Question

One mole of ideal monoatomic gas was taken through reversible isochoric heating from 100 K to 1000 K. Calculate $\Delta S_{system},\Delta S_{surr}$, and $\Delta S_{total}$ when the process carried out irreversibly (one step).

• A. $\Delta S_{sys}=\frac{3}{2}Rln10;\Delta S_{surr}=-\frac{3}{2}R\left(0.9\right);\Delta S_{total}=\frac{3}{2}R\left(1.403\right)$
• B. $\Delta S_{sys}=-\frac{3}{2}Rln10;\Delta S_{surr}=\frac{3}{2}R\left(0.9\right);\Delta S_{total}=-\frac{3}{2}R\left(1.403\right)$
• C. $\Delta S_{sys}=-\frac{3}{2}Rln10;\Delta S_{surr}=-\frac{3}{2}R\left(0.9\right);\Delta S_{total}=\frac{3}{2}R\left(3.203\right)$
• D. None of these

Answer 1

Answer: (A)

$\Delta S_{sys}=\frac{3}{2}Rln10;\Delta S_{surr}=-\frac{3}{2}R\left(0.9\right);\Delta S_{total}=\frac{3}{2}R\left(1.403\right)$

An isochoric process is a constant volume process. Since entropy is a point functions, change of entropy between two points would be some irrespective of the process being reversible or irreversible or of different path.
$\therefore {\left(△S_{sys}\right)}_{irrev.onestep}={\left(△S_{sys}\right)}_{rev}={\int }_1^2\frac{dQ}{T}$
Here, $△S_{sys}={\int }_{100}^{1000}\frac{nCrdT}{T}$ Given $n=1,Cr=\frac{3}{R}$ (monoatomic)
$\therefore △S_{sys}=\frac{3}{2}R{\left[lnT\right]}_{100}^{1000}=\frac{3}{2}Rln10$
Now, $(△S{)}_{sys}+(△{S)}_{surrouding}={\left(△S\right)}_{Total}>0$ for irreversible process
$(△S{)}_{surrounding}=\frac{-△Q}{T_{surrounding}}$
$=-\frac{3}{2}R\frac{(1000-100)}{1000}=-\frac{3}{2}R\left(0.9\right)$
$(△S{)}_{total}=(△S{)}_{sys}+(△{S)}_{surrouding}=\frac{3}{2}R(ln10-0.9)$
$=\frac{3}{2}R\left(1.403\right)$