Prove that de=pdv

The First Law of Thermodynamics states that the change in internal energy ($de$) of a system is equal to the heat added to the system ($dq$) plus the work done on the system ($dw_{on}$).

$$de = dq + dw_{on}$$

For a reversible process involving a change in volume, the infinitesimal work done on the system ($dw_{on}$) by its surroundings is given by:

$$dw_{on} = p dv$$

(Note: This convention defines work done on the system as positive. In some contexts, work done by the system is defined as $dw_{by} = p dv$, in which case $dw_{on} = -p dv$.)

An adiabatic process is a thermodynamic process where no heat is exchanged between the system and its surroundings. Therefore, the infinitesimal change in heat ($dq$) is zero.

$$dq = 0$$

Substitute the expressions for $dq$ and $dw_{on}$ into the First Law of Thermodynamics:

$$de = 0 + p dv$$ $$de = p dv$$

This equation holds true for a reversible adiabatic process where work done on the system is considered positive.

The final answer is $\boxed{de = pdv}$.