Question

The change in entropy of 2 moles of an ideal gas upon isothermal expansion at $250K$ from $20litre$ until the pressure becomes $1atm$ Take $R=0.08Latmmo{l}^{-1}{K}^{-1}$ ln 2= 0.3), is :

• A. $1.382cal/K$
• B. $-1.2cal/K$
• C. $1.2cal/K$
• D. $2.77cal/K$

Answer 1

The correct option is C $1.2cal/K$

Given,
$\text{Temperature,}T=243.6K$
$n=2$ mol;
$\text{Initial volume},V_1=20L$
$\text{Final pressure},P_2=1atm$
$\text{Gas constant},R=0.08Latmmo{l}^{-1}{K}^{-1}$
$P_1=\frac{2\times 0.08\times 250}{20}=2atm\approx 2atm$

We know,
$\Delta S=nRln\left(\frac{P_1}{P_2}\right)$
Where, $\Delta S$ = change in entropy
Again, $R=2cal{K}^{-1}mo{l}^{-1}$

$\therefore \Delta S=2\times 2\times$ln $\left(2\right)$
$\Rightarrow \Delta S=4\times 0.3$
$=1.2cal/K$