# Solid & liquid entropy change

1. Dec 9, 2015

### son hong chang

usually, entropy change in solid and liquid is formulated as Cp(or Cv)ln(T2/T1) or Q/T(integral)+ Sgen.
so, considering the former and the latter equations, Do their entropy changes include irreversibility due to Sgen?

And next question,
reservoir has no change in temperature and volume. therefore, its'entropy change is zero.
(According to this formula: dS=Cvln(T2/T1)+Rln(V2/V1) )
but when we solve many matters in thermodynamics especially in entropy part, we think that the reservoir exchanging the heat with another part has the entropy change expressed as Q/T not zero.
why is that?

Last edited: Dec 9, 2015
2. Dec 10, 2015

### PietKuip

Because the reservoir has infinite heat capacity?
(Just my guess that this confused you.)

3. Dec 10, 2015

### Staff: Mentor

The first equation is correct. I'm not sure about the second equation, unless T represents the temperature at the boundary with the surroundings and the integration is over time. The change in entropy from and initial thermodynamic equilibrium state to a final thermodynamic equilibrium state of a system depends only on the initial and final states, and not on the path between the two states.
The formula you wrote is for an ideal gas, which would not normally be used as a reservoir. But if you apply $ΔS=mC_v\ln{(T_2/T_1)}$ to a reservoir that has a very high value of $mC_v$, you find that it reduces to Q/T. Here's how:

$$ΔS=mC_v\ln{(T_2/T_1)}=mC_v\ln{\left(1+\frac{(T_2-T_1)}{T_1}\right)}$$
If $mC_v$ is very large (as is the case of a reservoir, where it is infinite), the temperature change is very small. So we can expand the natural log term in a Taylor series and retain only the first term in the expansion. We thereby obtain:
$$ΔS\rightarrow mC_v\frac{(T_2-T_1)}{T_1}$$
But, $$mC_v(T_2-T_1)=Q$$Therefore,
$$ΔS\rightarrow \frac{Q}{T_1}$$

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