# How can we calculate #DeltaS# for an isothermal process?

##### 1 Answer

Well, there are two ways I can think of. Both show that entropy increases as volume increases at constant temperature.

**METHOD 1**

You can always start from the **first law of thermodynamics**:

#DeltaE = q + w#

It is simplest to consider a reversible process, i.e. the 100% efficient process:

#DeltaE = q_(rev) + w_(rev)# ,where:

#q_(rev)# is thereversible heat flow.#w_(rev) = -int PdV# is thereversible workdone from the perspective of thesystem.#DeltaE# is the change in internal energy.

For an isothermal process, all ideal gases have

#q_(rev) = -w_(rev) = int PdV#

Now, consider the **definition of the entropy**:

#dS = (deltaq_(rev))/T# ,where

#delta# indicates a path function and#d# indicates a state function. Both are differentials and can be integrated.

Thus, by finding

#int_((1))^((2)) dS = int_((1))^((2)) 1/T deltaq_(rev)#

#= DeltaS = q_(rev)/T#

For ideal gases, we know the form of the pressure.

#P = (nRT)/V#

#=> q_(rev) = nRT int_(V_1)^(V_2) 1/V dV#

#= nRTln(V_2/V_1)#

So, we have:

#color(blue)(DeltaS("const. T") = nRln(V_2/V_1))#

**METHOD 2**

Write the total differential of the entropy as a function of temperature and volume:

#dS(T,V) = ((delS)/(delT))_VdT + ((delS)/(delV))_TdV#

The constant temperature process, or isothermal process, is the second term, so we seek only:

#DeltaS = cancel(int_(T_1)^(T_2) ((delS)/(delT))_VdT)^(0)# #+# #int_(V_1)^(V_2) ((delS)/(delV))_TdV#

#dA = -Sul(dT) - Pul(dV)#

Since all thermodynamic functions (*the cross-derivatives here are equal*:

#((delS)/(delV))_T = ((delP)/(delT))_V#

So, we have:

#DeltaS ("const. T") = int_(V_1)^(V_2) ((delP)/(delT))_VdV#

And for ideal gases, we again know the form of the pressure:

#P = (nRT)/V#

So its partial derivative with respect to temperature at constant volume is

#color(blue)(DeltaS("const. T")) = nRint_(V_1)^(V_2) 1/VdV#

#= color(blue)(nRln(V_2/V_1))#

just as we got before.