If I put the weight back on the piston, the system will again achieve its initial state.

No, it won't. In the end, the pressure will be the same, but the temperature and therefore the volume will be higher.

Firstly, in a real system there will be friction due to gas viscosity and piston/cylinder interaction.

But even in an ideal system, after the first weight is removed, the gas expands against only the pressure of the remaining weight(rather than the gas's own higher pressure) and therefore does less work than it otherwise could. $\Delta U_1 = -W_1$. When the weight is replaced, the compression is against the gas's own pressure, and more work would be required to return the piston to its orginial position than the gas orginially did. $\Delta U_2$ would need to be of greater magnitude than $\Delta U_1$ to get back to the starting pistion position.

If I put the weight back on the piston, the system will again achieve its initial state.

No, it won't. In the end, the pressure will be the same, but the temperature and therefore the volume will be higher.

Firstly, in a real system there will be friction due to gas viscosity and piston/cylinder interaction.

But even in an ideal system, after the first weight is removed, the gas expands against only the pressure of the remaining weight(rather than the gas's own higher pressure) and therefore does less work than it otherwise could. $\Delta U_1 = -W_1$. When the weight is replaced, the compression is against the gas's own pressure, and more work would be required to return the piston to its orginial position than the gas orginially did. $\Delta U_2$ would need to be of greater magnitude than $\Delta U_1$ to get back to the starting pistion position.

see section 3.3.2 and 3.3.3 of the following reference:

http://sites.tufts.edu/andrewrosen/files/2012/11/Thermo-Review.pdf

If I put the weight back on the piston, the system will again achieve its initial state.

No, it won't. In the end, the pressure will be the same, but the temperature and therefore the volume will be higher.

Firstly, in a real system there will be friction due to gas viscosity and piston/cylinder interaction.

But even in an ideal system, after the first weight is removed, the gas expands against a constant pressure (rather than its own higher pressure) and therefore does less work than it otherwise could. $\Delta U_1 = -W = P\Delta V = nR\Delta T$. When the weight it replaced, the compression is against the gas's own pressure, and more work would be required to return the piston to its orginial position then the gas orginially did. $\Delta U_2$ would need to be of greater magnitude than $\Delta U_1$ to get back to the starting pistion position.

If I put the weight back on the piston, the system will again achieve its initial state.

No, it won't. In the end, the pressure will be the same, but the temperature and therefore the volume will be higher.

Firstly, in a real system there will be friction due to gas viscosity and piston/cylinder interaction.

But even in an ideal system, after the first weight is removed, the gas expands against only the pressure of the remaining weight(rather than the gas's own higher pressure) and therefore does less work than it otherwise could. $\Delta U_1 = -W_1$. When the weight is replaced, the compression is against the gas's own pressure, and more work would be required to return the piston to its orginial position than the gas orginially did. $\Delta U_2$ would need to be of greater magnitude than $\Delta U_1$ to get back to the starting pistion position.