Two paths in the p-V plane

Problem

Same initial state $(p_1,V_1)$ and final state $(p_2,V_2)$, with $p_1>p_2$ and $V_2>V_1$. Path 1 goes through $(p_1,V_2)$ (high isobar then isochore down); path 2 goes through $(p_2,V_1)$ (isochore down then low isobar). Compare the work done along the two paths and comment on why $\Delta U$ is instead the same.

Solution

The work is the area under the isobaric leg (the isochores do no work, $\Delta V = 0$).

Path 1. The expansion happens along the high-pressure isobar $p_1$, from $V_1$ to $V_2$:

$L_1 = p_1(V_2-V_1)$

Path 2. The expansion happens along the low-pressure isobar $p_2$:

$L_2 = p_2(V_2-V_1)$

Comparison. Since $p_1 > p_2$ and $V_2 > V_1$, path 1 does more work:

$L_1 > L_2$

Why $\Delta U$ is the same. Internal energy is a state variable: it depends only on the initial state A and the final state B, not on the route followed. Both paths start at $(p_1,V_1)$ and end at $(p_2,V_2)$, so $\Delta U$ is IDENTICAL. Only the split between heat and work differs.

$\ev{L_1=p_1(V_2-V_1)>L_2=p_2(V_2-V_1);\ \Delta U\ \text{equal}}$

Links

Topics: Thermodynamics Concepts: Thermodynamic processes · Internal energy Methods: P-V and ln P - ln V diagram Objects: Ideal gas