# Why don't we consider VΔP when we define Q?

We only define $Q = \Delta U + W_\text{exp}$ (expansion work = $-P\Delta V$). If heat can cause $\Delta U$ and work, why work is defined only as expansion work in the first place where there are other forms of works, such as isochoric work ($=V\Delta P$)?

Let's say, there is a container with ideal gas $T, P, V$ and $V$ can't be changed by this container. I give it heat therefore $T, P$ changes to $2T$, $2P$. In this constant volume process, expansion work, Wexp is 0. That's why $Q_v = \Delta U$. In the mean time (in this constant volume process), $\Delta H = Q_v + V\Delta P$ and $V\Delta P$ is work done in constant volume process. My question is that why can't we just say this heat $Q = \Delta + V\Delta P$? Do we just ignore the non-expansion work? or is it included in $\Delta U$? or because the definition of $Q$ is $\Delta U+P\Delta V$ originally?

One more thing. I googled it all day and someone wrote that $V\Delta P$ is work in flow process. I can't imagine how matter can flow in the constant volume container, and even though it can flow, what is the relationship between $V\Delta P$? They didn't explain it why so my question get bigger and bigger.

• Do you mean to ask why not $\Delta U = q+ P\Delta V + V\Delta P$ ? – MollyCooL Mar 27 '18 at 12:03
• What the heck is isochoric work? Work is force times displacement (over which the force acts), not a change in force times no displacement. – Chet Miller Mar 27 '18 at 12:08
• @ChesterMiller While I agree with you, do note that in the OP's previous question, the +2 score answer told the OP that $V\Delta P$ is "isochoric work". That answer got no downvotes or comments saying it is wrong. So, probably that's where the OP got this notion from. – Gaurang Tandon Mar 27 '18 at 12:22
• @MollyCooL In a sense, yes. Could you explain why this ignore VΔP in that? – Zillai Mar 27 '18 at 13:18
• For what kind of process? Isochoric , isothermal or anything else? – MollyCooL Mar 27 '18 at 13:21

Let's start by the fundamental ideas. Assume, you have a container with volume $V$ with a piston on it of cross sectional area $A$ and container is filled with gas.
Suppose, you push the container by $\mathrm dx$ amount. So, the gas is compressed and work is done on the gas.
By the definition of work classically, $$\mathrm dW=F\,\mathrm dx$$ Now if we use the fact that $P=\frac FA$, and $\mathrm dV=A\,\mathrm dx$, we will have $$\mathrm dW=\frac FAA\,\mathrm dx= P\,\mathrm dV$$ So, the classical definition of work only tells us the that work done on the gas should be $$\int_{V_1}^{V_2}P\,\mathrm dV=P\,\Delta V$$ if $P$ is constant.
So, the definition tells us there is no $V\,\Delta P$ term in the expresion for work. And, therfore, in the first law of thermodynamics, i.e. $\Delta Q = \Delta U - W_\text{on the system}$, also, that expression doesn't occur from work, and it is not also included in $\Delta U$ either.
But if you consider enthalpy change for a system, there $V\,\mathrm dP$ term will occur as $$\mathrm dH=\mathrm dU+\mathrm d(PV)=\mathrm dU+P\,\mathrm dV+V\,\mathrm dP$$

• Actually, that last equation is not mathematically correct, unless it involves differentials rather than deltas. – Chet Miller Mar 27 '18 at 13:00
• After edit: Change “ VdeltaP term will occur” to VdP... – MollyCooL Mar 27 '18 at 13:26
• Thank you for your answers. But maybe my writing was not clear. I meant like a steel container of which volume can't be changed. Let's say I fixed your container's piston so that it can't be moved and put heat in it. volume can't be changed so only pressure get higher, followed by temperture higher. In this case, no expansion, no Work, -PdV=0 AND WHAT ABOUT -VdP? shouldn't we consider it because even if it's not work like others say, it's definitely other form of energy. enthalpy consider it. so why not Q? this is why I suggested this form of energy might be included in ΔU. – Zillai Mar 27 '18 at 13:50
• Enthalpy and $Q$ are different quantities. Enthalpy is a state function and it is defined as $H = U + PV$. But heat is not a state function, we can only measure its change i.e. $\Delta$Q, but can't measure it in a particular state of a system. On the other hand, We can measure enthalpy of a system at a particular state. So, better not to compare enthalpy with heat. – Soumik Das Mar 27 '18 at 14:17
• Isn’t it like we cannot measure enthalpy at a particular state and hence only the change? – MollyCooL Mar 27 '18 at 14:20

As an answer to your comment, it is not that we ignore the P$\Delta V$ term. Let us consider the first law statement, $$\mathrm{d}U = \mathrm{d}Q+ \mathrm{d}W$$ And we have, $\mathrm{d}W = P\mathrm{d}V$. Notice that I use differentials instead of delta terms.

Considering any process, Intgerating the expression and setting appropriate limits, we actually don’t get $\Delta W = P\Delta V$. This is because the pressure is not constant and cannot be taken out of the integral. Hence the $V \Delta P$ is technically not ignored.(This was also explained in Soumik’s answer). Only for an isobaric process $V\Delta P$ term can be ignored. For all other processes involving an ideal gas, if we work with $PV = nRT$, differentiating we get, $$P\Delta V + V\Delta P = nR\Delta T$$ And incase of an isochoric process, we get the following relation, $$V\Delta P = nR\Delta T$$