# Entropy Change for a Reversible, Isothermal Process

In general, it is not possible to evaluate ΔS directly by integrating dQ/T.
However, in the special case of an

,

#### isothermal

process it is possible and easy too!
Because temperature is constant, T can be pulled out of the intergral:
Internally Reversible, Isothermal Processes
This is particularly useful for determining the change in entropy of a

#### thermal reservoir

or for the two reversible, isothermal steps in the

#### Carnot Cycle

.
Constant T
T1 = T2
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### Ch 7, Lesson B, Page 4 - Entropy Change for a Reversible, Isothermal Process

• Although we are not often able to use the definition of entropy to directly evaluate ΔS, there is at least one type of process where we can get away with this.
• In an internally reversible, isothermal process, our definition of entropy can be used AND T pops out of the integral because it is constant.
• That leaves with just T times the integral of dQ on the RHS and the integral of dQ is just Q.
• It doesn’t get much easier than that !
• So, for an internally reversible, isothermal process, ΔS is just Q over T.
• That’s cool, but when can we use it ?
• We can use it for two of the 4 steps in the Carnot Cycle !  You’ve gotta love that because it only leaves two other steps that we need to consider.
• But we will most frequently use this result when we are evaluating the change in entropy of a thermal reservoir.
• Thermal reservoirs are internally reversible because there really isn’t anything going on inside of them at all.  They are isothermal by definition.
• So, DS for a thermal reservoir is simply Q over T of the reservoir.
• We are left with two problems.
• 1 – How do we evaluate ΔS for internally reversible, but NON isothermal processes ?
• 2 - How do we evaluate ΔS for irreversible processes ?