temperature of the apparatus was measured. The stopcock was then opened and the gas expanded irreversibly into the vacuum.
Because the surroundings were not affected during the expansion into a vacuum, w was equal to zero. The gas expanded rapidly so there was little opportunity for heat to be transferred to or from the surroundings, and q therefore vanished to a good approximation. From the first law, ΔU was equal to zero. If a change in temperature of the gas occurred, heat would be transferred to or from the surroundings after the expansion was complete, and the final temperature of the surroundings
Figure 5
Figure 5 The Apparatus for the Joule Experiment (Schematic).

would differ from the initial temperature. If the heat capacity of the apparatus and the heat capacity of the gas are known, the change in temperature of the gas could be calculated.
The Joule experiment was carried out several times with various volumes for the second chamber. The ratio ΔT/ΔV would be determined for each experiment and extrapolated to zero value of ΔV, where ΔV is the final volume of the gas minus its initial volume. This extrapolation is equivalent to taking the mathematical limit, so the result is a partial derivative, called the Joule coefficient and denoted by μJ:
Equation 13
The Joule coefficient is related to (∂U/∂V)T ,n by use of the cycle rule,
Equation 14
Joule was unsuccessful in his attempt to measure the Joule coefficient because the changes in temperature that occurred were too small to be measured by his thermometers, even though he used pressures up to 22 atm. Later versions of the experiment with better apparatus have given nonzero values of (∂U/∂V)T,n for real gases.
There are better ways than the Joule experiment to determine values of (∂U/∂V)T,n. Once values for CV and for (∂U/∂V)T,n are obtained, ΔU can be calculated for any process that begins with one equilibrium state and ends with another equilibrium state.
Figure 6
Figure 6 The Curve Representing the Path for the ΔU Line Integral.

A change in internal energy for a nonisothermal process can be calculated by carrying out a line integral of dU.

Example 16

Calculate ΔU for a process that takes 1.000 mol of argon from T  298.15K and V 2.000 L to T  373.15K and V  20.000 L. Does the result depend on whether the process is reversible?

Solution

Since U is a state function, we can choose any path with the same initial state and final state to calculate ΔU.We choose to integrate along the reversible path shown in Figure 6, consisting of an isothermal expansion and a constant-volume change in temperature. The first step of the path is that of the previous example, so ΔU1, the change in energy for that part, is equal to 67 J mol-1. For the second step
Solution
Because U is a state function, the result does not depend on whether the actual process is reversible, so long as the initial and final states are metastable or equilibrium states.

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