Question

To find ΔU for an ideal gas that starts at (P1, V1, T1) and goes to (P2, V2, T2), it's stated that there are two possible paths.

ΔU = ΔU(isothermal) + ΔU(isobaric)
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 (P1
to P2 at constant T) followed by an isobaric path (T1 to T2 at constant P). From my understanding, the gas is now at P2 and T2, but this says nothing about getting the volume from V1 to V2.

Answer 1

The equation ΔU = ΔU(isothermal) + ΔU(isochoric) represents the change in internal energy of an ideal gas using two different paths: an isothermal path and an isochoric path. The isothermal path assumes constant temperature, while the isochoric path assumes constant volume. Therefore, ΔU only depends on the changes in temperature and volume along these paths, and not the changes in pressure.

Step-by-step explanation:

The equation ΔU = ΔU(isothermal) + ΔU(isochoric) represents the change in internal energy (ΔU) of an ideal gas using two different paths.

The isothermal path assumes the temperature remains constant and the work done is given by W = P(V₂ - V₁) which only considers the change in volume.

The isochoric path assumes the volume remains constant and the work done is zero as there is no change in volume.

Therefore, ΔU = ΔU(isothermal) + ΔU(isochoric).

In your example, where the gas goes from (P₁, V₁, T₁) to (P₂, V₂, T₂), the first path is an isothermal path (P₁ to P₂ at constant T) and the second path is an isochoric path (T₁ to T₂ at constant V). It is important to note that the gas is already at the final volume (V₂) after the first path, so only the change in temperature (T₁ to T₂) needs to be considered for the second path.