Question

The entropy change involved in the isothermal reversible expansion of $1$ mole of an ideal gas from a volume of $20\,L$ to $200\,L$ at $300\,K$ is ____ $JK^{- 1}.$
Give answer after rounding off to the nearest integer value.

Answer 1

We know that work done in isothermal reversible expansion,
$q_{\text{rev}}=-w_{\max}=2.303\times nRT \log_{10}\frac{V_{2}}{V_{1}}$
Given that,
$n=1,R=8.314\,JK^{- 1}\,mol^{- 1},T=300\,K$
$V_{1}=20\,L,V_{2}=200\,L$
Putting the values in above equation.
$\therefore q_{\text{rev}}=2.303\times 1\times 8.314\times 300\times \left(\log\right)_{10}\left(\frac{200}{20}\right)$
$=2.303\times 1\times 8.314\times 300\times \left(\log\right)_{10}\left(\right.10\left.\right)$
$=2.303\times 1\times 8.314\times 300$
According to third law of thermodynamics,
$\Delta S=\frac{q_{\text{rev}}}{T}$
$\therefore \Delta S=\frac{2 . 303 \times 1 \times 8 . 314 \times 300}{300}=19.15\,JK^{- 1}$
So the Entropy change is $\cong19JK^{- 1}$