Q. Ten moles of an ideal gas at constant temperature 500 K is compressed from 50 L to 5 L. Work done in the process is (Given, $R = 8.31\ J - mol^{- 1} - K^{- 1}$)
A) $- 1.2 \times 10^{4}J$
B) $- 2.4 \times 10^{4}\ J$
C) $- 4.8 \times 10^{4}\ J$
D) $- 9.6 \times 10^{4}J$
Explanation
(d) Work done at constant temperature (ie, isothermal process),
$W = 2.3nRT\log_{10}\left( \frac{V_{2}}{V_{1}} \right)$
$= 2.3 \times 10 \times 8.31 \times 500 \times \log_{10}\left( \frac{5}{50} \right)$
$= - 9.6 \times 10^{4}J$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.