Can the entropy of an ideal gas change during an isothermal process?

Question

30175 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-04T06:37:10

Yes.

#DeltaS_T = nRln(V_2/V_1)#,

i.e. at constant temperature, expanding gases increase in entropy.

Yes, #DeltaS# is not a function of only temperature, so it is not zero.

An isothermal process has #DeltaT = 0#, but one can write a total differential for the entropy as a function of #T# and #V#:

#dS(T,V) = ((delS)/(delT))_VdT + ((delS)/(delV))_TdV##" "" "bb((1))#

In this case, one could say that at constant temperature, #dT = 0#, so we simplify #(1)# down to:

#dS_T = ((delS)/(delV))_TdV##" "" "bb((2.1))#

The natural variables associated with this partial derivative are #T# and #V#, which are found in the Helmholtz Maxwell relation:

#dA = -SdT - PdV# #" "" "bb((3))#

For any state function, the cross-derivatives are equal, so from #(3)#, we rewrite #(2.1)# using the relation:

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

Therefore, in terms of a partial derivative that uses the ideal gas law, we get:

#dS_T = ((delP)/(delT))_VdV# #" "" "bb((2.2))#

The right-hand side of #(2.2)# from the ideal gas law gives:

#((delP)/(delT))_V = (del)/(delT)[(nRT)/V]_V = (nR)/V#

Therefore, the change in entropy of an ideal gas at a specific, constant temperature is (by integrating #(2.2)#):

#color(blue)(DeltaS_T) = int_((1))^((2)) dS_T = nR int_(V_1)^(V_2) 1/VdV#

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

So if the gas expands in the isothermal process, then yes, it will have increased entropy.