Question

Can a process on an ideal gas be both adiabatic and isothermal?

Answer 1

According to the first law of thermodynamics, change in internal energy, $∆$U is equal to the difference between heat supplied to the gas, $∆$Q and the work done on the gas,​$∆$W, such that $\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$​. In an adiabatic process, $∆$Q = 0 and in an isothermal process, change in temperature, $∆$T = 0. Therefore,
$$\begin{gathered} \mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W \\ \Rightarrow \mathrm{\Delta }Q=n{C}_v\mathrm{\Delta }T+\mathrm{\Delta }W \\ \Rightarrow 0=n{C}_v\left(0\right)+\mathrm{\Delta }W \\ \Rightarrow \mathrm{\Delta }W=0, \end{gathered}$$
where C_v is the heat capacity at constant volume.
This shows that if the process is adiabatic as well as isothermal, no work will be done. So, a process on an ideal gas cannot be both adiabatic and isothermal.