Question

An ideal gas contained in a piston-cylinder assembly is compressed isothermally in an internally reversible process.

(a) Determine if the entropy change of the gas is greater than, equal to or less than zero, justify your answer
(b) Determine if for the same change of state, the entropy change for an irreversible process is greater than, equal to or less than part (a)

Answer 1

a)
`\Delta S<0`

b) entropy of the sistem equal to a), entropy of the universe grater than a).

Step-by-step explanation:

a) The change of entropy for a reversible process:

`\delta S=(\delta Q)/(T)`

`\Delta S=(Q)/(T)`

The energy balance:

`\delta U=[tex]\delta Q- \delta W`

If the process is isothermical the U doesn't change:

`0=[tex]\delta Q- \delta W`

`\delta Q= \delta W`

`Q= W`

The work:

`W=\int_(V1)^(V2)P*dV`

If it is an ideal gas:

`P=(n*R*T)/(V)`

`W=\int_(V1)^(V2)(n*R*T)/(V)*dV`

Solving:

`W=n*R*T*ln(V2/V1)`

Replacing:

`\Delta S=(n*R*T*ln(V2/V1))/(T)`

`\Delta S=n*R*ln(V2/V1)}`

Given that it's a compression: V2<V1 and ln(V2/V1)<0. So:

`\Delta S<0`

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.