Adiabatic Expansion of Ideal Gas

[AndyDude]

If I found out that work is done by a system and hence positive (dV>0), and I am asked to determine the work done on the gas (in the system). The answer would be the magnitude of work done by system but now negative since it is being done on the gas?

 

Does this make any sense?

[avenir001]

taking the perspective of the system, work done by the system is negative (and leads to a decrease in the internal energy of the system)...and work done on the system is positive (and increases the system's internal energy).

 

remember w = - S PdV (where S stands for the integral sign )

now the gas is your system and it's expanding adiabatically...which means dV is positive...so w is negative...which means that, by expanding, the system has done work (ie work by the system on the surroundings) and has lost energy as a result.

 

now if u put a weight on the piston and compress the system, dV is gonna be negative (ie volume is decreasing)...and w is gonna be positive, meaning that work is being done on the system, leading to an increase in its internal energy.

 

hope this helps.

[AndyDude]

Hey,

Thanks for the reply. It does make sense now. It's the Kaplan book that made things more confusing for me.

It defines internal energy as U=Q-W; W>0 if it's done by the system & vice-versa.

Of course your definition would have U=Q+W whereby W= -PdV.

Thanks again.