Search any question & find its solution
Question:
Answered & Verified by Expert
One mole of an ideal gas is expanded isothermally and reversibly to half of its initial pressure. $\Delta S$ for the process in $\mathrm{J} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ is
$[\ln 2=0.693$ and $R=8.314, \mathrm{~J} /(\mathrm{mol} \mathrm{K})]$
Options:
$[\ln 2=0.693$ and $R=8.314, \mathrm{~J} /(\mathrm{mol} \mathrm{K})]$
Solution:
1741 Upvotes
Verified Answer
The correct answer is:
$5.76$
$5.76$
For isothermal process $(\Delta T=0)$
$$
\begin{aligned}
\Delta S & =R \ell \mathrm{n} \frac{P_1}{P_2}=8.314 \ell \mathrm{n} 2 \\
& =8.314 \times 0.693=5.76 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$$
$$
\begin{aligned}
\Delta S & =R \ell \mathrm{n} \frac{P_1}{P_2}=8.314 \ell \mathrm{n} 2 \\
& =8.314 \times 0.693=5.76 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
\end{aligned}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.