Can the entropy of an ideal gas change during an isothermal process?
1 Answer
Yes.
#DeltaS_T = nRln(V_2/V_1)# ,i.e. at constant temperature, expanding gases increase in entropy.
Yes,
An isothermal process has
#dS(T,V) = ((delS)/(delT))_VdT + ((delS)/(delV))_TdV# #" "" "bb((1))#
In this case, one could say that at constant temperature,
#dS_T = ((delS)/(delV))_TdV# #" "" "bb((2.1))#
The natural variables associated with this partial derivative are
#dA = -SdT - PdV# #" "" "bb((3))#
For any state function, the cross-derivatives are equal, so from
#((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
#((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
#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.