To find ΔU for an ideal gas that starts at ($P_1$, $V_1$, $T_1$) and goes to ($P_2$, $V_2$, $T_2$), it's stated that there are two possible paths.

  1. ΔU = ΔU(isothermal) + ΔU(isobaric)
  2. ΔU = ΔU(isothermal) + ΔU(isochoric)

I don't understand why ΔU is not equal to ΔU(isothermal) + ΔU(isobaric) + ΔU(isochoric) since P, V, and T are all changing. For example, path 1 shows an isothermal path ($P_1$ to $P_2$ at constant T) followed by an isobaric path ($T_1$ to $T_2$ at constant P). From my understanding, the gas is now at $P_2$ and $T_2$, but this says nothing about getting the volume from $V_1$ to $V_2$.

  • 1
    $\begingroup$ From p, V, T, only 2 of 3 are independent. For p2 and T2, there is only 1 possible value of V2. $\endgroup$
    – Poutnik
    Commented Oct 3, 2022 at 18:38

1 Answer 1


$pV = nRT \implies p=f_1(T,V), V=f_2(T,p), T=f_3(p,V)$

The isothermic + isobaric path:

$p_1$, $T_1$, $V_1$ $\overset{V=c/p}\rightarrow$ $p_2$, $T_1$, $V_3$ $\overset{V=cT}\rightarrow$ $p_2$, $T_2$, $V_2$

The isothermic + isochoric path:

$p_1$, $T_1$, $V_1$ $\overset{V=c/p}\rightarrow$ $p_3$, $T_1$, $V_2$ $\overset{p=cT}\rightarrow$ $p_2$, $T_2$, $V_2$


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