Different paths, same state

Problem

One mole of gas is taken from state A to B by two paths: (1) an isotherm; (2) two reversible adiabatics with an isochore in between that brings it back to $T_A$. Discuss the total $\Delta U$, $Q$ and $L$ in the two cases.

Solution

Internal energy: a state variable. $U$ depends only on the state (for an ideal gas, only on $T$), not on the path. If A and B have the same temperature, then:

$\Delta U_1 = \Delta U_2 = 0$

IDENTICAL along both paths.

Heat and work: path-dependent.

Path 1 (isotherm). Since $\Delta U = 0$, the first law gives $Q = L$:

$Q_1 = L_1 = nRT\ln\!\frac{V_B}{V_A}$

Path 2 (adiabatic + isochoric + adiabatic). In the two adiabatics $Q = 0$ but the work $L_\text{adiab}\neq 0$ (at the expense of internal energy); in the isochore $L = 0$ and $Q = n c_V \Delta T$. The individual contributions of $Q$ and $L$ differ from those of path 1.

Conclusion. The final state is the same and $\Delta U$ coincides, but the “tally” of heat and work depends on the route taken: $Q$ and $L$ are not state variables.

$\ev{\Delta U\ \text{equal};\ Q,L\ \text{path-dependent}}$

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Topics: Thermodynamics Concepts: Thermodynamic processes · Internal energy Objects: Ideal gas