Our values are:
![m=14g\\T=298K\\P=1bar](https://tex.z-dn.net/?f=m%3D14g%5C%5CT%3D298K%5C%5CP%3D1bar)
a) For isothermal reversible expansion, We have that the change in the Internal Energy is 0, that is,
![\Delta U = 0](https://tex.z-dn.net/?f=%5CDelta%20U%20%3D%200)
We can solve for the Work made in the system.
![|q| = |W|= nRT ln(\frac{V_2}{V1} )](https://tex.z-dn.net/?f=%7Cq%7C%20%3D%20%7CW%7C%3D%20nRT%20ln%28%5Cfrac%7BV_2%7D%7BV1%7D%20%29)
![|W| = \frac{1}{2}*8.314*298*ln(2) = 858.66J](https://tex.z-dn.net/?f=%7CW%7C%20%3D%20%5Cfrac%7B1%7D%7B2%7D%2A8.314%2A298%2Aln%282%29%20%3D%20858.66J)
With work made it, this same quantity is necessary to add to the system through Heat. Then,
![\Delta S_{system} = \frac{Q_{rev}}{T} = \frac{858.66}{298} = 2.88 J/K](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bsystem%7D%20%3D%20%5Cfrac%7BQ_%7Brev%7D%7D%7BT%7D%20%3D%20%5Cfrac%7B858.66%7D%7B298%7D%20%3D%202.88%20J%2FK)
For Surrounding we apply the same relation,
![\Delta S_{surrounding} = \frac{Q_{surrounding}}{T} = \frac{-858.66}{T} = -2.88J/K](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bsurrounding%7D%20%3D%20%5Cfrac%7BQ_%7Bsurrounding%7D%7D%7BT%7D%20%3D%20%5Cfrac%7B-858.66%7D%7BT%7D%20%3D%20-2.88J%2FK)
Therefore, ![\Delta S_{total} = 0](https://tex.z-dn.net/?f=%5CDelta%20S_%7Btotal%7D%20%3D%200)
b) For an isothermal irreversible expansion
We consider the external presion as 0, by this way, the work done is also 0.
![P_{ext} = 0 \Rightarrow W=0](https://tex.z-dn.net/?f=P_%7Bext%7D%20%3D%200%20%5CRightarrow%20W%3D0)
Since isothermal irreversible expansion change in Internal Energy is 0, so
, since isothermal
Therefore
![\Delta S_{surr} = 0](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bsurr%7D%20%3D%200)
How the condition are the same, for the System the entropy is the same
But ![\Delta S{system} = 2.88 J/K](https://tex.z-dn.net/?f=%5CDelta%20S%7Bsystem%7D%20%3D%202.88%20J%2FK)
![\Delta S_{total} = 2.88 J/K](https://tex.z-dn.net/?f=%5CDelta%20S_%7Btotal%7D%20%3D%202.88%20J%2FK)
c) And an adiabatic reversible expansion
Since the process is Adiabatic,
![q = 0](https://tex.z-dn.net/?f=q%20%3D%200)
Therefore, ![\Delta S_{sys} =0 \Rightarrow \Delta S_{surr} = 0](https://tex.z-dn.net/?f=%5CDelta%20S_%7Bsys%7D%20%3D0%20%5CRightarrow%20%5CDelta%20S_%7Bsurr%7D%20%3D%200)