# How do you calculate #DeltaS_(sys)# for the isothermal expansion of 1.5 mol of an ideal gas from 20.0 L to 22.5L?

##### 1 Answer

I got

We begin by realizing that an isothermal expansion is one where

To get from

#DeltaS_"sys" = int_(V_1)^(V_2) ((delS)/(delV))_TdV#

We denote molar entropy as

#color(green)(DeltabarS_"sys" = int_(barV_1)^(barV_2) ((delbarS)/(delbarV))_TdbarV)#

From the Maxwell Relation for

#dbarA = -barSdT - PdbarV# ,

we use the **cross-derivative relationship** to get

#((delbarS)/(delbarV))_T = color(green)(((delP)/(delT))_barV)# ,

meaning that the change in entropy due to the change in volume at constant temperature is **equal** to the change in pressure due to the change in temperature at a constant volume.

Next, we would have to figure out what this new derivative gives for the ideal gas law. Recall that the ideal gas law is:

#PV = nRT#

or

#PbarV = RT#

#=> P = (RT)/(barV)#

The partial derivative of pressure with respect to temperature at *constant* molar volume is then:

#=> ((delP)/(delT))_barV = (del)/(delT)[(RT)/(barV)]_barV#

#= R/(barV)(del)/(delT)[T]_barV = R/(barV)((delT)/(delT))_barV#

#= R/(barV)cancel((dT)/(dT))#

#= R/(barV)#

So, what's left is to integrate the result by substituting

#color(blue)(DeltabarS_"sys") = int_(barV_1)^(barV_2) R/(barV)dbarV#

#= Rint_(barV_1)^(barV_2) 1/(barV)dbarV = R|[ln(barV)]|_(barV_1)^(barV_2)#

#= R[ln(barV_2) - ln(barV_1)] = Rln(barV_2/(barV_1)) = Rln((V_2"/"cancel(n))/(V_1"/"cancel(n)))#

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

Notice how the molar volumes have canceled out to give the regular volume.

Therefore, the **change in entropy** for the system at constant temperature is:

#color(blue)(DeltaS_"sys") = nDeltabarS_"sys" = nRln(V_2/V_1)#

#= ("1.5 mols")("8.314472 J/mol"cdot"K")ln(("22.5 L")/("20.0 L"))#

#=# #color(blue)("1.47 J/K")#