In the topic, CHANGE IN ENTROPY why we take q(rev.) = RTlnV2/V1 ?

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Answer 1 · 2018-03-13T21:51:30

Because the change in entropy is STATED to be isothermal...

#q_(rev,T) = nRTln(V_2/V_1)#

Isothermal changes in entropy, if #S = S(T,V)#, are given by:

#DeltaS_T = int_(V_1)^(V_2) ((delS)/(delV))_TdV#

The Helmholtz free energy Maxwell relation relates the natural variables #T# and #V#:

#dA = -SdT - PdV#

Since #A# is a state function, the cross-derivatives are equal:

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

As a result,

#DeltaS_T = int_(V_1)^(V_2) ((delP)/(delT))_VdV#

For ideal gases, #PV = nRT#. Therefore:

#DeltaS_T = int_(V_1)^(V_2) (del)/(delT)[(nRT)/V]_VdV#

#= nRint_(V_1)^(V_2) 1/VdV#

#= nR ln (V_2/V_1)#

Thus, the isothermal change in molar entropy is given by:

#(DeltaS_T)/n -= color(blue)(DeltabarS_T = Rln(V_2/V_1))#

And from the relations derived for Carnot's engine,

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

So, at constant temperature,

#color(blue)(q_(rev,T) = nRTln(V_2/V_1))#

in units of JOULES. Hence, there must be #n#...