# ΔU calculation for an ideal gas that starts at (P1, V1, T1) and goes to (P2, V2, T2)

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 $$V_2$$.

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

$$pV = nRT \implies p=f_1(T,V), V=f_2(T,p), T=f_3(p,V)$$
$$p_1$$, $$T_1$$, $$V_1$$ $$\overset{V=c/p}\rightarrow$$ $$p_2$$, $$T_1$$, $$V_3$$ $$\overset{V=cT}\rightarrow$$ $$p_2$$, $$T_2$$, $$V_2$$
$$p_1$$, $$T_1$$, $$V_1$$ $$\overset{V=c/p}\rightarrow$$ $$p_3$$, $$T_1$$, $$V_2$$ $$\overset{p=cT}\rightarrow$$ $$p_2$$, $$T_2$$, $$V_2$$