Question

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

Open in App
Solution

## 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, $\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W\phantom{\rule{0ex}{0ex}}⇒\mathrm{\Delta }Q=n{C}_{v}\mathrm{\Delta }T+\mathrm{\Delta }W\phantom{\rule{0ex}{0ex}}⇒0=n{C}_{v}\left(0\right)+\mathrm{\Delta }W\phantom{\rule{0ex}{0ex}}⇒\mathrm{\Delta }W=0,$ where Cv 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.

Suggest Corrections
0
Related Videos
The First Law of Thermodynamics
PHYSICS
Watch in App
Explore more