$dS=(\frac{\text{d}\text{e}\text{l}\text{S}}{\text{d}\text{e}\text{l}\text{T}}{)}_{V}dT+(\frac{\text{d}\text{e}\text{l}\text{S}}{\text{d}\text{e}\text{l}\text{V}}{)}_{T}dV=\frac{C_V}{T}dT+(\frac{\text{d}\text{e}\text{l}\text{S}}{\text{d}\text{e}\text{l}\text{V}}{)}_{T}dV,$ please find $S$ for $1$ mole of ideal gas changes from $300K$ and $20L$ to $600K$ and $40L.($where $C_V=\frac{3}{2}R,R=8\mathrm{.}314\frac{J}{m}$olK,$ln2=0\mathrm{.}693,ln3=1\mathrm{.}099,ln7=1\mathrm{.}946,$ and the state function of ideal gas is $PV=nRT)($hint: you should use one of Maxwell relations to solve this calculation.$)()$

Integrals : $($a$){\displaystyle \underset{0}{\overset{}{}}}x{e}^{-a{x}^2}dx.($b$)I={\displaystyle \underset{0}{\overset{}{}}}{\displaystyle \underset{0}{\overset{}{}}}x{e}^{-a{x}^2}y{e}^{-y^2}dxdy.$