Is the work done by the system equal to negative of the work done by the surroundings on the system?

If an ideal gas expands into its surroundings through a massless piston, is the work done by the gas on the surrounds equal to the negative of the work done by surroundings on the system?




As Karsten Theis said in his comment, $w\_{sys} = –w\_{surr}$, always. But it can be hard to think about work flow. Thus, to help understand this result more intuitively, consider conservation of energy:


$$\Delta U\_{sys} = –\Delta U\_{surr}$$.


If the system is closed (no movement of matter between system and surroundings), the only way we can affect the internal energy of the system is through the flow of heat or work. Hence:


$$\Delta U = q + w$$


And thus:


$$q\_{sys} + w\_{sys} = -q\_{surr} - w\_{surr}$$.


Now, while it may be difficult to think about work, it should be intuitively obvious that any heat that flows into the system must have flown out of the surroundings, and visa versa. Thus we have:


$$q\_{sys} = -q\_{surr}$$


This in turn leaves us with:


$$w\_{sys} = -w\_{surr}$$


From this we can see that work flow across a system boundary is just like heat flow across a system boundary: Whatever goes out of the system has to go into the surroundings, and visa versa.


This can be extended to an open system, to show that the above statement is generally true.


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