# Adiabatic processes and the First Law

The first law of thermodynamics in my notes is : $\Delta E=\Delta Q +\Delta W$. Then later in my notes for an adiabatic process: $\Delta Q \implies dE=-pdV$. Then for a monatomic gas $E=\frac{3}{2}Nk_BT=\frac{3}{2}pV$. Then $$dE=\frac{3}{2}pdV+\frac{3}{2}Vdp=-pdV$$ which gives $$\frac{5}{2}pdV=\frac{-3}{2}Vdp$$ which can be solved to give $$V=p^{-3/5}.const$$ and $$\Delta W= - \int^{V_2}_{V_1}cV^{-5/3}dV$$

I dont see why we dont just write $\Delta E=\Delta Q +\Delta W$ and $\Delta Q=0$ gives $\Delta E=\Delta W$. But apparently this is the case for an isochoric process, so I doubt it the case for both.

I may also have a problem with my understanding of the difference between $\Delta$ and $\delta$. I have read and understood https://math.stackexchange.com/questions/750328/when-are-delta-x-delta-x-dx-and-text%C4%91x-exactly-the-same-when-ar.

I dont see why we dont just write $\Delta E=\Delta Q +\Delta W$ and $\Delta Q=0$ gives $\Delta E=\Delta W$.
That's exactly what you do. $\delta Q = 0$ is the definition of an adiabatic process. Pressure and volume are well defined in the case of a quasi-static adiabatic process, in which change the work done on the system (and hence the change in energy) is $dE = \delta W = -p\,dV$.
Isochoric means constant volume. In the case of a quasi-static process, work is given by $\delta W = p\,dV$. This is zero for a constant volume process, so in the case of a quasi-static isochoric process, we have $dE = \delta Q$.