Answer:
a) [tex]\Delta S<0[/tex]
b) entropy of the sistem equal to a), entropy of the universe grater than a).
Explanation:
a) The change of entropy for a reversible process:
[tex]\delta S=\frac{\delta Q}{T}[/tex]
[tex]\Delta S=\frac{Q}{T}[/tex]
The energy balance:
[tex]\delta U=[tex]\delta Q- \delta W[/tex]
If the process is isothermical the U doesn't change:
[tex]0=[tex]\delta Q- \delta W[/tex]
[tex]\delta Q= \delta W[/tex]
[tex]Q= W[/tex]
The work:
[tex]W=\int_{V1}^{V2}P*dV[/tex]
If it is an ideal gas:
[tex]P=\frac{n*R*T}{V}[/tex]
[tex]W=\int_{V1}^{V2}\frac{n*R*T}{V}*dV[/tex]
Solving:
[tex]W=n*R*T*ln(V2/V1)[/tex]
Replacing:
[tex]\Delta S=\frac{n*R*T*ln(V2/V1)}{T}[/tex]
[tex]\Delta S=n*R*ln(V2/V1)}[/tex]
Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:
[tex]\Delta S<0[/tex]
b) The entropy change of the sistem will be equal to the calculated in a), but the change of entropy of the universe will be 0 in a) (reversible process) and in b) has to be positive given that it is an irreversible process.
