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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.

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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 $dE = \delta W = -p\,dV$.

But apparently this is the case for an isochoric process, so I doubt it the case for both.

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 $dE = \delta Q$.

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