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The first law is given by:

$$ \Delta U = Q + W $$

Because of the conservation of energy, we also have:

$$ \Delta U_{universe} = \Delta U_{system} + \Delta U_{surroundings} = 0 $$

This implies that:

$$ Q_{sys} + W_{sys} + Q_{surr} + W_{surr} = 0 $$

Because $ Q_{surr} = -Q_{sys} $

$$ W_{sys} = -W_{surr} $$

But this obviously can't be true. Imagine we have a gas within a frictionless piston-cilinder system. Let's say the Gas has an internal pressure $2P_0$ and the surroundings has an "internal" pressure of $P_0$. The piston will move due to the pressure difference. If now the piston moves a very small amount, the absolute value of the work done by the system on the surroundings is obviously bigger than the absolute value of the work done by the surroundings on the system. This then implies that

$$ W_{sys} \not= -W_{surr} $$

Where am I wrong?

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  • $\begingroup$ If the pressures are different, then the piston must have mass (otherwise it would obtain infinite acceleration). This means that it too will gain kinetic energy - that is your 'missing' energy. If, however, you wait for the piston come to rest again once the pressures have equalised, the relation becomes true again. Good question! $\endgroup$ – James Wirth May 8 at 11:05
  • $\begingroup$ @JamesWirth Good question, and good answer. Make it an answer so that the OP can accept it. $\endgroup$ – garyp May 8 at 12:12
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    $\begingroup$ Wait why is Work system not equal to work surroundings? $\endgroup$ – Schwarz Kugelblitz May 8 at 12:59
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We take your example where the gas inside is at a pressure of $2P_0$ and the surroundings are at $P_0$. As I stated in my comment, the piston must be massive in order for this scenario to make sense!

The unbalanced force on the piston causes it to accelerate; if the piston moves an incremental distance $\delta x$, then the work done on the piston is

$W = 2P_0 \delta x - P_0 \delta x = P_0 \delta x = \Delta T_{piston}$

This is the kinetic energy gained by the piston! From Newton's third law, you can also see that the work done on the internal gas by the piston is $-2P_0 \delta x$ and that done on the surroundings by the piston is $P_0 \delta x$. Since we're ignoring transfers of heat, these works are the changes in internal energy of the system and surroundings respectively. Accordingly, the total change in energy of the universe is $P_0 \delta x - 2P_0 \delta x + P_0 \delta x = 0$. Energy is conserved!

However, that's not to say that the equation you state isn't valid for a massive piston under certain circumstances. Consider letting the piston go and waiting until it comes to rest again. Let's say the work done by the internal gas on the piston is $W_1$ and the work done by the external gas on the piston is $W_2$. We then have, using the work energy theorem,

$W_1 + W_2 = 0 \implies W_1 = -W_2$

Now, again through the use of Newton's third law, the work done by the piston on the internal gas is $-W_1$, and the work done by the piston on the surroundings is $-W_2$, which is also equal to $W_1$.

We see that the relation $W_{sys} = -W_{surr}$ is recovered! The key is to apply it between equilibrium states of the piston, if the piston is massive.

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$$ W_{sys} \not= -W_{surr} $$

Where am I wrong?

You are wrong in using two different pressures when calculating the work done by the system and the surroundings.

If the piston is massless ($m=0$), then the pressure at the boundary between the system and surroundings has to be the same, otherwise per Newton's second law

$$a=\frac{F_{net}}{m}=∞$$

When calculating work, the pressure is always the external pressure (pressure of surroundings). So, for your example, the pressure used to calculate work is $P_0$. At the boundary the pressure of the gas is also $P_0$, but there will pressure gradients in the gas due to disequilibrium making the process irreversible. If the process is carried out infinitely slowly so that the gas is always in equilibrium with the surroundings, then the external pressure used to calculate work is the same as pressure throughout the gas.

If the piston has mass, then the mass of the piston must be included as part of the system (or surroundings) when calculating the work done by/on the system or surroundings.

Hope this helps.

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" the absolute value of the work done by the system on the surroundings is obviously bigger than the absolute value of the work done by the surroundings on the system "

No it's not. Whatever work is done BY the system is by definition the work done ON the surroundings. Why and How would it be greater in any case. It would simply violate energy conservation law. Instead of system and surrounding consider an isolated gaseous chamber with a piston in between.

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  • $\begingroup$ Read the answers above, they give a correct explanation. $\endgroup$ – Dabruh May 8 at 15:53

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