Our values are:
[tex]m=14g\\T=298K\\P=1bar[/tex]
a) For isothermal reversible expansion, We have that the change in the Internal Energy is 0, that is,
[tex]\Delta U = 0[/tex]
We can solve for the Work made in the system.
[tex]|q| = |W|= nRT ln(\frac{V_2}{V1} )[/tex]
[tex]|W| = \frac{1}{2}*8.314*298*ln(2) = 858.66J[/tex]
With work made it, this same quantity is necessary to add to the system through Heat. Then,
[tex]\Delta S_{system} = \frac{Q_{rev}}{T} = \frac{858.66}{298} = 2.88 J/K[/tex]
For Surrounding we apply the same relation,
[tex]\Delta S_{surrounding} = \frac{Q_{surrounding}}{T} = \frac{-858.66}{T} = -2.88J/K[/tex]
Therefore, [tex]\Delta S_{total} = 0[/tex]
b) For an isothermal irreversible expansion
We consider the external presion as 0, by this way, the work done is also 0.
[tex]P_{ext} = 0 \Rightarrow W=0[/tex]
Since isothermal irreversible expansion change in Internal Energy is 0, so
[tex]\Delta U = q = 0[/tex], since isothermal
Therefore
[tex]\Delta S_{surr} = 0[/tex]
How the condition are the same, for the System the entropy is the same
But [tex]\Delta S{system} = 2.88 J/K[/tex]
[tex]\Delta S_{total} = 2.88 J/K[/tex]
c) And an adiabatic reversible expansion
Since the process is Adiabatic,
[tex]q = 0[/tex]
Therefore, [tex]\Delta S_{sys} =0 \Rightarrow \Delta S_{surr} = 0[/tex]
The mass of nitrogen = 14kg
Temperature = 298 k
Pressure = 1 bar
0.5*8.314*298*ln2
= 858.7 Joules
Next we have to solve for the system through the heat
858.7/298
= 2.8814J/K
This is the same as the temperature that is in the surrounding. We conclude that the change in isothermal expansion = 0
b. Isothermal change in the equation = 0
This means that the process is the same as in a above.
c. Given that this is an adiabatic expansion then q = 0.
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