# Is heat $\delta Q$ an exact differential for an isochoric process (ideal gases)?

Generally speaking, heat and work are path-dependent, thus $$\delta Q$$ and $$\delta W$$ are not exact differentials. By first law of thermodynamics, we know that $$dU=\delta Q - \delta W$$ but $$\delta W=0$$ for an isochoric process, that yields $$dU=\delta Q$$. Does this make work an exact differential in this specific situation? Am I neglecting something? This sounds weird to me.

• So basically given your comment below this is not a thermodynamics question and you want to know if mathematically you can write $dU=dQ$? – Bob D Apr 3 at 11:20
• Yes, that was part of the question. I might have misexplained myself below. What I meant in that comment was that I knew that during an isochoric reversible process for an ideal gas, heat equals the variation of internal energy. The answers perfectly fit my question. – Feynman_00 Apr 3 at 20:07

The reason why heat and work are path dependent and therefore not exact differentials is that, unlike internal energy, they are not system properties. While there can be a change in internal energy of a system, there is no “change” in the work or heat of a system because a system does not “possess” work or heat.

Consequently $$\delta$$ means an amount of energy transfer to or from a system in the form of heat and work, as opposed to $$d$$ which means a change in the amount of internal energy possessed by the system. The fact that an isochoric process eliminates energy transfer in the form of work does make heat a system property. It simply means the change in internal energy is due solely to energy transfer by heat. Similarly, the fact that an adiabatic process eliminates energy transfer in the form of heat does make work a property of the system. It simply means the change in internal energy is due solely to energy transfer in the form of work.

The selection of a specific path between two states does not make heat or work a system property.

Hope this helps.

Let's assume that the system is in equilibrium throughout the isochoric process and closed. We say the process has to be quasi-static. Under these conditions the work done on the system is $$\delta W=-pdV+\sum_iy_iX_i$$ where the $$X_i$$ represent different work variables, arising from different physical interactions. For example the work on a closed homogenous system done by electrical or magnetic field is $$\delta W=-pdV+\vec{E}d\vec{P}+\vec{H}d\vec{M}$$ Let's assume the only work coordinate in question is the volume $$V$$ due to compression of the system. And let's furthermore assume that we only move in the thermodynamic subspace spanned by $$V=\textrm{const.}$$ Then we indeed have

$$\delta W=0$$

And by the first law of thermodynamics $$dU=\delta Q$$ As the LHS is integrable the RHS has to be as well. So the state function $$Q$$ exists for systems undergoing quasi static processes on the subspace defined by $$V=const.$$ under the ssumptions that no other work coordinates are involved. But I don't see any reason how this would be useful.

• Yes. The relation between heat and energy is clear from a physical standpoint. I've just started working with inexact differentials, so my question was more about the mathematical meaning of that equation in differential form. – Feynman_00 Apr 3 at 9:59
• I don't think this is correct. You are assuming that the path is reversible. I can, however, do irreversible work on the system without changing its volume. – By Symmetry Apr 3 at 10:07
• @BySymmetry The argument only requires conservation of energy. Can you expand on your remark? – TheoreticalMinimum Apr 3 at 11:10
• To me since the work calculated only pertains to the isochoric path the work depends on that path. The selection of a particular path doesn’t alter the path dependency character of work or heat. Just my opinion – Bob D Apr 3 at 11:28
• @By Symmetry what kind of irreversible work? Stirrer work? – Bob D Apr 3 at 11:35