# What is the entropy change for the isothermal expansion of #"0.75 g"# of #"He"# from #"5.0 L"# to #"12.5 L"#?

##### 1 Answer

I got

I assume you aren't given an equation and have to derive it.

If we treat entropy as *a function of the temperature and volume* (that is,

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

Since we are looking at the **change in entropy due to a change in volume**, we only consider

Another key step is to relate this to a derivative we are more familiar with. Recall that the **Helmholtz free energy** *a function of temperature and volume*, and that its **Maxwell relation** is:

#dA = -SdT - PdV = -(SdT + PdV)#

Since

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

You should get to know this pretty well. The righthand side is a derivative we are familiar with. The ideal gas law can then be used.

#PV = nRT#

#=> P = (nRT)/V#

Now, the derivative is feasible in terms of an equation we know:

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

#= (nR)/V(dT)/(dT)#

#= (nR)/V#

This therefore gives:

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

Next, we can integrate this over the two volumes to get the change in entropy. By moving the

#int_(S_1)^(S_2) dS = DeltaS = int_(V_1)^(V_2) (nR)/VdV#

The integral of

#DeltaS = nRint_(V_1)^(V_2) 1/VdV = nR|[ln|V|]|_(V_1)^(V_2)#

#= nR(lnV_2 - lnV_1)#

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

So, given

#color(blue)(DeltaS) = 0.75 cancel"g He" xx cancel("1 mol")/(4.0026 cancel"g He") xx "8.314472 J/"cancel"mol"cdot"K" xx ln("12.5 L"/"5.0 L")#

#=# #color(blue)("1.428 J/K")#